Ch 7 · Q–

Class 11 Science Exercise NCERT Solutions Olympiad Board Exam

Chapter 7

Redox Reactions

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

30 Questions

65–95 min Ideal time

Q1 Now at

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NUMERIC3 marks

Assign oxidation number to the underlined elements in each of the following species:

(a) NaH₂PO₄ (P)
(b) NaHSO₄ (S)
(c) H₄P₂O₇ (P)
(d) K₂MnO₄ (Mn)
(e) CaO₂ (O)
(f) NaBH₄ (B)
(g) H₂S₂O₇ (S)
(h) KAl(SO₄)₂.12H₂O (S)

📘 Concept & Theory ›

Oxidation number (oxidation state) represents the apparent charge on an atom assuming complete transfer of electrons. It is an important tool for identifying oxidation and reduction processes and balancing redox reactions.

To determine oxidation numbers, the following standard rules are generally applied:

Elements in their free state have oxidation number 0.
Alkali metals (Group 1) always have oxidation number +1.
Alkaline earth metals (Group 2) always have oxidation number +2.
Hydrogen is usually +1, but it is −1 in metal hydrides.
Oxygen is usually −2, but it is −1 in peroxides.
The algebraic sum of oxidation numbers of all atoms in a neutral compound is zero.
The algebraic sum of oxidation numbers in an ion equals the charge on the ion.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify atoms having fixed oxidation numbers.
Assign their standard oxidation numbers.
Assume the oxidation number of the required element to be x.
Apply the rule that the sum of oxidation numbers equals the overall charge.
Solve the resulting equation to obtain the oxidation number.

✏️ Solution Complete Solution ›

Step-by-step Solution · 26 steps

(a) NaH₂PO₄
Let the oxidation number of phosphorus be x.
Known oxidation numbers:
- Na = +1
- H = +1
- O = −2
Since the compound is neutral, \[\begin{aligned} (+1)+2(+1)+x+4(-2)&=0\\ 1+2+x-8&=0\\ x-5&=0\\ x&=+5 \end{aligned}\] Oxidation number of phosphorus = +5
(b) NaHSO₄
Let the oxidation number of sulphur be x.
Known oxidation numbers:
- Na = +1
- H = +1
- O = −2
Since compound is neutral \[ \begin{aligned} (+1)+(+1)+x+4(-2)&=0\\ 1+1+x-8&=0\\ x-6&=0\\ x&=+6 \end{aligned} \] Oxidation number of sulphur = +6
(c) H₄P₂O₇
Let oxidation number of each phosphorus atom be x.
Known oxidation numbers:
- H = +1
- O = −2
\[\begin{aligned} 4(+1)+2x+7(-2)&=0\\ 4+2x-14&=0\\ 2x-10&=0\\ 2x&=10\\ x&=+5 \end{aligned}\] Oxidation number of phosphorus = +5
(d) K₂MnO₄
Let oxidation number of manganese be x.
Known oxidation numbers:
- K = +1
- O = −2
\[\begin{aligned} 2(+1)+x+4(-2)&=0\\ 2+x-8&=0\\ x-6&=0\\ x&=+6 \end{aligned} \] Oxidation number of manganese = +6
(e) CaO₂
This compound is a peroxide.
In peroxides, oxygen has oxidation number −1, not −2.
Verification \[\begin{aligned} (+2)+2(-1)=0\\ 2-2=0 \end{aligned}\] Oxidation number of oxygen = −1
(f) NaBH₄
NaBH₄ contains the tetrahydroborate ion.
Here hydrogen is bonded to boron, which is less electronegative than hydrogen.
Therefore, hydrogen behaves as hydride and has oxidation number −1.
Let oxidation number of boron be x. \[\begin{aligned} (+1)+x+4(-1)&=0\\ 1+x-4&=0\\ x-3&=0\\ x&=+3 \end{aligned}\] Oxidation number of boron = +3
(g) H₂S₂O₇
Let oxidation number of each sulphur atom be x.
Known oxidation numbers:
- H = +1
- O = −2
\[\begin{aligned} 2(+1)+2x+7(-2)&=0\\ 2+2x-14&=0\\ 2x-12&=0\\ 2x&=12\\ x&=+6 \end{aligned}\] Oxidation number of sulphur = +6
(h) KAl(SO₄)₂.12H₂O
Water of crystallization does not affect the oxidation number of sulphur.
Consider only the sulphate ion.
Let oxidation number of sulphur be x.
In sulphate ion, \[\begin{aligned} x+4(-2)&=-2\\ x-8&=-2\\ x&=+6 \end{aligned}\] Oxidation number of sulphur = +6

💡 Answer Final Answer ›

Species	Element	Oxidation Number
NaH₂PO₄	P	+5
NaHSO₄	S	+6
H₄P₂O₇	P	+5
K₂MnO₄	Mn	+6
CaO₂	O	−1
NaBH₄	B	+3
H₂S₂O₇	S	+6
KAl(SO₄)₂.12H₂O	S	+6

🎯 Exam Significance Exam Significance ›

Oxidation number is one of the most frequently tested concepts in CBSE Class 11 and Class 12 Chemistry.
Questions based on assigning oxidation numbers appear regularly in school examinations.
This concept forms the foundation for balancing redox equations.
JEE Main and NEET frequently include oxidation-number based objective questions.
Understanding special cases such as peroxides and metal hydrides helps avoid common mistakes in competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Always assign oxidation numbers with the help of standard rules.
Hydrogen is generally +1 but becomes −1 in metal hydrides such as NaBH₄.
Oxygen is generally −2 but becomes −1 in peroxides such as CaO₂.
Water of crystallization does not change the oxidation number of atoms in the main compound.
The sum of oxidation numbers in a neutral compound is always zero.
Careful application of these rules allows quick determination of unknown oxidation states in complex compounds.

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📘 Concept & Theory Concept Required ›

Most compounds contain atoms of the same element in identical chemical environments, so each atom has the same oxidation number. However, in some compounds, atoms of the same element exist in different bonding environments. Such compounds may have either:

Fractional (average) oxidation numbers.
More than one oxidation state for the same element.
Different oxidation numbers for different atoms of the same element.

In such cases, oxidation numbers should be interpreted carefully based on the molecular structure rather than by simply applying algebraic rules.

🗺️ Solution Roadmap Step-by-step Plan ›

Assign known oxidation numbers.
Apply oxidation-number rules wherever possible.
If all atoms are chemically equivalent, calculate the average oxidation number.
If atoms occupy different environments, determine the oxidation number of each atom separately.
Explain the reason (rationalisation) for the obtained result.

✏️ Solution Complete Solution ›

Step-by-step Solution · 40 steps

(a) KI₃
KI₃ consists of
K⁺ + I₃⁻
The triiodide ion contains three iodine atoms sharing an overall charge of −1.
herefore, the average oxidation number of iodine is \[\begin{aligned} 3x&=-1\\ x&=-\frac{1}{3} \end{aligned}\]
Average oxidation number of iodine = −1/3
Rationalisation: This is an average oxidation number. The three iodine atoms are bonded together in the triiodide ion, and the negative charge is delocalized over the entire ion rather than localized on one atom. Therefore, each iodine is assigned an average oxidation number of −1/3.
(b) H₂S₄O₆
Let the average oxidation number of sulphur be x.
Known oxidation numbers:
- Hydrogen = +1
- Oxygen = −2
\[\begin{aligned} 2(+1)+4x+6(-2)&=0\\ 2+4x-12&=0\\ 4x&=10\\ x&=+2.5 \end{aligned}\] Average oxidation number of sulphur = +2.5
Rationalisation:
In tetrathionic acid, the four sulphur atoms are not equivalent.
- Two terminal sulphur atoms have oxidation number +5.
- Two inner sulphur atoms have oxidation number 0.
The average oxidation number is therefore \[\frac{(+5)+(+5)+0+0}{4}=+2.5\]
Hence, +2.5 is only an average value and not the oxidation number of every sulphur atom.
(c) Fe₃O₄
Let the average oxidation number of iron be x.
Known oxidation number:
- Oxygen = −2
\[\begin{aligned} 3x+4(-2)&=0\\ 3x-8&=0\\ x&=\frac{8}{3}\\ &=+2.67 \end{aligned}\] Average oxidation number of iron = +8/3 (or +2.67)
Rationalisation:
Fe₃O₄ is actually a mixed oxide.
It may be represented as \[\mathrm{FeO\cdot Fe_2O_3}\]
Thus,
- One Fe atom has oxidation number +2.
- Two Fe atoms have oxidation number +3.
The average oxidation number becomes \[\frac{2+3+3}{3}=\frac{8}{3}\] Therefore, iron exists in two different oxidation states
(d) CH₃CH₂OH
The two carbon atoms are present in different chemical environments.
Hence, their oxidation numbers are calculated separately. First carbon (CH₃–)
- Each C–H bond contributes −1 to carbon.
- Three C–H bonds give
\[-1-1-1=-3\]
C–C bond contributes zero.
Oxidation number = −3
Second carbon (–CH₂OH)
- Two C–H bonds contribute −2.
- One C–O bond contributes +1.
- One C–C bond contributes 0.
\[-2+1=-1\]
Oxidation number = −1
Rationalisation:
The two carbon atoms are attached to different groups, so they possess different oxidation numbers.
(e) CH₃COOH
The two carbon atoms again occupy different environments.
First carbon (CH₃–)
- Three C–H bonds contribute −3.
- One C–C bond contributes 0.
Oxidation number = −3
Second carbon (COOH)
- One C=O bond contributes +2.
- One C–O bond contributes +1.
- One C–C bond contributes 0.
\[+2+1=+3\]
Oxidation number = +3
Rationalisation:
The carboxyl carbon is bonded to highly electronegative oxygen atoms, making its oxidation number positive, whereas the methyl carbon remains highly reduced.

💡 Answer Final Answer ›

Species	Element	Oxidation Number	Remarks
KI₃	I	−1/3	Average oxidation number in I₃⁻
H₂S₄O₆	S	+2.5	Average value; actual sulphur atoms are +5 and 0
Fe₃O₄	Fe	+8/3 (+2.67)	Average value; Fe exists as Fe²⁺ and Fe³⁺
CH₃CH₂OH	C	−3, −1	Two carbon atoms have different oxidation numbers
CH₃COOH	C	−3, +3	Two carbon atoms are in different oxidation states

🎯 Exam Significance Exam Significance ›

Questions involving average oxidation numbers are frequently asked in CBSE board examinations.
Mixed oxidation states are important for understanding redox chemistry and disproportionation reactions.
JEE Main and NEET often test oxidation numbers in compounds such as Fe₃O₄, KI₃, and organic molecules.
Being able to assign oxidation numbers in organic compounds is useful in later chapters dealing with alcohols, aldehydes, carboxylic acids, and oxidation reactions.
Recognizing when oxidation numbers are average values helps avoid common conceptual mistakes.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Fractional oxidation numbers represent average oxidation states and do not imply fractional charge on an individual atom.
Mixed oxides such as Fe₃O₄ contain atoms in more than one oxidation state.
Atoms of the same element can have different oxidation numbers when they occupy different chemical environments.
In organic compounds, oxidation number depends on the atoms directly bonded to carbon.
Always interpret oxidation numbers with the actual molecular structure in mind rather than relying solely on algebraic calculations.

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📘 Concept & Theory Theory / Concept ›

A redox reaction is a chemical reaction in which oxidation and reduction occur simultaneously. This involves the transfer of electrons from one species to another.

The following definitions are useful while identifying a redox reaction:

Oxidation is an increase in oxidation number or loss of electrons.
Reduction is a decrease in oxidation number or gain of electrons.
If one element is oxidized and another is reduced in the same reaction, the reaction is a redox reaction.

🗺️ Solution Roadmap Step-by-step Plan ›

Assign oxidation numbers to the important elements in reactants and products.
Compare the oxidation numbers before and after the reaction.
Identify the species undergoing oxidation and reduction.
Conclude whether the reaction is a redox reaction.

✏️ Solution Complete Solution ›

Step-by-step Solution · 31 steps

(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g)

Assign oxidation numbers.

Species	Element	Oxidation Number
CuO	Cu	+2
H₂	H	0
Cu	Cu	0
H₂O	H	+1

Observe the changes.
- Copper changes from +2 → 0.
- Hydrogen changes from 0 → +1.
Reduction: \[\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\]
Oxidation: \[\mathrm{H_2 \rightarrow 2H^+ +2e^-}\]
Conclusion:
Copper is reduced while hydrogen is oxidized. Therefore, the reaction is a redox reaction.
(b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

Assign oxidation numbers.

Species	Element	Oxidation Number
Fe₂O₃	Fe	+3
CO	C	+2
Fe	Fe	0
CO₂	C	+4

Observe the changes.
- Iron changes from +3 → 0.
- Carbon changes from +2 → +4.
Reduction: \[\mathrm{Fe^{3+}+3e^- \rightarrow Fe}\]
Oxidation: \[\mathrm{C^{+2}\rightarrow C^{+4}+2e^-}\]
Conclusion:
Iron oxide is reduced while carbon monoxide is oxidized. Hence, this is a redox reaction.
c) 4BCl₃(g) + 3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s)
Assign oxidation numbers.
In BCl₃,
- Cl = −1
- B = +3
In B₂H₆, hydrogen is bonded to boron. Hydrogen is more electronegative than boron, so H has oxidation number −1.
Let oxidation number of boron be \(x\). \[\begin{aligned} 2x+6(-1)&=0\\ 2x&=6\\ x&=+3 \end{aligned}\]
Thus, boron remains +3.
Now consider hydrogen.
- In LiAlH₄, hydrogen exists as hydride (−1).
- In B₂H₆, hydrogen is also −1.
Similarly, Li remains +1, Al remains +3 and Cl remains −1.
Observation:
No element undergoes any change in oxidation number.
Conclusion:
This reaction is not a redox reaction. It is a transfer of hydride ions (a metathesis/complex formation reaction) without electron transfer.

Note: Although this question appears in the NCERT exercise under redox reactions, oxidation numbers remain unchanged for all elements. Hence, according to oxidation-number criteria, it is not a redox reaction.
(d) 2K(s) + F₂(g) → 2KF(s)

Assign oxidation numbers.

Element	Reactant	Product
K	0	+1
F	0	−1

Oxidation: \[\mathrm{K\rightarrow K^+ +e^-}\]
Reduction: \[\mathrm{F_2+2e^- \rightarrow 2F^-}\]
Conclusion:
Potassium loses electrons while fluorine gains electrons. Therefore, the reaction is a redox reaction.
(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
Determine oxidation number of nitrogen.
In NH₃ \[\begin{aligned} x+3(+1)&=0\\ x&=-3 \end{aligned}\]
In NO, \[\begin{aligned} x+(-2)&=0\\ x&=+2 \end{aligned}\]
Thus nitrogen changes from \[-3\rightarrow +2\]
Oxygen changes from \[0\rightarrow -2\]
Oxidation \[\mathrm{N^{-3}\rightarrow N^{+2}}\]
Reduction \[\mathrm{O_2\rightarrow O^{2-}}\]
Conclusion:
Nitrogen is oxidized and oxygen is reduced. Therefore, this is a redox reaction.

💡 Answer Final Answer ›

Reaction	Oxidized Species	Reduced Species	Redox?
(a)	H₂	CuO	Yes
(b)	CO	Fe₂O₃	Yes
(c)	None	None	No (oxidation numbers remain unchanged)
(d)	K	F₂	Yes
(e)	NH₃	O₂	Yes

🎯 Exam Significance Exam Significance ›

Identifying oxidation and reduction using oxidation numbers is a fundamental skill tested in CBSE Board examinations.
JEE Main and NEET frequently include conceptual questions asking students to determine whether a reaction is redox or non-redox.
Mixed conceptual questions involving oxidation-number changes and electron transfer are common in competitive examinations.
Reaction (c) is particularly important because it demonstrates that not every reaction involving hydrides is a redox reaction. Students should verify oxidation-number changes before concluding.
These concepts serve as the foundation for balancing redox reactions and understanding electrochemistry in higher classes.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

A reaction is classified as a redox reaction only when oxidation and reduction occur simultaneously.
An increase in oxidation number indicates oxidation, while a decrease indicates reduction.
Electron transfer accompanies every redox reaction.
Always assign oxidation numbers before identifying oxidation and reduction.
Do not assume every chemical reaction is a redox reaction; verify changes in oxidation numbers carefully.

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📘 Concept & Theory Theory / Concept ›

A reaction is called a redox reaction if oxidation and reduction occur simultaneously. This is identified by assigning oxidation numbers to the atoms before and after the reaction.

In this reaction, fluorine converts water into hypofluorous acid (HOF) while itself getting reduced to hydrogen fluoride (HF). Interestingly, the oxygen atom present in water undergoes oxidation, whereas fluorine undergoes reduction.

This reaction demonstrates that although fluorine is the most electronegative element, it can only exhibit an oxidation number of −1 in its compounds and therefore always undergoes reduction.

🗺️ Solution Roadmap Step-by-step Plan ›

Assign oxidation numbers to fluorine and oxygen in reactants.
Assign oxidation numbers to the same elements in products.
Compare the oxidation numbers.
Identify oxidation and reduction.
Conclude whether the reaction is a redox reaction.

✏️ Solution Complete Solution ›

Step-by-step Solution · 18 steps

Assign oxidation numbers in the reactants.
Water (H₂O)
- Hydrogen = +1
- Oxygen = −2
Fluorine molecule (F₂)
Since fluorine is in its elemental state, \[\boxed{\text{Oxidation number of F}=0}\]
Assign oxidation numbers in the products.
Hydrogen fluoride (HF)
- Hydrogen = +1
- Fluorine = −1
Hypofluorous acid (HOF)
Let the oxidation number of oxygen be x.
Known oxidation numbers:
- Hydrogen = +1
- Fluorine = −1
Since HOF is a neutral molecule, \[\begin{aligned} (+1)+x+(-1)&=0\\ x&=0 \end{aligned}\]
Therefore,
- Hydrogen = +1
- Oxygen = 0
- Fluorine = −1

Compare oxidation numbers.

Element	Reactant	Product	Change
Fluorine	0	−1	Reduction
Oxygen	−2	0	Oxidation

Identify oxidation and reduction.
Reduction of fluorine: \[\mathrm{F_2+2e^- \rightarrow 2F^-}\]
Oxidation number changes as \[0\rightarrow -1\]
Oxidation of oxygen:
Oxygen changes from \[-2\rightarrow 0\] This corresponds to a loss of electrons (oxidation).
Conclusion

Since oxygen is oxidized while fluorine is reduced, oxidation and reduction occur simultaneously.

Therefore,

H₂O(s) + F₂(g) → HF(g) + HOF(g)

is a redox reaction.

💡 Answer Final Answer ›

In this reaction,

Fluorine undergoes reduction because its oxidation number changes from 0 to −1.
Oxygen undergoes oxidation because its oxidation number changes from −2 to 0.

Since oxidation and reduction occur simultaneously, the reaction is a redox reaction.

🎯 Exam Significance Exam Significance ›

This is one of the most important conceptual questions from the NCERT textbook and is frequently asked in CBSE examinations.
JEE Main and NEET often test oxidation-number calculations in unusual compounds such as HOF.
The question reinforces the concept that fluorine always has an oxidation number of −1 in its compounds.
Students learn that oxygen need not always possess an oxidation number of −2; its value depends on the compound.
The problem strengthens the ability to identify oxidation and reduction by systematic oxidation-number analysis.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Always assign oxidation numbers before deciding whether a reaction is redox.
Fluorine has oxidation number 0 in F₂ but −1 in all its compounds.
Oxygen in HOF has oxidation number 0, not −2.
A change in oxidation number confirms oxidation or reduction.
A reaction is classified as a redox reaction only when both oxidation and reduction occur simultaneously.

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📘 Concept & Theory Theory / Concept ›

Oxidation number is assigned using standard oxidation-number rules. However, in compounds containing peroxide linkages or atoms in different bonding environments, assigning the usual oxidation number (such as oxygen = −2) may lead to an incorrect result. Therefore, the molecular structure must always be considered while determining oxidation numbers. This question emphasizes an important principle:

First determine the oxidation number using standard rules.
Then verify whether any atom is present in a special structural environment (for example, peroxide oxygen).
If a special bond exists, revise the calculation accordingly.

🗺️ Solution Roadmap Step-by-step Plan ›

Assign oxidation numbers of atoms having fixed values.
Assume the oxidation number of the required element as x.
Use the overall charge of the molecule or ion.
Suggest the molecular structure.
Explain why a direct calculation may give an incorrect (fallacious) result.

✏️ Solution Complete Solution ›

Step-by-step Solution · 15 steps

(a) H₂SO₅ (Peroxomonosulphuric acid or Caro's acid)
If all oxygen atoms are incorrectly assumed to have oxidation number −2: \[\aligned2(+1)+x+5(-2)=0\\ 2+x-10=0\\ x=+8 \end{aligned}\]
This result is impossible because sulphur does not exhibit an oxidation state of +8.
Consider the actual structure. H₂SO₅ contains one peroxide linkage (–O–O–).
- Three oxygen atoms bonded normally have oxidation number −2.
- Two peroxide oxygen atoms have oxidation number −1 each.
Therefore, \[\begin{aligned} 2(+1)+x+3(-2)+2(-1)&=0\\ 2+x-6-2&=0\\ x&=+6 \end{aligned}\]
Oxidation number of sulphur = +6
Suggested Structure \[ \begin{array}{ccccc} &&&&O\\ &&&&||\\ HO&-&S&-&O&-&O&-&H\\ &&&&||\\ &&&&O \end{array} \] The O—O bond represents the peroxide linkage.
(b) Cr₂O₇²⁻ (Dichromate ion)
Let the oxidation number of chromium be x.
Oxygen has oxidation number −2. \[\begin{aligned} 2x+7(-2)&=-2\\ 2x-14&=-2\\ 2x&=12\\ x&=+6 \end{aligned}\] Oxidation number of chromium = +6
Suggested Structure \[ \begin{array}{cccccc} &&O&&&&O\\ &&||&&&&||\\ ^-O&-&Cr&-&O&-&Cr&-&O^-\\ &&||&&&&||\\ &&O&&&&O \end{array} \] The two chromium atoms are connected through one bridging oxygen atom.
(c) NO₃⁻ (Nitrate ion)
Let the oxidation number of nitrogen be x.
Oxygen has oxidation number −2. \[\begin{aligned} x+3(-2)&=-1\\ x-6&=-1\\ x&=+5 \end{aligned}\]
Oxidation number of nitrogen = +5
Suggested Structure \[ \begin{array}{ccc} &&O\\ &&||\\ ^-O&-&N&-&O^- \end{array} \]
In reality, nitrate ion exhibits resonance, and all three N–O bonds become equivalent due to electron delocalization.
Accounting for the Fallacy
The fallacy arises if we mechanically assign oxidation number −2 to every oxygen atom.

In H₂SO₅, two oxygen atoms are connected through an O–O peroxide bond. Oxygen in a peroxide has oxidation number −1, not −2.

If all oxygen atoms are taken as −2, sulphur appears to have oxidation number +8, which is chemically impossible.

Therefore, oxidation-number calculations must always be consistent with the actual molecular structure.

💡 Answer Final Answer ›

Compound/Ion	Oxidation Number	Important Structural Feature
H₂SO₅	S = +6	Contains one peroxide (O–O) linkage
Cr₂O₇²⁻	Cr = +6	Two Cr atoms joined by one bridging oxygen
NO₃⁻	N = +5	Trigonal planar ion with resonance

🎯 Exam Significance Exam Significance ›

This is one of the most concept-based NCERT questions on oxidation numbers.
Questions involving peroxide compounds are frequently asked in CBSE Board, JEE Main, and NEET.
Students should remember that oxygen does not always possess oxidation number −2.
Dichromate and nitrate ions are important species in inorganic chemistry and analytical chemistry.
Understanding the relationship between molecular structure and oxidation number helps solve advanced redox problems accurately.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Always verify whether oxygen is present as oxide, peroxide, or superoxide before assigning its oxidation number.
Sulphur in Caro's acid has oxidation number +6 because two oxygen atoms are peroxide oxygens (−1 each).
Chromium in dichromate ion has oxidation number +6.
Nitrogen in nitrate ion has oxidation number +5.
Never rely solely on algebraic calculations; always consider the actual molecular structure.

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📘 Concept & Theory Theory / Concept ›

The chemical formula of an ionic compound is obtained by combining cations and anions in such a ratio that the total positive charge equals the total negative charge. The Roman numeral written after the metal name indicates its oxidation state (or valency) in that compound.

To write the correct chemical formula, it is essential to know the charges of common ions.

Ion	Charge
Hg²⁺	+2
Ni²⁺	+2
Sn⁴⁺	+4
Tl⁺	+1
Fe³⁺	+3
Cr³⁺	+3
Cl⁻	−1
SO₄²⁻	−2
O²⁻	−2

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the cation and its oxidation state.
Identify the anion and its charge.
Balance the total positive and negative charges.
Write the simplest whole-number ratio as the chemical formula.

✏️ Solution Complete Solution ›

Step-by-step Solution · 26 steps

(a) Mercury(II) chloride
Mercury(II) ion is \[\mathrm{Hg^{2+}}\]
Chloride ion is \[\mathrm{Cl^-}\]
Two chloride ions are required to balance one Hg²⁺. \[(+2)+2(-1)=0\]
Formula \[\boxed{\mathrm{HgCl_2}}\]
(b) Nickel(II) sulphate
Nickel(II) ion \[\mathrm{Ni^{2+}}\]
Sulphate ion \[\mathrm{SO_4^{2-}}\]
The charges are already balanced. \[(+2)+(-2)=0\]
Formula \[\boxed{\mathrm{NiSO_4}}\]
(c) Tin(IV) oxide
Tin(IV) ion \[\mathrm{Sn^{4+}}\]
Oxide ion \[\mathrm{O^{2-}}\]
Two oxide ions are required. \[(+4)+2(-2)=0\]
Formula \[\boxed{\mathrm{SnO_2}}\]
(d) Thallium(I) sulphate
Thallium(I) ion \[\mathrm{Tl^+}\]
Sulphate ion \[\mathrm{SO_4^{2-}}\]
Two Tl⁺ ions balance one sulphate ion. \[2(+1)+(-2)=0\]
Formula \[\boxed{\mathrm{Tl_2SO_4}}\]
(e) Iron(III) sulphate
Iron(III) ion \[\mathrm{Fe^{3+}}\]
Sulphate ion \[\mathrm{SO_4^{2-}}\]
The least common multiple of 3 and 2 is 6.
Therefore, combine
- 2 Fe³⁺ ions → +6
- 3 SO₄²⁻ ions → −6
\[2(+3)+3(-2)=0\]
Formula \[\boxed{\mathrm{Fe_2(SO_4)_3}}\]
Chromium(III) ion \[\mathrm{Cr^{3+}}\]
Oxide ion \[\mathrm{O^{2-}}\]
LCM of 3 and 2 is 6.
- 2 Cr³⁺ ions → +6
- 3 O²⁻ ions → −6
\[2(+3)+3(-2)=0\]
Formula \[\boxed{\mathrm{Cr_2O_3}}\]

💡 Answer Final Answer ›

S.No.	Compound	Chemical Formula
1	Mercury(II) chloride	HgCl₂
2	Nickel(II) sulphate	NiSO₄
3	Tin(IV) oxide	SnO₂
4	Thallium(I) sulphate	Tl₂SO₄
5	Iron(III) sulphate	Fe₂(SO₄)₃
6	Chromium(III) oxide	Cr₂O₃

🎯 Exam Significance Exam Significance ›

Writing chemical formulas from oxidation states is a fundamental skill tested in CBSE Board examinations.
Questions on nomenclature and formula writing frequently appear in JEE Main and NEET.
The concept strengthens understanding of oxidation numbers, valency, and charge neutrality.
Many inorganic chemistry reactions require students to know the correct formulas of common salts and oxides.
Mastery of this topic helps in balancing chemical equations and predicting reaction products.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

The Roman numeral specifies the oxidation state of the metal ion.
The total positive and negative charges must always balance in a neutral compound.
Use the lowest whole-number ratio of ions while writing chemical formulas.
Polyatomic ions such as sulphate remain unchanged while balancing charges.
Parentheses are used when more than one polyatomic ion is present in the formula, for example, Fe₂(SO₄)₃.

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Q7 →

📘 Concept & Theory Theory / Concept ›

The oxidation number of an element depends on the electronegativity of the atoms bonded to it. Carbon and nitrogen exhibit a wide range of oxidation states because they form stable compounds with both electropositive and electronegative elements.

As the oxidation number increases, the element undergoes oxidation. As the oxidation number decreases, the element undergoes reduction.

This question tests your understanding of assigning oxidation numbers and recognizing compounds representing different oxidation states of an element.

🗺️ Solution Roadmap Step-by-step Plan ›

Recall the possible oxidation states of carbon and nitrogen.
Choose one common compound representing each oxidation state.
Verify the oxidation number by applying standard oxidation-number rules.

✏️ Solution Complete Solution ›

Step-by-step Solution · 2 steps

(A) Carbon: Oxidation States from −4 to +4

Oxidation Number	Representative Compound	Verification
−4	CH₄ (Methane)	\[ x+4(+1)=0 \] \[ x=-4 \]
−3	C₂H₆ (Ethane)	Each carbon has oxidation number \[ -3 \]
−2	C₂H₄ (Ethene)	Each carbon has oxidation number \[ -2 \]
−1	C₂H₂ (Ethyne)	Each carbon has oxidation number \[ -1 \]
0	HCHO (Formaldehyde)	\[ x+2(+1)+(-2)=0 \] \[ x=0 \]
+1	CH₃CHO (Acetaldehyde)	Carbonyl carbon has oxidation number \[ +1 \]
+2	HCOOH (Formic acid)	Carbon has oxidation number \[ +2 \]
+3	CH₃COOH (Acetic acid)	Carboxyl carbon has oxidation number \[ +3 \]
+4	CO₂ (Carbon dioxide)	\[ x+2(-2)=0 \] \[ x=+4 \]

(B) Nitrogen: Oxidation States from −3 to +5

Oxidation Number	Representative Compound	Verification
−3	NH₃ (Ammonia)	\[ x+3(+1)=0 \] \[ x=-3 \]
−2	N₂H₄ (Hydrazine)	Each nitrogen has oxidation number \[ -2 \]
−1	NH₂OH (Hydroxylamine)	\[ x+3(+1)+(-2)=0 \] \[ x=-1 \]
0	N₂ (Nitrogen gas)	Elemental state \[ 0 \]
+1	N₂O (Nitrous oxide)	Average oxidation number \[ +1 \]
+2	NO (Nitric oxide)	\[ x+(-2)=0 \] \[ x=+2 \]
+3	HNO₂ (Nitrous acid)	\[ (+1)+x+2(-2)=0 \] \[ x=+3 \]
+4	NO₂ (Nitrogen dioxide)	\[ x+2(-2)=0 \] \[ x=+4 \]
+5	HNO₃ (Nitric acid)	\[ (+1)+x+3(-2)=0 \] \[ x=+5 \]

🎯 Exam Significance Exam Significance ›

This question develops the ability to assign oxidation numbers in inorganic and organic compounds.
CBSE Board examinations frequently ask students to identify oxidation states of carbon and nitrogen in common compounds.
JEE Main and NEET often test trends in oxidation states and the oxidizing or reducing behavior associated with them.
Knowledge of oxidation-state variation is essential for understanding redox reactions, electrochemistry, p-block chemistry, and environmental chemistry.
Many reactions involving carbon and nitrogen compounds can be predicted by comparing their oxidation states.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Carbon exhibits oxidation states from −4 to +4.
Nitrogen exhibits oxidation states from −3 to +5.
Oxidation number increases as atoms bond with more electronegative elements such as oxygen.
Oxidation number decreases as atoms bond with electropositive elements such as hydrogen.
Assign oxidation numbers systematically using standard rules before drawing conclusions.

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Q8 →

📘 Concept & Theory Theory / Concept ›

Whether a substance acts as an oxidising agent or a reducing agent depends upon the oxidation state of the central atom (or the atoms involved) and its ability to either gain or lose electrons.

An oxidising agent accepts electrons and undergoes reduction.
A reducing agent donates electrons and undergoes oxidation.
A substance having an intermediate oxidation state can often undergo both oxidation and reduction, enabling it to act as both an oxidising and a reducing agent.
A substance in its highest oxidation state generally cannot be oxidized further and therefore behaves only as an oxidising agent.

🗺️ Solution Roadmap Step-by-step Plan ›

Determine the oxidation state of the important element in each compound.
Identify whether the oxidation state is intermediate or maximum.
Explain whether the substance can undergo oxidation, reduction, or both.

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps

(A) Sulphur dioxide (SO₂)
Calculate the oxidation number of sulphur. \[\begin{aligned} x+2(-2)&=0\\ x&=+4 \end{aligned}\]
Sulphur exhibits oxidation states ranging from \[-2,\;0,\;+2,\;+4,\;+6\]
The oxidation state +4 is an intermediate oxidation state. Therefore, sulphur can
- be oxidized from +4 to +6, or
- be reduced from +4 to lower oxidation states.
Hence, SO₂ can act as both an oxidising agent and a reducing agent.
(B) Hydrogen peroxide (H₂O₂)
In hydrogen peroxide, oxygen has oxidation number -1
Oxygen can undergo
- oxidation from −1 to 0 (forming O₂), or
- reduction from −1 to −2 (forming H₂O).
Therefore, H₂O₂ behaves both as an oxidising agent and as a reducing agent.
Examples:
As an oxidising agent \[\mathrm{H_2O_2 + 2I^- +2H^+ \rightarrow I_2 +2H_2O}\] As a reducing agent \[\mathrm{H_2O_2 + O_3 \rightarrow H_2O +2O_2}\]
(C) Ozone (O₃)
In ozone, oxygen is in the elemental form. \[\text{Oxidation number of oxygen}=0\]
The oxidation state of oxygen cannot increase beyond zero in ordinary chemical reactions because oxygen is one of the most electronegative elements.
Instead, oxygen readily gains electrons to form oxide ions having oxidation number -2
Therefore, ozone undergoes reduction and behaves only as an oxidising agent

For example,\[\mathrm{O_3+2I^-+H_2O\rightarrow I_2+O_2+2OH^-}\]

(D) Nitric acid (HNO₃)
Calculate the oxidation number of nitrogen. \[\begin{aligned}(+1)+x+3(-2)&=0\\ x&=+5\end{aligned}\]

The oxidation state +5 is the maximum oxidation state of nitrogen.
Nitrogen cannot be oxidized further.
However, it can be reduced to lower oxidation states such as

+4 in NO₂

+2 in NO

0 in N₂

−3 in NH₃

Therefore, nitric acid only accepts electrons and behaves exclusively as an oxidising agent.

💡 Answer Final Answer ›

Compound Oxidation State Nature Reason

SO₂ S = +4 Oxidising and Reducing Intermediate oxidation state; can be oxidized or reduced.

H₂O₂ O = −1 Oxidising and Reducing Oxygen can be oxidized to 0 or reduced to −2.

O₃ O = 0 Only Oxidising Oxygen readily undergoes reduction to −2 but cannot be oxidized further under normal conditions.

HNO₃ N = +5 Only Oxidising Nitrogen is already in its maximum oxidation state and cannot undergo further oxidation.

Compound	Oxidation State	Nature	Reason
SO₂	S = +4	Oxidising and Reducing	Intermediate oxidation state; can be oxidized or reduced.
H₂O₂	O = −1	Oxidising and Reducing	Oxygen can be oxidized to 0 or reduced to −2.
O₃	O = 0	Only Oxidising	Oxygen readily undergoes reduction to −2 but cannot be oxidized further under normal conditions.
HNO₃	N = +5	Only Oxidising	Nitrogen is already in its maximum oxidation state and cannot undergo further oxidation.

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked conceptual questions from the NCERT textbook.

CBSE Board examinations often ask students to explain why certain substances behave as oxidising agents, reducing agents, or both.

JEE Main and NEET frequently test oxidation-state reasoning instead of direct memorization.

The question connects oxidation numbers with the chemical behavior of important industrial chemicals such as sulphur dioxide, hydrogen peroxide, ozone, and nitric acid.

Understanding this concept is essential for balancing redox reactions and predicting reaction products.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Substances with intermediate oxidation states can often behave as both oxidising and reducing agents.

SO₂ contains sulphur in the +4 oxidation state, allowing both oxidation and reduction.

Hydrogen peroxide contains oxygen in the −1 oxidation state, enabling it to undergo oxidation as well as reduction.

Ozone behaves only as an oxidising agent because oxygen is readily reduced from 0 to −2.

Nitric acid behaves only as an oxidising agent because nitrogen is already in its highest oxidation state (+5).

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Q9 →

Q9

NUMERIC3 marks

Consider the reactions:
(a) 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(aq) + 6O₂(g)

(b) O₃(g) + H₂O₂(l) → H₂O(l) + 2O₂(g)

Why is it more appropriate to write these reactions as:

(a) 6CO₂(g) + 12H₂O(l) → C₆H₁₂O₆(aq) + 6H₂O(l) + 6O₂(g)

(b) O₃(g) + H₂O₂(l) → H₂O(l) + O₂(g) + O₂(g)

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

📘 Concept & Theory Theory / Concept ›

In many redox reactions, the balanced chemical equation does not reveal the actual source of atoms in the products. Although the equation satisfies the laws of conservation of mass and charge, it does not indicate which reactant contributes a particular atom.

To determine the actual reaction pathway, chemists use isotopic tracer techniques. Stable or radioactive isotopes are introduced into one reactant, and their movement is monitored throughout the reaction.

This approach has been widely used to establish the mechanism of photosynthesis and many oxidation-reduction reactions.

🗺️ Solution Roadmap Step-by-step Plan ›

Examine reaction (a) carefully.

Identify the actual source of oxygen evolved.

Examine reaction (b).

Identify the origin of each oxygen molecule produced.

Suggest the appropriate experimental technique.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps
(a) Photosynthesis
The commonly written equation is \[\mathrm{6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2}\]

This equation is chemically balanced, but it does not indicate the origin of oxygen atoms.

Experimental studies using isotopic tracers have shown that the oxygen gas released during photosynthesis comes entirely from water and not from carbon dioxide.

Therefore, the reaction is written as
\[\mathrm{6CO_2+12H_2O\rightarrow C_6H_{12}O_6+6H_2O+6O_2}\]

This representation clearly shows that:

Water acts both as a reactant and as a product.

The oxygen liberated during photosynthesis originates from water molecules.

Some water molecules participate in the reaction, while others are regenerated.

Reaction between ozone and hydrogen peroxide
The reaction is usually written as \[\mathrm{O_3+H_2O_2\rightarrow H_2O+2O_2}\]

Although balanced, this equation does not indicate which oxygen molecule comes from ozone and which comes from hydrogen peroxide.

Therefore, it is better represented as
\[\mathrm{O_3+H_2O_2\rightarrow H_2O+O_2+O_2}\]

This notation emphasizes that:

One oxygen molecule originates from ozone.

The other oxygen molecule originates from hydrogen peroxide.

The two oxygen molecules are chemically identical but arise from different reactants.

Technique Used to Investigate These Reactions

The most suitable technique is the isotopic tracer technique.

In this method, oxygen atoms are replaced with the stable isotope ¹⁸O.

For example:

If H₂¹⁸O is used in photosynthesis, the evolved oxygen gas contains ¹⁸O.

If C¹⁸O₂ is used, the evolved oxygen gas does not contain ¹⁸O.

This proves that the oxygen released during photosynthesis originates from water and not from carbon dioxide.

Similarly, isotopic labeling of ozone or hydrogen peroxide can identify the origin of oxygen molecules formed in reaction (b).

🎯 Exam Significance Exam Significance ›

This question is one of the most important NCERT conceptual questions and has been repeatedly asked in CBSE Board examinations.

JEE Main and NEET frequently test the concept that oxygen evolved during photosynthesis originates from water.

The isotopic tracer technique is an important experimental method used in chemistry, biology, and biochemistry.

This question develops an understanding of reaction mechanisms rather than simply balancing equations.

Knowledge of isotope labeling is useful in studying metabolic pathways, enzyme mechanisms, and industrial reaction mechanisms.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

A balanced equation does not always reveal the actual source of atoms in the products.

During photosynthesis, the oxygen released comes from water, not carbon dioxide.

In the reaction between ozone and hydrogen peroxide, the two oxygen molecules originate from different reactants.

Isotopic labeling with ¹⁸O is the standard method for tracing oxygen atoms.

The isotopic tracer technique is a powerful tool for investigating reaction mechanisms.

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Q10 →

Q10

NUMERIC2 marks

The compound AgF₂ is an unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?

📘 Concept & Theory Theory / Concept ›

The oxidising power of a substance depends on its tendency to gain electrons (i.e., undergo reduction). A species containing an element in a high oxidation state generally has a strong tendency to accept electrons and get reduced to a more stable oxidation state.

Silver usually exhibits the oxidation state +1, which is highly stable. Although silver can exist in the +2 oxidation state in compounds such as AgF₂, this state is much less stable. Consequently, Ag²⁺ readily accepts an electron to convert into the more stable Ag⁺ ion.

🗺️ Solution Roadmap Step-by-step Plan ›

Determine the oxidation state of silver in AgF₂.

Compare it with the most stable oxidation state of silver.

Explain why Ag²⁺ readily undergoes reduction.

Relate this behavior to its strong oxidising nature.

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps
Determine the oxidation number of silver.

In AgF₂, fluorine always has an oxidation number of −1.
Let the oxidation number of silver be x\[\begin{aligned}x+2(-1)&=0\\x&=+2\end{aligned}\]

Thus, silver is present as \[\boxed{\mathrm{Ag^{2+}}}\]

Compare with the stable oxidation state.

Silver most commonly exists as \[\boxed{\mathrm{Ag^+}}\]

The +1 oxidation state is considerably more stable than the +2 oxidation state because of its favorable electronic configuration and lower energy.

Explain the reduction process.

The Ag²⁺ ion readily accepts one electron and gets reduced to Ag⁺.
\[ \mathrm{Ag^{2+}+e^- \rightarrow Ag^+} \]
Since Ag²⁺ has a strong tendency to gain an electron, it readily oxidizes other substances by accepting electrons from them.

Draw the conclusion.

Because Ag²⁺ is unstable and has a strong tendency to convert into the more stable Ag⁺ state, AgF₂ acts as a very powerful oxidising agent.

🎯 Exam Significance Exam Significance ›

This is a frequently asked NCERT conceptual question in CBSE Board examinations.

JEE Main and NEET often test the relationship between oxidation state, stability, and oxidising strength.

The question reinforces the principle that unstable higher oxidation states generally exhibit strong oxidising properties.

Understanding oxidation-state stability helps explain the chemistry of transition elements.

This concept is useful in studying redox reactions, electrochemistry, and coordination chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Silver usually exhibits the stable oxidation state of +1.

In AgF₂, silver exists in the less stable +2 oxidation state.

Ag²⁺ readily gains an electron to form Ag⁺.

A strong tendency to undergo reduction makes AgF₂ a powerful oxidising agent.

Higher, unstable oxidation states generally possess greater oxidising power than lower, stable oxidation states.

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Q11 →

Q11

NUMERIC3 marks

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

📘 Concept & Theory Theory / Concept ›

The extent of a redox reaction depends upon the relative amounts of the oxidising agent and the reducing agent.

An oxidising agent accepts electrons and oxidizes another substance.

A reducing agent donates electrons and reduces another substance.

If the reducing agent is present in excess, the oxidising agent receives more electrons and is reduced to a lower oxidation state.

Conversely, if the oxidising agent is present in excess, the reducing agent loses more electrons and is oxidized to a higher oxidation state.

Thus, the final oxidation state of the products depends upon the relative quantities of the reactants.

🗺️ Solution Roadmap Step-by-step Plan ›

Understand the effect of excess oxidising or reducing agent.

Study reactions that produce different products depending upon the reaction conditions.

Compare the oxidation states of the products formed.

Conclude how the amount of reactants influences the oxidation state.

✏️ Solution Complete Solution ›

Step-by-step Solution · 14 steps
Phosphorus reacts with chlorine

(i) Chlorine is limited (phosphorus in excess)
\[\mathrm{P_4+6Cl_2\rightarrow4PCl_3}\]

Oxidation state of phosphorus:

In P₄ = 0

In PCl₃ = +3

The oxidation stops at the lower oxidation state +3.

(ii) Chlorine is in excess
\[\mathrm{P_4+10Cl_2\rightarrow4PCl_5}\]

Oxidation state of phosphorus:

In PCl₅ = +5

The stronger oxidising conditions produce the higher oxidation state.

Iron reacts with oxygen
Limited oxygen supply \[\mathrm{2Fe+O_2\rightarrow2FeO}\]

Oxidation state of iron: +2

Excess oxygen supply \[\mathrm{4Fe+3O_2\rightarrow2Fe_2O_3}\]

Oxidation state of iron +3
Greater availability of oxygen causes further oxidation of iron.

Carbon burns in oxygen
Oxygen is limited \[\mathrm{2C+O_2\rightarrow2CO}\]

Oxidation state of carbon: +2

Oxygen is in excess \[\mathrm{C+O_2\rightarrow CO_2}\]

Oxidation state of carbon +4
With excess oxygen, carbon undergoes complete oxidation.

General Justification
These examples clearly show that:

When the reducing agent is in excess, oxidation is incomplete and products with lower oxidation states are obtained.

When the oxidising agent is in excess, oxidation proceeds further, producing compounds with higher oxidation states.

Therefore, the oxidation state of the final product depends upon the relative amounts of oxidising and reducing agents used in the reaction.

🎯 Exam Significance Exam Significance ›

This is a highly important conceptual question from the NCERT textbook and is frequently asked in CBSE Board examinations.

JEE Main and NEET often test how reaction conditions influence the oxidation state of products.

Students should remember that the quantity of reactants can determine whether oxidation is partial or complete.

The concept is widely applicable in metallurgy, industrial chemistry, corrosion, combustion, and inorganic chemistry.

Understanding oxidation-state changes helps predict reaction products in unfamiliar chemical reactions.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Excess reducing agent generally results in products having lower oxidation states.

Excess oxidising agent generally produces products having higher oxidation states.

The oxidation state of the product depends on the availability of electrons during the reaction.

Partial oxidation yields lower oxidation states, whereas complete oxidation yields higher oxidation states.

Always examine the reaction conditions before predicting the final oxidation product.

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Q12 →

Q12

NUMERIC3 marks

How do you count for the following observations ?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why ?

📘 Concept & Theory Theory / Concept ›

The oxidising ability of a reagent depends not only on its oxidation potential but also on the reaction medium. Acidic, alkaline and alcoholic media often alter the reaction pathway, the reaction rate and the stability of the products.

Similarly, whether concentrated sulphuric acid acts merely as an acid or also as an oxidising agent depends upon the reducing strength of the anion present.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the oxidation and reduction taking place.

Explain the role of the reaction medium.

Determine whether concentrated sulphuric acid acts only as an acid or also as an oxidising agent.

Write the balanced chemical equations.

✏️ Solution Complete Solution ›

Step-by-step Solution · 17 steps
(a) Oxidation of Toluene to Benzoic Acid
Concept:

Potassium permanganate is a powerful oxidising agent in acidic, alkaline and neutral media. During oxidation of the side chain of aromatic hydrocarbons, the oxidation proceeds more smoothly in alkaline (or alcoholic alkaline) potassium permanganate.

Initially, the methyl group of toluene is oxidized to the potassium salt of benzoic acid (potassium benzoate). On subsequent acidification, potassium benzoate is converted into benzoic acid.

The alkaline medium also minimizes undesirable side reactions and prevents further oxidation of the aromatic ring.

Explanation

The methyl group attached to the benzene ring undergoes oxidation.

Potassium permanganate is reduced from manganese(+7) to manganese dioxide, MnO₂.

Acidification converts potassium benzoate into benzoic acid.

Balanced Redox Equation
The oxidation in alkaline medium may be represented as \[\mathrm{C_6H_5CH_3+2KMnO_4\rightarrow C_6H_5COOK+2MnO_2+KOH+H_2O}\]

After acidification, \[\mathrm{C_6H_5COOK+HCl\rightarrow C_6H_5COOH+KCl}\]

Overall reaction \[\boxed{\mathrm{C_6H_5CH_3\xrightarrow[\text{alkaline/alcoholic }KMnO_4]{}C_6H_5COOH}}\]

Conclusion
Therefore, alkaline (or alcoholic alkaline) potassium permanganate is preferred because it efficiently oxidizes the side-chain methyl group into the carboxyl group while minimizing side reactions.

(b) Reaction of Concentrated Sulphuric Acid with Chlorides and Bromides
Concept

Concentrated sulphuric acid behaves as

a strong acid, and

a strong oxidising agent.

Whether oxidation occurs depends upon the reducing ability of the halide ion.

Case I: Chloride
Chloride ion is a weak reducing agent.

Therefore, concentrated sulphuric acid acts only as an acid.
\[\mathrm{NaCl+H_2SO_4\rightarrow NaHSO_4+HCl}\]
Since chloride cannot reduce sulphuric acid, no oxidation occurs.

Hence, only colourless hydrogen chloride gas is evolved.

Case II: Bromide
Bromide ion is a much stronger reducing agent than chloride.

Initially, hydrogen bromide is produced.
\[\mathrm{NaBr+H_2SO_4\rightarrow NaHSO_4+HBr}\]
The HBr formed immediately reduces concentrated sulphuric acid.

Oxidation half-reaction
\[\mathrm{2Br^-\rightarrow Br_2+2e^-}\]

Reduction half-reaction
\[\mathrm{H_2SO_4\rightarrow SO_2}\]

Overall reaction \[\mathrm{2HBr+H_2SO_4\rightarrow Br_2+SO_2+2H_2O}\]

The reddish-brown vapours observed are due to bromine.

Conclusion
Because bromide is a stronger reducing agent than chloride, it reduces concentrated sulphuric acid to sulphur dioxide while itself being oxidized to bromine.

🎯 Exam Significance Exam Significance ›

This question is one of the most frequently asked conceptual questions from the NCERT textbook.

CBSE Board examinations often ask students to explain the different behaviour of chloride and bromide ions with concentrated sulphuric acid.

JEE Main and NEET regularly test oxidation of alkyl side chains using potassium permanganate.

The reaction of concentrated sulphuric acid with halides is a standard topic in p-block chemistry and redox chemistry.

Students should understand how reaction medium influences the oxidising power of potassium permanganate.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Alkaline (or alcoholic alkaline) KMnO₄ selectively oxidizes the side chain of toluene to benzoic acid.

Potassium benzoate is formed first and is converted into benzoic acid upon acidification.

Concentrated sulphuric acid acts only as an acid with chlorides.

Bromide ions reduce concentrated sulphuric acid and are oxidized to bromine.

The reducing power of halide ions increases in the order

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Q13 →

Q13

NUMERIC3 marks

Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:
(a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)

(b) HCHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH⁻(aq) → 2Ag(s) + HCOO⁻(aq) + 4NH₃(aq) + 2H₂O(l)

(c) HCHO(l) + 2Cu²⁺(aq) + 5OH⁻(aq) → Cu₂O(s) + HCOO⁻(aq) + 3H₂O(l)

(d) N₂H₄(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l)

(e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

📘 Concept & Theory Theory / Concept ›

To identify the oxidised substance, reduced substance, oxidising agent and reducing agent, the oxidation numbers of the important elements in the reactants and products are compared.

Oxidation → Increase in oxidation number.

Reduction → Decrease in oxidation number.

Oxidising agent → Gets reduced.

Reducing agent → Gets oxidised.

🗺️ Solution Roadmap Step-by-step Plan ›

Assign oxidation numbers.

Identify the increase or decrease in oxidation number.

Determine oxidation and reduction.

Identify the oxidising and reducing agents.

✏️ Solution Complete Solution ›

Step-by-step Solution · 19 steps
(a) 2AgBr + C₆H₆O₂ → 2Ag + 2HBr + C₆H₄O₂
Oxidation number changes

Silver: +1 → 0 (Reduction)

Hydroquinone (C₆H₆O₂) is oxidized to benzoquinone (C₆H₄O₂).

Result

Quantity Substance

Oxidised C₆H₆O₂ (Hydroquinone)

Reduced AgBr (Ag⁺)

Oxidising Agent AgBr

Reducing Agent C₆H₆O₂

(b) HCHO + 2[Ag(NH₃)₂]⁺ → 2Ag + HCOO⁻
Oxidation number of carbon
In formaldehyde (HCHO) \[\begin{aligned} x+2(+1)+(-2)&=0\\ x&=0 \end{a;igned}\]

In formate ion (HCOO⁻) \[\begin{aligned} (+1)+x+2(-2)&=-1\\ x&=+2 \end{aligned\]

Carbon changes \[0\rightarrow +2\]

Silver changes \[+1\rightarrow0\]

Result

Quantity Substance

Oxidised HCHO

Reduced [Ag(NH₃)₂]⁺

Oxidising Agent Tollens' reagent

Reducing Agent HCHO

(c) HCHO + 2Cu²⁺ + 5OH⁻ → Cu₂O + HCOO⁻
Carbon again changes \[0\rightarrow +2\]

Copper changes \[+2\rightarrow +1\]

since Cu₂O contains Cu⁺.
Result

Quantity Substance

Oxidised HCHO

Reduced Cu²⁺

Oxidising Agent Cu²⁺ (Fehling's/Benedict's reagent)

Reducing Agent HCHO

(d) N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
Nitrogen
In hydrazine \[2x+4(+1)=0\] \[x=-2\]

In N₂ - 0

Thus, nitrogen changes \[-2\rightarrow0\]

Oxygen in H₂O₂ \[-1\rightarrow-2\]

Result

Quantity Substance

Oxidised N₂H₄

Reduced H₂O₂

Oxidising Agent H₂O₂

Reducing Agent N₂H₄

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
This is the discharge reaction of a lead-acid storage battery.

Oxidation numbers

Pb(s) = 0

Pb in PbO₂ = +4

Pb in PbSO₄ = +2

Changes:

Pb: 0 → +2 (Oxidation)

Pb: +4 → +2 (Reduction)

Result

Quantity Substance

Oxidised Pb(s)

Reduced PbO₂

Oxidising Agent PbO₂

Reducing Agent Pb(s)

💡 Answer Final Answer ›

Reaction Oxidised Substance Reduced Substance Oxidising Agent Reducing Agent

(a) Hydroquinone (C₆H₆O₂) AgBr AgBr Hydroquinone

(b) HCHO [Ag(NH₃)₂]⁺ Tollens' reagent HCHO

(c) HCHO Cu²⁺ Cu²⁺ HCHO

(d) N₂H₄ H₂O₂ H₂O₂ N₂H₄

(e) Pb PbO₂ PbO₂ Pb

🎯 Exam Significance Exam Significance ›

This is one of the most important NCERT exercises for identifying oxidation and reduction using oxidation numbers.

CBSE Board examinations frequently ask students to identify oxidising and reducing agents in redox reactions.

JEE Main and NEET commonly include questions based on Tollens' reagent, Fehling's solution, hydrogen peroxide, and lead-acid battery reactions.

Understanding oxidation-number changes is essential for balancing redox equations and studying electrochemistry.

The lead-acid battery reaction is also an important application-based topic in electrochemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

The substance whose oxidation number increases is oxidised and acts as the reducing agent.

The substance whose oxidation number decreases is reduced and acts as the oxidising agent.

Tollens' reagent oxidizes aldehydes to carboxylates while depositing metallic silver.

Fehling's reagent oxidizes aldehydes and is reduced to red Cu₂O.

Lead-acid batteries operate through simultaneous oxidation of Pb and reduction of PbO₂.

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Q14 →

Q14

NUMERIC3 marks

Consider the reactions: \[\mathrm{2S_2O_3^{2-}(aq)+I_2(s)\rightarrow S_4O_6^{2-}(aq)+2I^-(aq)}\] \[\mathrm{S_2O_3^{2-}(aq)+2Br_2(l)+5H_2O(l)\rightarrow 2SO_4^{2-}(aq)+4Br^-(aq)+10H^+(aq)}\] Why does the same reductant, thiosulphate react differently with iodine and bromine ?

📘 Concept & Theory Theory / Concept ›

The product obtained in a redox reaction depends not only on the reducing agent but also on the strength of the oxidising agent. A weak oxidising agent causes only partial oxidation of the reducing agent, whereas a strong oxidising agent can oxidize it completely to its highest oxidation state.

Among the halogens, the oxidising power follows the order
\[\boxed{\mathrm{F_2>Cl_2>Br_2>I_2}}\]
Thus, bromine is a stronger oxidising agent than iodine.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the oxidising agents in both reactions.

Compare their oxidising strengths.

Determine the oxidation products of thiosulphate.

Explain why different products are formed.

✏️ Solution Complete Solution ›

Step-by-step Solution · 12 steps
Reaction (1): Thiosulphate with Iodine
\[\mathrm{2S_2O_3^{2-}+I_2\rightarrow S_4O_6^{2-}+2I^-}\]

Iodine is reduced. \[\mathrm{I_2+2e^-\rightarrow 2I^-}\]

Thiosulphate is oxidized only partially.

Two thiosulphate ions combine to form tetrathionate ion. \[\mathrm{2S_2O_3^{2-}\rightarrow S_4O_6^{2-}+2e^-}\]

No sulphur atom reaches its maximum oxidation state.
This is a mild oxidation.

Reaction (2): Thiosulphate with Bromine
\[\mathrm{S_2O_3^{2-}+2Br_2+5H_2O\rightarrow 2SO_4^{2-}+4Br^-+10H^+}\]

Bromine is reduced. \[\mathrm{Br_2+2e^-\rightarrow 2Br^-}\]

Thiosulphate is completely oxidized.
Sulphur is converted into sulphate ion, where sulphur has oxidation number: +6

This represents complete oxidation.

Why Does Thiosulphate Behave Differently?

Iodine is a comparatively weak oxidising agent. It can oxidize thiosulphate only to tetrathionate.

Bromine is a much stronger oxidising agent. Therefore, it oxidizes thiosulphate completely to sulphate ions.

Thus, the different oxidation products are due to the different oxidising strengths of iodine and bromine.

Comparison

Property Iodine Bromine

Oxidising strength Weaker Stronger

Extent of oxidation of S₂O₃²⁻ Partial Complete

Main sulphur product S₄O₆²⁻ SO₄²⁻

🎯 Exam Significance Exam Significance ›

This is one of the most important NCERT conceptual questions on redox chemistry.

CBSE Board examinations frequently test the relationship between oxidising strength and reaction products.

JEE Main and NEET often ask questions based on the relative oxidising powers of halogens.

Students should remember that the extent of oxidation depends upon the strength of the oxidising agent.

Thiosulphate is widely used in iodometric titrations, making this concept important in analytical chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Iodine is a weaker oxidising agent than bromine.

Weak oxidising agents generally cause partial oxidation.

Strong oxidising agents can oxidize substances completely to their highest oxidation states.

Thiosulphate is oxidized to tetrathionate by iodine but to sulphate by bromine.

The oxidising power of halogens decreases in the order

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Q15 →

Q15

NUMERIC3 marks

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

📘 Concept & Theory Theory / Concept ›

The oxidising or reducing nature of a substance depends on its tendency to gain or lose electrons.

An oxidising agent readily gains electrons and gets reduced.

A reducing agent readily loses electrons and gets oxidised.

Among the halogens, oxidising power depends upon the ease with which the halogen molecule accepts electrons to form halide ions.
\[ \mathrm{X_2+2e^- \rightarrow 2X^-} \]
Among hydrohalic acids, reducing power depends upon how readily the halide ion loses electrons.
\[ \mathrm{2X^- \rightarrow X_2+2e^-} \]
The trends are:

Oxidising power of halogens
\[ \boxed{\mathrm{F_2>Cl_2>Br_2>I_2}} \]
Reducing power of hydrohalic acids (or halide ions)
\[ \boxed{\mathrm{HF

🗺️ Solution Roadmap Step-by-step Plan ›

Compare the electron-accepting tendency of halogens.

Explain why fluorine is the strongest oxidising agent.

Compare the electron-donating tendency of hydrohalic acids.

Explain why HI is the strongest reducing agent.

Support the explanation with balanced chemical equations.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps
(A) Fluorine is the Best Oxidising Agent
Fluorine has the highest electronegativity among all elements and possesses a very high standard reduction potential.

Therefore, fluorine readily gains electrons according to \[\mathrm{F_2+2e^- \rightarrow 2F^-}\] As a result, fluorine can oxidize all other halide ions.

Example 1
\[\mathrm{F_2+2Cl^- \rightarrow 2F^-+Cl_2}\]

Example 2
\[\mathrm{F_2+2Br^- \rightarrow 2F^-+Br_2}\]

Example 3 \[\mathrm{F_2+2I^- \rightarrow 2F^-+I_2}\]

These reactions show that fluorine oxidizes chloride, bromide and iodide ions to their respective halogens.

Since no other halogen can oxidize fluoride ion, fluorine is the strongest oxidising agent among the halogens.

(B) Hydroiodic Acid is the Best Reducing Agent
Hydroiodic acid dissociates completely in water to give iodide ions. \[\mathrm{HI\rightarrow H^++I^-}\]

Iodide ion has the largest ionic size among the halide ions and holds its outermost electron least strongly.
Hence, it readily loses electrons. \[\mathrm{2I^- \rightarrow I_2+2e^-}\]

Therefore, HI acts as the strongest reducing agent among hydrohalic acids. Example 1
Hydroiodic acid reduces concentrated sulphuric acid. \[\mathrm{8HI+H_2SO_4\rightarrow H_2S+4I_2+4H_2O}\]

Example 2
Hydroiodic acid reduces ferric ions. \[\mathrm{2Fe^{3+}+2I^-\rightarrow 2Fe^{2+}+I_2}\]

These reactions demonstrate that iodide ion is readily oxidized to iodine while reducing other substances.

🎯 Exam Significance Exam Significance ›

This is one of the most important conceptual questions from the NCERT chapter on redox reactions.

CBSE Board examinations frequently ask students to compare the oxidising power of halogens and the reducing power of hydrohalic acids.

JEE Main and NEET often test these periodic trends using reaction-based questions.

The question links periodic properties with oxidation-reduction behavior and standard reduction potentials.

Understanding these trends helps in predicting displacement reactions among halogens and halide ions.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Fluorine has the highest oxidising power among all halogens.

Fluorine can oxidize Cl⁻, Br⁻ and I⁻ ions.

Hydroiodic acid is the strongest reducing hydrohalic acid because iodide ions lose electrons most readily.

The oxidising power of halogens decreases down the group.

The reducing power of hydrohalic acids increases down the group.

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Q16 →

Q16

NUMERIC3 marks

Why does the following reaction occur ? \(\mathrm{XeO_6^{4-}(aq)+2F^-(aq)+6H^+(aq)\rightarrow XeO_3(g)+F_2(g)+3H_2O(l)}\) What conclusion about the compound \(\mathrm{Na_4XeO_6}\) (of which \(\mathrm{XeO_6^{4–}}\) is a part) can be drawn from the reaction.

📘 Concept & Theory Theory / Concept ›

This reaction is an example of a redox reaction in which one species is oxidized while another is reduced. To understand why the reaction occurs, we compare the oxidation numbers of xenon and fluorine before and after the reaction.

The reaction also demonstrates that a compound may simultaneously contain both oxidising and reducing properties depending on the oxidation state of its constituent elements.

🗺️ Solution Roadmap Step-by-step Plan ›

Determine the oxidation number of xenon in XeO₆⁴⁻.

Determine the oxidation number of xenon in XeO₃.

Determine the oxidation-number change of fluorine.

Identify oxidation and reduction.

Draw the conclusion regarding Na₄XeO₆.

✏️ Solution Complete Solution ›

Step-by-step Solution · 13 steps
Oxidation number of xenon in XeO₆⁴⁻

Let the oxidation number of xenon be x. \[\begin{aligned}x+6(-2)&=-4\\ x-12&=-4\\ x&=+8\end{aligned}\]

Thus, xenon is present in its highest oxidation state, +8.

Oxidation number of xenon in XeO₃ \[\begin{aligned} x+3(-2)&=0\\ x&=+6 \end{aligned}\]

Xenon changes from \[+8\rightarrow+6\]

Hence, xenon undergoes reduction.

Oxidation number of fluorine

In fluoride ion, \[\begin{aligned} \mathrm{F^-} \end{aligned}\]

the oxidation number is: -1

In fluorine molecule, \(\mathrm{F_2}\) the oxidation number is: 0

Therefore, fluorine changes from \[-1\rightarrow0\] Hence, fluoride ion undergoes oxidation.

Identify oxidation and reduction

Species Oxidation Number Change Process

Xe +8 → +6 Reduction

F −1 → 0 Oxidation

Since oxidation and reduction occur simultaneously, the reaction is a redox reaction.

"Conclusion regarding Na₄XeO₆

In Na₄XeO₆, xenon exists in the extremely high oxidation state of +8, which is highly unstable.

Because xenon readily accepts electrons and is reduced to the more stable +6 oxidation state, the xenate(VIII) ion acts as a very strong oxidising agent.

The reaction shows that even fluoride ions, which are normally difficult to oxidize because fluorine is the most electronegative element, are oxidized to fluorine gas.

This is possible only because XeO₆⁴⁻ possesses exceptionally strong oxidising power.

🎯 Exam Significance Exam Significance ›

This question highlights the relationship between oxidation state and oxidising power.

It illustrates that elements in their highest oxidation state generally behave as strong oxidising agents.

The question is important for understanding the chemistry of noble gases, especially xenon compounds.

JEE Main and NEET frequently test oxidation-state calculations involving noble gas compounds.

The reaction emphasizes that fluoride ions are oxidized only by exceptionally powerful oxidising agents.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Xenon has oxidation number +8 in XeO₆⁴⁻.

Xenon is reduced to +6 in XeO₃.

Fluoride ions are oxidized to fluorine gas.

XeO₆⁴⁻ is an exceptionally strong oxidising agent.

Na₄XeO₆ contains xenon in its highest oxidation state and is therefore highly unstable.

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Q17

NUMERIC3 marks

Consider the following reactions:

\(\mathrm{H_3PO_2(aq)+4AgNO_3(aq)+2H_2O(l)\rightarrow H_3PO_4(aq)+4Ag(s)+4HNO_3(aq)}\)

\(\mathrm{H_3PO_2(aq)+2CuSO_4(aq)+2H_2O(l)\rightarrow H_3PO_4(aq)+2Cu(s)+2H_2SO_4(aq)}\)

\(\mathrm{C_6H_5CHO(l)+2[Ag(NH_3)_2]^+(aq)+3OH^-(aq)\rightarrow C_6H_5COO^-(aq)+2Ag(s)+4NH_3(aq)+2H_2O(l)}\)

\)\mathrm{C_6H_5CHO(l)+2Cu^{2+}(aq)+5OH^-(aq)\rightarrow \text{No reaction}}\)

📘 Concept & Theory Theory / Concept ›

The reducing ability of a substance depends upon its ease of oxidation. Some reducing agents are strong enough to reduce both silver(I) and copper(II) ions, whereas weaker reducing agents may reduce only silver(I).

Similarly, aldehydes exhibit different behavior toward Tollens' reagent and Fehling's solution depending upon whether they are aliphatic or aromatic.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the reducing agent in each reaction.

Compare the oxidising strengths of Ag⁺ and Cu²⁺.

Explain the behavior of hypophosphorous acid.

Explain the different behavior of benzaldehyde toward Tollens' reagent and Fehling's solution.

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps
Part (A): Reactions of Hypophosphorous Acid
Oxidation state of phosphorus in H₃PO₂

Let the oxidation number of phosphorus be x. \[\begin{aligned} x+3(+1)+2(-2)=0\\ x+3-4=0\\ x=+1 \end{aligned}\]

In phosphoric acid (H₃PO₄) \[\begin{aligned} x+3(+1)+4(-2)=0\\ x+3-8=0\\ x=+5 \end{aligned}\]

Thus phosphorus undergoes oxidation \[+1\rightarrow+5\] Hence, H₃PO₂ acts as a reducing agent.

Reaction (a)
Silver ion is reduced. \[\mathrm{Ag^+ +e^-\rightarrow Ag}\] Therefore, AgNO₃ acts as the oxidising agent.

Reaction (b)
Copper(II) ion is reduced. \[\mathrm{Cu^{2+}+2e^-\rightarrow Cu}\] Hence, CuSO₄ also acts as an oxidising agent.

Since H₃PO₂ reduces both Ag⁺ and Cu²⁺, it is a strong reducing agent.

Part (B): Reactions of Benzaldehyde
Reaction (c): With Tollens' Reagent
Benzaldehyde contains the aldehyde group.

Tollens' reagent is a mild oxidising agent.

It oxidizes benzaldehyde into benzoate ion.
\[\mathrm{C_6H_5CHO\rightarrow C_6H_5COO^-}\]
Silver ions are reduced.
\[\mathrm{Ag^+\rightarrow Ag}\] Hence, a silver mirror is produced.

Reaction (d): With Fehling's Solution
Fehling's solution contains alkaline Cu²⁺ ions.

Aromatic aldehydes such as benzaldehyde do not reduce Fehling's solution.

The aldehyde group is directly attached to the benzene ring, which stabilizes it by resonance and makes oxidation comparatively difficult under the conditions of Fehling's test.

Therefore, no red precipitate of Cu₂O is formed.

Final Explanation

Reaction Reason

(a) H₃PO₂ is oxidized from P(+1) to P(+5) and reduces Ag⁺ to Ag.

(b) H₃PO₂ also reduces Cu²⁺ to Cu because it is a strong reducing agent.

(c) Benzaldehyde reduces Tollens' reagent and gives a silver mirror.

(d) Benzaldehyde does not reduce Fehling's solution because aromatic aldehydes are resistant to oxidation under alkaline Cu²⁺ conditions.

🎯 Exam Significance Exam Significance ›

This question combines concepts of oxidation number, reducing agents, and qualitative organic analysis.

CBSE Board examinations frequently ask students to distinguish between Tollens' reagent and Fehling's solution.

JEE Main and NEET often test why aromatic aldehydes respond differently from aliphatic aldehydes.

Hypophosphorous acid is a frequently asked example of a phosphorus compound exhibiting strong reducing properties.

Understanding these reactions helps in identifying functional groups and predicting reaction products.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

H₃PO₂ is a strong reducing agent because phosphorus is oxidized from +1 to +5.

Tollens' reagent oxidizes both aliphatic and aromatic aldehydes.

Fehling's solution generally oxidizes aliphatic aldehydes but not aromatic aldehydes.

Benzaldehyde gives a positive Tollens' test but a negative Fehling's test.

Silver ions are more readily reduced than Cu²⁺ ions under these reaction conditions.

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Q18 →

Q18

NUMERIC3 marks

Balance the following redox reactions by the ion-electron (half-reaction) method.

\(\mathrm{MnO_4^- + I^- \rightarrow MnO_2 + I_2}\) (Basic medium)

\(\mathrm{MnO_4^- + SO_2 \rightarrow Mn^{2+} + HSO_4^-}\) (Acidic medium)

\(\mathrm{H_2O_2 + Fe^{2+} \rightarrow Fe^{3+} + H_2O}\) (Acidic medium)

\(\mathrm{Cr_2O_7^{2-} + SO_2 \rightarrow Cr^{3+} + SO_4^{2-}}\) (Acidic medium)

📘 Concept & Theory Theory / Concept ›

The ion-electron (half-reaction) method is a systematic procedure used to balance redox reactions, particularly those occurring in acidic or basic media.

The steps are:

Separate oxidation and reduction half-reactions.

Balance atoms other than oxygen and hydrogen.

Balance oxygen atoms using H₂O.

Balance hydrogen atoms using H⁺ (acidic medium).

For basic medium, neutralize H⁺ by adding OH⁻ to both sides.

Balance charge using electrons.

Multiply the half-reactions so that the number of electrons lost equals the number of electrons gained.

Add the half-reactions and cancel common species.

🗺️ Solution Roadmap Step-by-step Plan ›

Write oxidation and reduction half-reactions.

Balance O, H and charge.

Equalize the number of electrons.

Add the half-reactions.

Cancel identical species and write the final balanced equation.

✏️ Solution Complete Solution ›

Step-by-step Solution · 23 steps
(a) MnO₄⁻ + I⁻ → MnO₂ + I₂ (Basic Medium)
Reduction Half-Reaction

\[\mathrm{MnO_4^- \rightarrow MnO_2}\]

Balance oxygen: \[\mathrm{MnO_4^- \rightarrow MnO_2+2H_2O}\]

Balance hydrogen: \[\mathrm{4H^+ + MnO_4^- \rightarrow MnO_2+2H_2O}\]

Balance charge: \[\mathrm{3e^-+4H^++MnO_4^- \rightarrow MnO_2+2H_2O}\]

Convert into basic medium:

Add \(4OH^-\) to both sides. \[\mathrm{3e^-+2H_2O+MnO_4^- \rightarrow MnO_2+4OH^-}\]

Oxidation Half-Reaction \[\mathrm{2I^- \rightarrow I_2+2e^-}\]

Multiply: Reduction ×2 and Oxidation ×3 \[\mathrm{6e^-+4H_2O+2MnO_4^- \rightarrow 2MnO_2+8OH^-}\] \mathrm{6I^- \rightarrow 3I_2+6e^-}

Final Balanced Equation \[\boxed{\mathrm{2MnO_4^-+6I^-+4H_2O\rightarrow 2MnO_2+3I_2+8OH^-}}\]

(b) MnO₄⁻ + SO₂ → Mn²⁺ + HSO₄⁻ (Acidic Medium)
Reduction Half-Reaction \[\mathrm{MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O}\]

Oxidation Half-Reaction \[\mathrm{SO_2+2H_2O\rightarrow HSO_4^-+3H^++2e^-}\]

Multiply Reduction ×2 and Oxidation ×5

Add and simplify.

Final Balanced Equation \[\boxed{\mathrm{2MnO_4^-+5SO_2+2H_2O+H^+\rightarrow 2Mn^{2+}+5HSO_4^-}}\]

(c) H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O (Acidic Medium)
Reduction Half-Reaction \[\mathrm{H_2O_2+2H^++2e^-\rightarrow 2H_2O }\]

Oxidation Half-Reaction \[\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-}\]

Multiply oxidation half by 2.

Final Balanced Equation \[\boxed{\mathrm{H_2O_2+2Fe^{2+}+2H^+\rightarrow 2Fe^{3+}+2H_2O}}\]

(d) Cr₂O₇²⁻ + SO₂ → Cr³⁺ + SO₄²⁻ (Acidic Medium)
Reduction Half-Reaction \[\mathrm{Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O}\]

Oxidation Half-Reaction \[\mathrm{SO_2+2H_2O\rightarrow SO_4^{2-}+4H^++2e^-}\]

Multiply oxidation half by 3. and Add and simplify.

Final Balanced Equation \[\boxed{\mathrm{Cr_2O_7^{2-}+3SO_2+2H^+\rightarrow 2Cr^{3+}+3SO_4^{2-}+H_2O}}\]

🎯 Exam Significance Exam Significance ›

The ion-electron method is the standard procedure prescribed by NCERT for balancing redox reactions.

CBSE Board examinations frequently ask students to balance reactions occurring in acidic and basic media.

JEE Main and NEET regularly include questions involving permanganate, dichromate, hydrogen peroxide, and iron redox systems.

Understanding half-reaction balancing is essential for electrochemistry and qualitative inorganic analysis.

Always balance atoms first, then charge, and finally electrons.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Use H₂O to balance oxygen atoms.

Use H⁺ in acidic medium and OH⁻ in basic medium.

Electrons are used only to balance charge.

The total number of electrons lost equals the total number of electrons gained.

The final balanced equation must satisfy conservation of both mass and charge.

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Q19 →

Q19

NUMERIC3 marks

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

\(\mathrm{P_4(s)+OH^-(aq)\rightarrow PH_3(g)+H_2PO_2^-(aq)}\)

\(\mathrm{N_2H_4(l)+ClO_3^-(aq)\rightarrow NO(g)+Cl^-(aq)}\)

\(\mathrm{Cl_2O_7(g)+H_2O_2(aq)\rightarrow ClO_2^-(aq)+O_2(g)+H^+}\)

📘 Concept & Theory Theory / Concept ›

The ion-electron (half-reaction) method and the oxidation number method are two standard procedures for balancing redox equations.

In basic medium:

Balance oxygen atoms using H₂O.

Balance hydrogen atoms using OH⁻.

Balance charge using electrons.

Finally, cancel common species to obtain the balanced equation.

The oxidation number method gives the same final equation by equating the total increase and decrease in oxidation numbers.

✏️ Solution Complete Solution ›

Step-by-step Solution · 29 steps
(a) P₄ + OH⁻ → PH₃ + H₂PO₂⁻
Oxidation Numbers

In elemental phosphorus \[\mathrm{P_4}\]

Oxidation number of phosphorus is = 0

In phosphine \(\mathrm{PH_3}\) \[x+3(+1)=0\Rightarrow x=-3\]

In hypophosphite ion \(\mathrm{H_2PO_2^-}\) \[\begin{aligned}x+2(+1)+2(-2)&=-1\\x&=+1\end{aligned}\]

Thus phosphorus undergoes simultaneous oxidation and reduction (disproportionation).

Balanced Equation \[\boxed{\mathrm{P_4+3OH^-+3H_2O\rightarrow PH_3+3H_2PO_2^-}}\]

Oxidising Agent: P₄
Reducing Agent: P₄

Since the same substance undergoes oxidation and reduction, phosphorus acts as both oxidising and reducing agent.

(b) N₂H₄ + ClO₃⁻ → NO + Cl⁻
Oxidation Numbers

For nitrogen in hydrazine \[2x+4(+1)=0\Rightarrow x=-2\]

In nitric oxide \[x+(-2)=0\Rightarrow x=+2\]

Nitrogen changes \[-2\rightarrow+2\]

For chlorine:

In chlorate ion \[x+3(-2)=-1\Rightarrow x=+5\]

In chloride ion = -1

Chlorine changes \[+5\rightarrow-1\]

Balanced Equation \[\boxed{\mathrm{3N_2H_4+2ClO_3^-\rightarrow 6NO+2Cl^-+6H_2O}}\]

Oxidising Agent: ClO₃⁻

Reducing Agent: N₂H₄

(c) Cl₂O₇ + H₂O₂ → ClO₂⁻ + O₂
Note: Although the question mentions basic medium, the given reaction contains H⁺. Therefore, the balanced equation obtained in acidic medium is first written and then converted to basic medium by adding equal numbers of OH⁻ ions to both sides.

Oxidation Numbers

In Cl₂O₇ \[2x+7(-2)=0\Rightarrow x=+7\]

In chlorite ion \[x+2(-2)=-1\Rightarrow x=+3\]

Chlorine changes \[+7\rightarrow+3\]

Oxygen in hydrogen peroxide changes \[-1\rightarrow0\]

Balanced Equation (Basic Medium) \[\boxed{\mathrm{Cl_2O_7+4H_2O_2+2OH^-\rightarrow 2ClO_2^-+4O_2+5H_2O}}\]

Oxidising Agent: Cl₂O₇

Reducing Agent: H₂O₂

🎯 Exam Significance Exam Significance ›

This exercise integrates oxidation number calculations with the ion-electron balancing method.

Reaction (a) is a classic example of a disproportionation reaction, where the same element undergoes oxidation and reduction simultaneously.

Hydrazine is an important reducing agent frequently tested in JEE Main and NEET.

Hydrogen peroxide can behave as either an oxidising or reducing agent depending on the reaction partner.

Balancing redox reactions in basic medium is a high-weightage topic in CBSE Board examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

The ion-electron method and oxidation number method always produce the same balanced equation.

P₄ undergoes disproportionation in alkaline medium.

Hydrazine is a powerful reducing agent.

Hydrogen peroxide acts as a reducing agent when reacting with stronger oxidising agents such as Cl₂O₇.

Always identify oxidation-number changes before balancing a redox equation.

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Q20 →

Q20

NUMERIC2 marks

What sorts of information can you draw from the following reaction? \[\mathrm{(CN)_2(g)+2OH^-(aq)\rightarrow CN^-(aq)+CNO^-(aq)+H_2O(l)}\]

📘 Concept & Theory A redox reaction can provide valuable information about the oxidation states of the elements involved and the roles played by different reactants. In some reactions, the same element present in one oxidation state undergoes both oxidation and reduction simultaneously. Such reactions are called disproportionation reactions. To identify this behavior, we first calculate the oxidation number of carbon in each species. ›

🗺️ Solution Roadmap Step-by-step Plan ›

Determine the oxidation number of carbon in cyanogen, cyanide ion and cyanate ion.

Identify whether oxidation and reduction occur simultaneously.

Determine the oxidising and reducing agents.

State the type of redox reaction.

✏️ Solution Complete Solution ›

Step-by-step Solution · 13 steps
Oxidation number of carbon in cyanogen, (CN)₂

In one CN group, nitrogen is more electronegative than carbon and is assigned an oxidation number of −3.

Let the oxidation number of carbon be x. \[x+(-3)=0\] \[x=+3\]

Therefore, carbon in cyanogen has oxidation number \[\boxed{+3}\]

Oxidation number of carbon in cyanide ion, CN⁻ \[x+(-3)=-1\] \[x=+2\]

Carbon changes from \[+3\rightarrow+2\]

This represents reduction

Oxidation number of carbon in cyanate ion, CNO⁻

Let the oxidation number of carbon be x. \[x+(-3)+(-2)=-1\] \[x=+4\]

Carbon changes from \[+3\rightarrow+4\]

This represents oxidation.

Interpretation

Thus, one carbon atom in cyanogen is reduced from +3 to +2, while another carbon atom is oxidized from +3 to +4.

Hence, the same compound undergoes both oxidation and reduction simultaneously.

This type of reaction is known as a disproportionation reaction.

Oxidation Number Summary

Species Oxidation Number of Carbon Process

(CN)₂ +3 Initial state

CN⁻ +2 Reduction

CNO⁻ +4 Oxidation

🎯 Exam Significance Exam Significance ›

This question tests the ability to calculate oxidation numbers in polyatomic ions.

It is a standard NCERT example illustrating a disproportionation reaction.

CBSE Board examinations frequently ask students to identify disproportionation reactions.

JEE Main and NEET often include questions based on oxidation-number changes rather than direct balancing.

Understanding disproportionation reactions is useful in electrochemistry, inorganic chemistry and environmental chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Carbon has oxidation number +3 in cyanogen.

Carbon is reduced to +2 in cyanide ion.

Carbon is oxidized to +4 in cyanate ion.

The same compound undergoes oxidation and reduction simultaneously.

Cyanogen acts as both the oxidising agent and the reducing agent.

The reaction is an example of disproportionation.

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Q21 →

Q21

NUMERIC3 marks

The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ions. Write a balanced ionic equation for the reaction.

📘 Concept & Theory Theory / Concept ›

A disproportionation reaction is a special type of redox reaction in which the same element present in one oxidation state undergoes both oxidation and reduction simultaneously.

Here, manganese exists initially in the +3 oxidation state.

One Mn³⁺ ion is reduced to Mn²⁺.

Another Mn³⁺ ion is oxidized to MnO₂, where manganese has oxidation state +4.

Thus, Mn³⁺ acts as both the oxidising agent and the reducing agent.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the oxidation half-reaction.

Write the reduction half-reaction.

Balance oxygen using H₂O.

Balance hydrogen using H⁺.

Balance charge using electrons.

Add the two half-reactions.

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps
Reduction Half-Reaction

Mn³⁺ is reduced to Mn²⁺ \[\mathrm{Mn^{3+}+e^-\rightarrow Mn^{2+}}\]

Oxidation Half-Reaction

Mn³⁺ is oxidized to MnO₂.

Balance oxygen by adding water. \[\mathrm{Mn^{3+}+2H_2O\rightarrow MnO_2}\]

Balance hydrogen \[\mathrm{Mn^{3+}+2H_2O\rightarrow MnO_2+4H^+}\]

Balance charge using electrons. \[\mathrm{Mn^{3+}+2H_2O\rightarrow MnO_2+4H^++e^-}\]

Add the Half-Reactions

The electrons cancel directly \[\mathrm{Mn^{3+}+Mn^{3+}+2H_2O\rightarrow Mn^{2+}+MnO_2+4H^+}\]

Balanced Ionic Equation \[\boxed{\mathrm{2Mn^{3+}+2H_2O\rightarrow Mn^{2+}+MnO_2+4H^+}}\]

🎯 Exam Significance Exam Significance ›

This is a classic NCERT example of a disproportionation reaction.

CBSE Board examinations frequently ask students to balance such ionic equations using the ion-electron method.

JEE Main and NEET often test the instability of Mn³⁺ ions in aqueous solution.

This reaction demonstrates how an intermediate oxidation state tends to undergo simultaneous oxidation and reduction.

Disproportionation reactions are important in inorganic chemistry, electrochemistry and coordination chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 4 points

Mn³⁺ is unstable in aqueous solution.

It undergoes disproportionation to form Mn²⁺ and MnO₂.

The same species acts as both oxidising and reducing agent.

The balanced ionic equation is

← Q20

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Q22 →

Q22

NUMERIC3 marks

Consider the elements: Cs, Ne, I and F
Answer the following:

Identify the element that exhibits only negative oxidation state.

Identify the element that exhibits only positive oxidation state.

Identify the element that exhibits both positive and negative oxidation states.

Identify the element which exhibits neither the negative nor the positive oxidation state.

📘 Concept & Theory Theory / Concept ›

The oxidation state of an element depends on its electronic configuration, electronegativity and ability to gain or lose electrons.

Highly electropositive metals generally exhibit only positive oxidation states.

Highly electronegative non-metals generally exhibit negative oxidation states.

Some elements can exhibit both positive and negative oxidation states depending upon the compound in which they occur.

Noble gases possess complete valence shells and generally do not exhibit oxidation states because they are chemically inert.

🗺️ Solution Roadmap Step-by-step Plan ›

Recall the periodic position of each element.

Identify its common oxidation states.

Select the element satisfying each condition.

✏️ Solution Complete Solution ›

Step-by-step Solution · 17 steps
Cesium (Cs)
Cesium belongs to Group 1 (alkali metals).

Electronic configuration: \[[Xe]\,6s^1\]

It readily loses one electron to form \[\mathrm{Cs^+}\]

Therefore, cesium exhibits only \[\boxed{+1}\] oxidation state

Fluorine (F)
Fluorine is the most electronegative element.

It always gains one electron to complete its octet.

Hence, fluorine exhibits only \[\boxed{-1}\] oxidation state in all its compounds.

It never exhibits positive oxidation states because no element is more electronegative than fluorine.

Iodine (I)
Iodine is less electronegative than fluorine, oxygen and chlorine.

Therefore, it can exhibit, negative oxidation state \[-1\] as in HI, KI

and positive oxidation states \[+1,\,+3,\,+5,\,+7\]

as in HOI, HIO₂, HIO₃ and HIO₄.

Thus iodine exhibits both positive and negative oxidation states.

Neon (Ne)
Neon is a noble gas with a completely filled outermost shell.

Electronic configuration: \[1s^22s^22p^6\]

It neither gains nor loses electrons under ordinary conditions.

Therefore, neon generally exhibits \[\boxed{0}\] oxidation state only.

🎯 Exam Significance Exam Significance ›

This question tests the periodic trends in oxidation states of elements.

CBSE Board examinations frequently ask students to identify oxidation states of representative elements.

JEE Main and NEET commonly include conceptual questions involving halogens, alkali metals and noble gases.

Fluorine is unique because it never exhibits a positive oxidation state.

Noble gases generally have oxidation number zero because of their complete valence-shell configuration.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Fluorine exhibits only the oxidation state −1.

Cesium exhibits only the oxidation state +1.

Iodine exhibits both positive and negative oxidation states.

Neon generally exhibits oxidation state 0 only.

Oxidation states are closely related to electronic configuration and electronegativity.

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Q23 →

Q23

NUMERIC3 marks

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

📘 Concept & Theory Theory / Concept ›

Chlorine is widely used as a disinfectant in water treatment plants because it destroys harmful microorganisms. However, excess dissolved chlorine imparts an unpleasant taste and odor and may be harmful to human health. Therefore, the residual chlorine is removed before the water is supplied for drinking.

Sulphur dioxide acts as a reducing agent. It reduces chlorine to harmless chloride ions, while itself being oxidized to sulphate.

Thus, this purification step is based on a redox reaction.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify oxidation and reduction.

Write the oxidation half-reaction.

Write the reduction half-reaction.

Combine the half-reactions to obtain the balanced redox equation.

Explain why this reaction is useful in water purification.

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps
Reduction of Chlorine
Chlorine gains electrons and is reduced to chloride ions. \[\mathrm{Cl_2+2e^-\rightarrow 2Cl^-}\]

Oxidation number of chlorine changes as \[0\rightarrow -1\]

Oxidation of Sulphur Dioxide
In sulphur dioxide, sulphur has oxidation number \[+4\]

In sulphate ion, sulphur has oxidation number \[+6\]

Therefore, sulphur undergoes oxidation.\[+4\rightarrow+6\]

Overall Balanced Reaction
Combining the oxidation and reduction half-reactions gives the balanced ionic equation: \[\boxed{\mathrm{Cl_2+SO_2+2H_2O\rightarrow 2Cl^-+SO_4^{2-}+4H^+}}\]

This equation is balanced with respect to atoms as well as charge.

Molecular Equation
The reaction is often represented in molecular form as \[\boxed{\mathrm{Cl_2+SO_2+2H_2O\rightarrow H_2SO_4+2HCl}}\]

Both equations describe the same redox process occurring in water.

Identification of Oxidising and Reducing Agents

Species Role Reason

Cl₂ Oxidising Agent Gets reduced from oxidation state 0 to −1.

SO₂ Reducing Agent Gets oxidized from oxidation state +4 to +6.

🎯 Exam Significance Exam Significance ›

This is an important application of redox reactions in environmental chemistry and water treatment.

CBSE Board examinations frequently ask the balanced equation and the role of sulphur dioxide in dechlorination.

JEE Main and NEET often test oxidation-number changes and the identification of oxidising and reducing agents.

The reaction demonstrates the practical use of sulphur dioxide as a reducing agent.

Questions based on chlorination and dechlorination are commonly asked in environmental and inorganic chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Chlorine is an effective disinfectant but excess chlorine is undesirable.

Sulphur dioxide removes excess chlorine through a redox reaction.

Chlorine is reduced from oxidation state 0 to −1.

Sulphur is oxidized from +4 in SO₂ to +6 in sulphate.

SO₂ acts as the reducing agent, while Cl₂ acts as the oxidising agent.

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Q24 →

Q24

NUMERIC3 marks

Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.

📘 Concept & Theory Theory / Concept ›

A disproportionation reaction is a redox reaction in which the same element in a particular oxidation state undergoes both oxidation and reduction simultaneously, producing two different oxidation states.

For disproportionation to occur, the element should possess an intermediate oxidation state. Such an oxidation state should be capable of being oxidized to a higher oxidation state as well as reduced to a lower oxidation state.

Elements that exhibit only one oxidation state, such as Na (+1), Mg (+2) or F (−1), generally do not undergo disproportionation.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify non-metals having multiple oxidation states.

Identify metals exhibiting more than one stable oxidation state.

Select elements capable of existing in intermediate oxidation states.

✏️ Solution Complete Solution ›

Step-by-step Solution · 15 steps
(a) Non-metals Showing Disproportionation
Many non-metals exhibit several oxidation states and therefore can undergo disproportionation reactions.

Important examples include:

Chlorine (Cl)

Bromine (Br)

Iodine (I)

Nitrogen (N)

Phosphorus (P)

Sulphur (S)

Typical examples are:

Chlorine \[\mathrm{Cl_2+2OH^-\rightarrow Cl^-+ClO^-+H_2O}\]

Here chlorine is simultaneously reduced from 0 to −1 and oxidized from 0 to +1.

Phosphorus

Here phosphorus undergoes simultaneous oxidation and reduction.

Sulphur

Sulphur in intermediate oxidation states such as sulphite can also undergo disproportionation under suitable conditions.

(b) Metals Showing Disproportionation
Metals capable of existing in intermediate oxidation states may undergo disproportionation.

Common examples are:

Copper (Cu)

Mercury (Hg)

Manganese (Mn)

Copper(I) \[\mathrm{2Cu^+\rightarrow Cu^{2+}+Cu}\]

Mercury(I) \[\mathrm{Hg_2^{2+}\rightarrow Hg^{2+}+Hg}\]

Manganese(III) \[\mathrm{2Mn^{3+}+2H_2O\rightarrow Mn^{2+}+MnO_2+4H^+}\]

These reactions occur because Cu(I), Hg(I) and Mn(III) are intermediate oxidation states and are thermodynamically unstable.

🎯 Exam Significance Exam Significance ›

This is an important NCERT conceptual question on disproportionation reactions.

CBSE Board examinations often ask students to identify elements capable of disproportionation.

JEE Main and NEET frequently test intermediate oxidation states of transition metals.

Students should remember that only elements having more than one accessible oxidation state can undergo disproportionation.

Representative examples such as Cl₂, Cu⁺, Hg₂²⁺ and Mn³⁺ are frequently used in competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Disproportionation occurs only when an element has an intermediate oxidation state.

Halogens except fluorine commonly undergo disproportionation.

Fluorine does not undergo disproportionation because it exhibits only the −1 oxidation state in its compounds.

Cu(I), Hg(I) and Mn(III) are well-known examples of metal ions that undergo disproportionation.

Understanding oxidation states helps predict whether disproportionation is possible.

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Q25 →

Q25

NUMERIC3 marks

In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

📘 Concept & Theory Theory / Concept ›

The first step in the Ostwald process is the catalytic oxidation of ammonia using oxygen over a platinum-rhodium catalyst.

The balanced chemical equation is
\[\boxed{\mathrm{4NH_3+5O_2\rightarrow 4NO+6H_2O}}\]
To calculate the maximum amount of product formed, we must first determine the limiting reagent.

The reactant that is completely consumed first limits the amount of product formed.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the balanced chemical equation.

Calculate the number of moles of NH₃ and O₂.

Identify the limiting reagent.

Calculate the moles of NO produced.

Convert moles of NO into mass.

✏️ Solution Complete Solution ›

Step-by-step Solution · 20 steps
Balanced Chemical Equation \[\mathrm{4NH_3+5O_2\rightarrow 4NO+6H_2O}\]

Calculate the Number of Moles

Ammonia
Molar mass of NH₃ \[14+3(1)=17\ \mathrm{g\,mol^{-1}}\]

\[\mathrm{Moles\ of\ NH_3=\frac{10.00}{17}=0.588\ mol}\]

Oxygen
Molar mass of O₂ \[32\ \mathrm{g\,mol^{-1}}\]

\[\mathrm{Moles\ of\ O_2=\frac{20.00}{32}=0.625\ mol}\]

Identify the Limiting Reagent

According to the balanced equation \[4\ mol\ NH_3\]

require \[5\ mol\ O_2\]

Therefore, oxygen required for 0.588 mol NH₃ is \[0.588\times\frac{5}{4}=0.735\ mol\]

Available oxygen \[0.625\ mol\]

Since \[0.625<0.735\]

oxygen is the limiting reagent.

Calculate the Moles of NO Produced

From the balanced equation \[5\ mol\ O_2\]

produce \[4\ mol\ NO\]

Therefore, \[\mathrm{Moles\ of\ NO=0.625\times\frac{4}{5}=0.500\ mol}\]

Calculate the Mass of NO

Molar mass of NO \[14+16=30\ \mathrm{g\,mol^{-1}}\]

\[\mathrm{Mass=0.500\times30=15.0\ g}\]

🎯 Exam Significance Exam Significance ›

This question combines the concepts of redox reactions with stoichiometry.

The Ostwald process is an important industrial process included in the NCERT syllabus.

CBSE Board examinations frequently ask questions involving limiting reagents and theoretical yield.

JEE Main and NEET regularly test mole calculations based on balanced chemical equations.

Students should always identify the limiting reagent before calculating the amount of product formed.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Number of moles of NH₃ = 0.588 mol.

Number of moles of O₂ = 0.625 mol.

Oxygen is the limiting reagent.

Maximum NO produced = 0.500 mol.

Maximum mass of NO = 15.0 g.

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Q26 →

Q26

NUMERIC3 marks

Using the standard electrode potentials given in Table 8.1, predict whether the following reactions are feasible:

Fe³⁺(aq) and I⁻(aq)

Ag⁺(aq) and Cu(s)

Fe³⁺(aq) and Cu(s)

Ag(s) and Fe³⁺(aq)

Br₂(aq) and Fe²⁺(aq)

📘 Concept & Theory Theory / Concept ›

The feasibility of a redox reaction can be predicted using the standard reduction potentials (E°).

A reaction is spontaneous when the standard cell potential is positive.
\[ \boxed{ E^\Theta_{\text{cell}} = E^\Theta_{\text{cathode}} - E^\Theta_{\text{anode}} } \]
If

\(E^\Theta_{\text{cell}}>0\), the reaction is feasible.

\(E^\Theta_{\text{cell}}<0\), the reaction is not feasible.

Standard Reduction Potentials Used

Half-cell Reaction \(E^\Theta\) (V)

\(\mathrm{Fe^{3+}+e^- \rightarrow Fe^{2+}}\) +0.77

\(\mathrm{Ag^++e^- \rightarrow Ag}\) +0.80

\(\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\) +0.34

\(\mathrm{Br_2+2e^- \rightarrow2Br^-}\) +1.07

\(\mathrm{I_2+2e^- \rightarrow2I^-}\) +0.54

✏️ Solution Complete Solution ›

Step-by-step Solution · 21 steps
(a) Fe³⁺(aq) and I⁻(aq)
Reduction: \[\mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}\qquadE^\Theta=+0.77\,V}\]

Oxidation: \[\mathrm{2I^-\rightarrow I_2+2e^-}\]

Using the reduction potential of iodine, \[E^\Theta_{\text{anode}}=+0.54\,V\]

\[E^\Theta_{\text{cell}}=0.77-0.54=+0.23\,V\]

Result:
The reaction is feasible.

(b) Ag⁺(aq) and Cu(s)
Reduction: \[E^\Theta=+0.80\,V\]

Oxidation of copper: \[E^\Theta_{\text{anode}}=+0.34\,V\]

\[E^\Theta_{\text{cell}}=0.80-0.34=+0.46\,V\]

Result:
The reaction is feasible.
Balanced equation: \[\mathrm{2Ag^++Cu\rightarrow 2Ag+Cu^{2+}}\]

(c) Fe³⁺(aq) and Cu(s)
Reduction:\[E^\Theta=+0.77\,V\]

Oxidation: \[E^\Theta=+0.34\,V\]

\[E^\Theta_{\text{cell}}=0.77-0.34=+0.43\,V\]

Result:
The reaction is feasible. \[\mathrm{2Fe^{3+}+Cu\rightarrow 2Fe^{2+}+Cu^{2+}}\]

(d) Ag(s) and Fe³⁺(aq)
Reduction:
\[\mathrm{Fe^{3+}\rightarrow Fe^{2+}}\]

Oxidation: \[\mathrm{Ag\rightarrow Ag^+}\]

\[E^\Theta_{\text{cell}}=0.77-0.80=-0.03\,V\]

Result:
The reaction is not feasible.

(e) Br₂(aq) and Fe²⁺(aq)
Reduction: \[E^\Theta=+1.07\,V\]

Oxidation: \[E^\Theta=+0.77\,V\]

\[E^\Theta_{\text{cell}}=1.07-0.77=+0.30\,V\]

Result:
The reaction is feasible. \[\mathrm{Br_2+2Fe^{2+}\rightarrow 2Br^-+2Fe^{3+}}\]

🎯 Exam Significance Exam Significance ›

This question introduces the use of standard electrode potentials to predict spontaneous redox reactions.

CBSE Board examinations frequently ask students to calculate \(E^\Theta_{\text{cell}}\) and determine reaction feasibility.

JEE Main and NEET regularly include conceptual questions based on electrochemical series.

A positive standard cell potential indicates a spontaneous redox reaction under standard conditions.

This concept forms the foundation of electrochemistry and galvanic cells.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

>Use the relation \[\boxed{E^\Theta_{\text{cell}}=E^\Theta_{\text{cathode}}-E^\Theta_{\text{anode}}}\]

If \(E^\Theta_{\text{cell}}>0\), the reaction is feasible.

If \(E^\Theta_{\text{cell}}<0\), the reaction is non-spontaneous.

Silver ion is a stronger oxidising agent than copper(II) ion.

Bromine is a stronger oxidising agent than Fe³⁺.

Electrochemical series helps predict the direction of electron transfer.

← Q25

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Q27 →

Q27

NUMERIC3 marks

Predict the products of electrolysis in each of the following:

An aqueous solution of AgNO₃ with silver electrodes.

An aqueous solution of AgNO₃ with platinum electrodes.

A dilute solution of H₂SO₄ with platinum electrodes.

An aqueous solution of CuCl₂ with platinum electrodes.

📘 Concept & Theory Theory / Concept ›

Electrolysis is the process in which electrical energy is used to bring about a non-spontaneous chemical reaction.

During electrolysis:

Cathode (−): Reduction takes place (gain of electrons).

Anode (+): Oxidation takes place (loss of electrons).

The products formed depend upon:

The nature of the electrolyte.

The discharge potentials of the ions.

Whether the electrodes are active (e.g., Ag, Cu) or inert (Pt, graphite).

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps
(i) Aqueous AgNO₃ using Silver Electrodes
At the Cathode (Reduction)
Silver ions gain electrons. \[\mathrm{Ag^++e^-\rightarrow Ag(s)}\]

Silver is deposited on the cathode.

At the Anode (Oxidation)
The silver anode dissolves. \[\mathrm{Ag(s)\rightarrowAg^++e^-}\]

Overall Result

Silver is deposited at the cathode.

An equal amount of silver dissolves from the anode.

The concentration of AgNO₃ remains practically unchanged.

(ii) Aqueous AgNO₃ using Platinum Electrodes
At the Cathode \[\mathrm{Ag^++e^-\rightarrow Ag(s)}\] Silver metal is deposited.

At the Anode
Nitrate ions are not oxidized.

Water undergoes oxidation. \[\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}\]

Overall Result

Silver is deposited at the cathode.

Oxygen gas is liberated at the anode.

The solution gradually becomes acidic because H⁺ ions are produced.

(iii) Dilute H₂SO₄ using Platinum Electrodes
At the Cathode
Hydrogen ions are reduced. \[\mathrm{2H^++2e^-\rightarrow H_2(g)}\]

At the Anode
Water is oxidized. \[\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}\]

Overall Reaction \[\mathrm{2H_2O\rightarrow 2H_2+O_2}\]

Overall Result

Hydrogen gas is evolved at the cathode.

Oxygen gas is evolved at the anode.

Sulphuric acid merely increases the conductivity and is not consumed.

(iv) Aqueous CuCl₂ using Platinum Electrodes
At the Cathode
Copper ions are reduced \[\mathrm{Cu^{2+}+2e^-\rightarrow Cu(s)}\]

At the Anode
Chloride ions are oxidized. \[\mathrm{2Cl^-\rightarrow Cl_2(g)+2e^-}\]

Overall Reaction \[\mathrm{CuCl_2\rightarrow Cu+Cl_2}\]

Overall Result

Copper metal is deposited on the cathode.

Chlorine gas is liberated at the anode.

💡 Answer Final Answer ›

Electrolyte Cathode Product Anode Product

AgNO₃, Ag electrodes Ag Ag⁺ (silver anode dissolves)

AgNO₃, Pt electrodes Ag O₂

Dilute H₂SO₄, Pt electrodes H₂ O₂

CuCl₂, Pt electrodes Cu Cl₂

🎯 Exam Significance Exam Significance ›

This question tests the principles governing electrolysis and electrode reactions.

CBSE Board examinations frequently ask students to predict products at the cathode and anode.

JEE Main and NEET often include conceptual questions involving inert and active electrodes.

Students should remember that active electrodes may participate in the reaction, whereas inert electrodes generally do not.

Electrolysis is closely related to electrochemistry, metallurgy and industrial chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Reduction always occurs at the cathode.

Oxidation always occurs at the anode.

Silver electrodes are active electrodes and dissolve during electrolysis.

Platinum is an inert electrode and does not participate in the reaction.

In aqueous CuCl₂, copper is deposited while chlorine gas is evolved.

Dilute sulphuric acid undergoes electrolysis to produce hydrogen and oxygen gases.

← Q26

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Q28 →

Q28

NUMERIC3 marks

Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.

📘 Concept & Theory Theory / Concept ›

The ability of a metal to displace another metal from its salt solution depends upon its reactivity or standard reduction potential (E°).

A metal having a more negative standard reduction potential is more reactive and can displace a metal having a higher (more positive) reduction potential from its salt solution.

Thus, metals higher in the electrochemical series are stronger reducing agents and readily displace metals lower in the series.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the standard reduction potentials of the given metals.

Arrange the metals in increasing order of reduction potential.

Determine the order of reactivity.

Predict the displacement order.

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps
Standard Reduction Potentials

Metal Half-cell Reaction \(E^\Theta\) (V)

Mg \(\mathrm{Mg^{2+}+2e^- \rightarrow Mg}\) −2.37

Al \(\mathrm{Al^{3+}+3e^- \rightarrow Al}\) −1.66

Zn \(\mathrm{Zn^{2+}+2e^- \rightarrow Zn}\) −0.76

Fe \(\mathrm{Fe^{2+}+2e^- \rightarrow Fe}\) −0.44

Cu \(\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\) +0.34

Arrange According to Reactivity

The more negative the reduction potential, the greater is the tendency of the metal to lose electrons.

Therefore, the order of decreasing reactivity is \[\boxed{\mathrm{Mg>Al>Zn>Fe>Cu}}\]

Interpretation

This means:

Magnesium can displace Al, Zn, Fe and Cu from their salt solutions.

Aluminium can displace Zn, Fe and Cu but not Mg.

Zinc can displace Fe and Cu but not Mg or Al.

Iron can displace Cu only.

Copper cannot displace any of the other four metals.

Magnesium displaces copper \[\mathrm{Mg+CuSO_4\rightarrow MgSO_4+Cu}\]

Zinc displaces iron \[\mathrm{Zn+FeSO_4\rightarrow ZnSO_4+Fe}\]

Iron displaces copper \[\mathrm{Fe+CuSO_4\rightarrow FeSO_4+Cu}\]

Copper does not displace zinc \[\mathrm{Cu+ZnSO_4\rightarrow \text{No reaction}}\]

🎯 Exam Significance Exam Significance ›

This question is based on the electrochemical series and standard electrode potentials.

CBSE Board examinations frequently ask students to predict displacement reactions.

JEE Main and NEET often test the relationship between reduction potential and metal reactivity.

A metal with a more negative reduction potential acts as a stronger reducing agent.

Knowledge of the electrochemical series is essential for understanding corrosion, metallurgy and electrochemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

More negative standard reduction potential means greater reactivity.

More reactive metals displace less reactive metals from their salt solutions.

The reactivity order is \[\boxed{\mathrm{Mg>Al>Zn>Fe>Cu}}\]

Magnesium is the strongest reducing agent among the given metals.

Copper is the least reactive metal in the given list.

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Q29 →

Q29

NUMERIC3 marks

Given the standard electrode potentials,
\(\mathrm{K^+/K\ =\ –2.93V,\ Ag^+/Ag\ =\ 0.80V}\), \(\mathrm{Hg^{2+}/Hg = 0.79V}\) \(\mathrm{Mg^{2+}/Mg = –2.37V. Cr^{3+}/Cr = –0.74V}\) arrange these metals in their increasing order of reducing power.

📘 Concept & Theory Theory / Concept ›

The reducing power of a metal is its tendency to lose electrons and undergo oxidation.

A metal having a more negative standard reduction potential loses electrons more readily and is therefore a stronger reducing agent.

Hence:

More positive \(E^\Theta\) → weaker reducing agent.

More negative \(E^\Theta\) → stronger reducing agent.

🗺️ Solution Roadmap Step-by-step Plan ›

List the given standard reduction potentials.

Compare their numerical values.

Arrange the metals from the highest \(E^\Theta\) to the lowest \(E^\Theta\).

This gives the increasing order of reducing power.

✏️ Solution Complete Solution ›

Step-by-step Solution · 4 steps
Standard Reduction Potentials

Metal \(E^\Theta\) (V)

Ag +0.80

Hg +0.79

Cr −0.74

Mg −2.37

K −2.93

Relate Reduction Potential to Reducing Power

Reducing power increases as the value of the standard reduction potential becomes more negative.

Therefore, the metals are arranged from the highest reduction potential to the lowest reduction potential.

💡 Answer Final Answer ›

The increasing order of reducing power is
\[\boxed{\mathrm{Ag < Hg < Cr < Mg < K}}\]
Thus,

Silver (Ag) is the weakest reducing agent.

Potassium (K) is the strongest reducing agent.

Explanation

Metal \(E^\circ\) (V) Reducing Power

Ag +0.80 Weakest

Hg +0.79 Very Weak

Cr −0.74 Moderate

Mg −2.37 Strong

K −2.93 Strongest

🎯 Exam Significance Exam Significance ›

This question tests the relationship between standard reduction potential and reducing power.

CBSE Board examinations frequently ask students to arrange metals according to reducing or oxidising strength.

JEE Main and NEET commonly include conceptual questions based on the electrochemical series.

Metals with more negative standard reduction potentials are stronger reducing agents because they lose electrons more readily.

This concept is fundamental to electrochemistry, metallurgy and displacement reactions.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 4 points

Reducing power increases as the standard reduction potential becomes more negative.

Potassium is the strongest reducing agent among the given metals.

Silver is the weakest reducing agent among the given metals.

The increasing order of reducing power is \[\boxed{\mathrm{Ag < Hg < Cr < Mg < K}}\]

← Q28

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Q30 →

Q30

NUMERIC2 marks

Depict the galvanic cell in which the reaction \(\mathrm{Zn(s)+2Ag^+(aq)\rightarrow Zn^{2+}(aq)+2Ag(s)}\) takes place, Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.

📘 Concept & Theory Theory / Concept ›

A galvanic (voltaic) cell converts chemical energy into electrical energy through a spontaneous redox reaction.

In every galvanic cell:

Oxidation occurs at the anode.

Reduction occurs at the cathode.

Electrons always flow through the external circuit from the anode to the cathode.

Ions move through the salt bridge to maintain electrical neutrality.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the species undergoing oxidation and reduction.

Determine the anode and cathode.

Write the cell notation.

Show the direction of electron flow and ion movement.

Write the half-cell reactions.

✏️ Solution Complete Solution ›

Step-by-step Solution · 18 steps
Identify Oxidation and Reduction

The overall reaction is \[\mathrm{Zn+2Ag^+\rightarrow Zn^{2+}+2Ag}\]

Zinc loses electrons. \[\mathrm{Zn\rightarrow Zn^{2+}+2e^-}\]

Therefore, zinc is oxidized.

Silver ions gain electrons. \[\mathrm{2Ag^++2e^-\rightarrow 2Ag}\] Therefore, silver ions are reduced.

Identify the Electrodes

Electrode Process

Zinc Electrode Oxidation (Anode)

Silver Electrode Reduction (Cathode)

Cell Representation \[\boxed{\mathrm{Zn(s)\;|\;Zn^{2+}(aq)\;||\;Ag^+(aq)\;|\;Ag(s)}}\]

The single vertical line represents the boundary between an electrode and its electrolyte.

The double vertical line represents the salt bridge.

Electrode Charges
At the zinc electrode, electrons are produced.

Therefore, the zinc electrode becomes
\[\boxed{\text{Negative}}\]

At the silver electrode, electrons are consumed.

Therefore, the silver electrode becomes
\[\boxed{\text{Positive}}\]

Current Carriers

In the external circuit

Electrons flow from the zinc electrode to the silver electrode.

In the salt bridge

Anions move towards the anode (zinc compartment).

Cations move towards the cathode (silver compartment).

Thus, electrical current inside the cell is carried by ions, whereas outside the cell it is carried by electrons.

Half-Cell Reactions

Anode (Oxidation) \[\boxed{\mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^-}}\]

Cathode (Reduction) \[\boxed{\mathrm{2Ag^+(aq)+2e^-\rightarrow 2Ag(s)}}\]

📊 Graph / Figure Graph / Figure ›

🎯 Exam Significance Exam Significance ›

This is one of the most important NCERT questions introducing galvanic cells.

CBSE Board examinations frequently ask students to identify the anode, cathode and direction of electron flow.

JEE Main and NEET commonly test cell notation, half-cell reactions and current carriers.

Students should remember that oxidation always occurs at the anode and reduction always occurs at the cathode.

The salt bridge completes the electrical circuit and maintains charge neutrality.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

The zinc electrode acts as the anode and is negatively charged.

The silver electrode acts as the cathode and is positively charged.

Electrons flow from Zn to Ag through the external wire.

Anions migrate towards the anode, while cations migrate towards the cathode through the salt bridge.

The cell notation is \[\boxed{\mathrm{Zn|Zn^{2+}||Ag^+|Ag}}\]

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Chapter Complete!

All 30 solutions for Redox Reactions covered.
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Quantity	Substance
Oxidised	C₆H₆O₂ (Hydroquinone)
Reduced	AgBr (Ag⁺)
Oxidising Agent	AgBr
Reducing Agent	C₆H₆O₂

Quantity	Substance
Oxidised	HCHO
Reduced	[Ag(NH₃)₂]⁺
Oxidising Agent	Tollens' reagent
Reducing Agent	HCHO

Quantity	Substance
Oxidised	HCHO
Reduced	Cu²⁺
Oxidising Agent	Cu²⁺ (Fehling's/Benedict's reagent)
Reducing Agent	HCHO

Quantity	Substance
Oxidised	N₂H₄
Reduced	H₂O₂
Oxidising Agent	H₂O₂
Reducing Agent	N₂H₄

Quantity	Substance
Oxidised	Pb(s)
Reduced	PbO₂
Oxidising Agent	PbO₂
Reducing Agent	Pb(s)

Reaction	Oxidised Substance	Reduced Substance	Oxidising Agent	Reducing Agent
(a)	Hydroquinone (C₆H₆O₂)	AgBr	AgBr	Hydroquinone
(b)	HCHO	[Ag(NH₃)₂]⁺	Tollens' reagent	HCHO
(c)	HCHO	Cu²⁺	Cu²⁺	HCHO
(d)	N₂H₄	H₂O₂	H₂O₂	N₂H₄
(e)	Pb	PbO₂	PbO₂	Pb

Property	Iodine	Bromine
Oxidising strength	Weaker	Stronger
Extent of oxidation of S₂O₃²⁻	Partial	Complete
Main sulphur product	S₄O₆²⁻	SO₄²⁻

Species	Oxidation Number Change	Process
Xe	+8 → +6	Reduction
F	−1 → 0	Oxidation

Reaction	Reason
(a)	H₃PO₂ is oxidized from P(+1) to P(+5) and reduces Ag⁺ to Ag.
(b)	H₃PO₂ also reduces Cu²⁺ to Cu because it is a strong reducing agent.
(c)	Benzaldehyde reduces Tollens' reagent and gives a silver mirror.
(d)	Benzaldehyde does not reduce Fehling's solution because aromatic aldehydes are resistant to oxidation under alkaline Cu²⁺ conditions.

Species	Oxidation Number of Carbon	Process
(CN)₂	+3	Initial state
CN⁻	+2	Reduction
CNO⁻	+4	Oxidation

Species	Role	Reason
Cl₂	Oxidising Agent	Gets reduced from oxidation state 0 to −1.
SO₂	Reducing Agent	Gets oxidized from oxidation state +4 to +6.

Half-cell Reaction	\(E^\Theta\) (V)
\(\mathrm{Fe^{3+}+e^- \rightarrow Fe^{2+}}\)	+0.77
\(\mathrm{Ag^++e^- \rightarrow Ag}\)	+0.80
\(\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\)	+0.34
\(\mathrm{Br_2+2e^- \rightarrow2Br^-}\)	+1.07
\(\mathrm{I_2+2e^- \rightarrow2I^-}\)	+0.54

Electrolyte	Cathode Product	Anode Product
AgNO₃, Ag electrodes	Ag	Ag⁺ (silver anode dissolves)
AgNO₃, Pt electrodes	Ag	O₂
Dilute H₂SO₄, Pt electrodes	H₂	O₂
CuCl₂, Pt electrodes	Cu	Cl₂

Metal	Half-cell Reaction	\(E^\Theta\) (V)
Mg	\(\mathrm{Mg^{2+}+2e^- \rightarrow Mg}\)	−2.37
Al	\(\mathrm{Al^{3+}+3e^- \rightarrow Al}\)	−1.66
Zn	\(\mathrm{Zn^{2+}+2e^- \rightarrow Zn}\)	−0.76
Fe	\(\mathrm{Fe^{2+}+2e^- \rightarrow Fe}\)	−0.44
Cu	\(\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\)	+0.34

Metal	\(E^\Theta\) (V)
Ag	+0.80
Hg	+0.79
Cr	−0.74
Mg	−2.37
K	−2.93

Metal	\(E^\circ\) (V)	Reducing Power
Ag	+0.80	Weakest
Hg	+0.79	Very Weak
Cr	−0.74	Moderate
Mg	−2.37	Strong
K	−2.93	Strongest

Electrode	Process
Zinc Electrode	Oxidation (Anode)
Silver Electrode	Reduction (Cathode)

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Redox Reactions

Explanation

Result

Why Does Thiosulphate Behave Differently?

"Conclusion regarding Na4XeO6

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Standard Reduction Potentials Used

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