What is the difference between oxidation and reduction in terms of oxidation state?

Oxidation is an increase in the oxidation state of an element caused by the loss of electrons. Reduction is a decrease in oxidation state caused by the gain of electrons.

How do you balance redox reactions using the ion-electron method?

Split reaction into oxidation and reduction half-reactions, balance atoms other than O and H, balance O using H₂O, balance H using H⁺ (or OH⁻ in basic medium), balance charges with electrons (e⁻), equalize electrons by multiplication, and add the half-reactions.

What is a disproportionation reaction with an example?

A disproportionation reaction is a redox reaction where the same element in a single oxidation state is simultaneously oxidized and reduced. Example: 2H₂O₂ (O at -1) → 2H₂O (O at -2) + O₂ (O at 0).

What is standard electrode potential (E°) and how does it predict feasibility?

Standard electrode potential measures the tendency of a chemical species to be reduced at 298 K, 1 atm, and 1 M concentration. For a cell reaction, E°cell = E°cathode - E°anode; a positive E°cell indicates thermodynamic feasibility (ΔG° < 0).

ALL UNITS

Unit 07 • Physical & Inorganic Chemistry

Redox Reactions

Where there is oxidation, there is always reduction. Classical and electronic models of electron transfer, comprehensive rules for assigning oxidation numbers, Stock notation, fractional oxidation state paradox ($C_3O_2, Br_3O_8, S_4O_6^{2-}, Fe_3O_4$), four types of redox reactions, balancing chemical equations via Oxidation Number and Half-Reaction (Ion-Electron) methods in acidic and basic media, redox titrations with self and external indicators, and electrochemical cells (Daniell Cell, standard electrode potentials $E^\circ$, activity series). Includes in-text solved problems 7.17.10, full NCERT exercises 7.17.30, quick revision matrix, and 3-level chapter tests.

Introduction: The Dual Nature of Redox Systems

Chemistry deals with varieties of matter and transformation of one kind into another. An overarching class of chemical changes is Redox Reactions (Reduction-Oxidation). A vast spectrum of biological and physical phenomena—respiration, photosynthesis, energy generation via fuel combustion, industrial extraction of metals and halogens, commercial dry and wet batteries, and corrosion—rest fundamentally on redox mechanisms.

$$\mathbf{\text{“Where there is oxidation, there is always reduction.”}}$$ $$\text{Oxidation and reduction are complementary halves of a unified chemical event.}$$

7.1 Classical Idea of Redox Reactions

Process	Classical Definition	Representative Chemical Equations
Oxidation	• Addition of oxygen • Addition of electronegative element • Removal of hydrogen • Removal of electropositive element	$2Mg + O_2 \rightarrow 2MgO$ (Addition of $O$) $Mg + Cl_2 \rightarrow MgCl_2$ (Addition of electronegative $Cl$) $2H_2S + O_2 \rightarrow 2S + 2H_2O$ (Removal of $H$ from $S$) $2K_4[Fe(CN)_6] + H_2O_2 \rightarrow 2K_3[Fe(CN)_6] + 2KOH$ (Removal of electropositive $K$)
Reduction	• Removal of oxygen • Removal of electronegative element • Addition of hydrogen • Addition of electropositive element	$2HgO \xrightarrow{\Delta} 2Hg + O_2$ (Removal of $O$) $2FeCl_3 + H_2 \rightarrow 2FeCl_2 + 2HCl$ (Removal of electronegative $Cl$) $CH_2=CH_2 + H_2 \rightarrow CH_3-CH_3$ (Addition of $H$) $2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$ (Addition of $Hg$)

7.1

In the reactions given below, identify the species undergoing oxidation and reduction: (i) H₂S(g) + Cl₂(g) → 2HCl(g) + S(s); (ii) 3Fe₃O₄(s) + 8Al(s) → 9Fe(s) + 4Al₂O₃(s); (iii) 2Na(s) + H₂(g) → 2NaH(s).

(i) $H_2S$ is oxidised because electropositive hydrogen is removed from sulphur (forming elemental $S$). $Cl_2$ is reduced because electropositive hydrogen is added to chlorine.
(ii) $Al$ is oxidised because oxygen is added to it (forming $Al_2O_3$). $Fe_3O_4$ is reduced because oxygen is removed from it (yielding elemental $Fe$).
(iii) $Na$ is oxidised because it forms $Na^+$ by losing electron density to more electronegative hydrogen; $H_2$ is reduced to hydride ($H^-$).

(i) H₂S oxidised, Cl₂ reduced; (ii) Al oxidised, Fe₃O₄ reduced; (iii) Na oxidised, H₂ reduced.

7.2 Redox in Terms of Electron Transfer Reactions

Reactions between elements to form ionic lattices explicitly involve electron exchanges: $$2Na_{(s)} + Cl_{2(g)} \rightarrow 2Na^+Cl^-_{(s)}$$

Oxidation (OIL): Loss of electron(s) $\implies \mathbf{Na \rightarrow Na^+ + e^-}$

Reduction (RIG): Gain of electron(s) $\implies \mathbf{Cl_2 + 2e^- \rightarrow 2Cl^-}$

Reducing Agent (Reductant): Electron donor (gets oxidised).
Oxidising Agent (Oxidant): Electron acceptor (gets reduced).

Electron Transfer Mechanism: Zinc Strip in Copper Nitrate Solution

7.2

Justify that the reaction 2Na(s) + H₂(g) → 2NaH(s) is a redox change.

Sodium hydride is an ionic compound composed of $Na^+$ and $H^-$ ions ($Na^+H^-$).
Splitting into two half-reactions: $$2Na_{(s)} \rightarrow 2Na^+ + 2e^- \quad (\text{Loss of electrons} \implies \text{Oxidation})$$ $$H_{2(g)} + 2e^- \rightarrow 2H^- \quad (\text{Gain of electrons} \implies \text{Reduction})$$ Since electron loss and electron gain occur simultaneously, the formation of $NaH$ is a redox change.

Sodium is oxidised to Na⁺; hydrogen is reduced to hydride H⁻. Overall change is redox.

7.3 Oxidation Number Concept and Operational Rules

Definition: Oxidation number denotes the electrical charge an atom appears to possess in a compound when all bonding electron pairs are allocated entirely to the more electronegative partner.

Rule No.	Chemical Environment	Assigned Oxidation State	Examples
Rule 1	Free or uncombined elemental state	Zero ($0$)	$H_2, O_2, Cl_2, P_4, S_8, Na, Mg, Al$ (all have $0$).
Rule 2	Monoatomic ions & fixed group metals	Charge on ion; Alkali metals $= \mathbf{+1}$; Alkaline earth $= \mathbf{+2}$; $Al = \mathbf{+3}$	$Na^+ (+1), Mg^{2+} (+2), Fe^{3+} (+3), Cl^- (-1)$.
Rule 3	Oxygen in compounds	• Normal oxides: $\mathbf{-2}$ • Peroxides ($-O-O-$): $\mathbf{-1}$ • Superoxides ($O_2^-$): $\mathbf{-1/2}$ • In $OF_2$: $\mathbf{+2}$; In $O_2F_2$: $\mathbf{+1}$	$H_2O, CO_2 (-2)$ $H_2O_2, Na_2O_2, BaO_2 (-1)$ $KO_2, RbO_2, CsO_2 (-1/2)$ $OF_2 (+2), O_2F_2 (+1)$
Rule 4	Hydrogen in compounds	• Non-metal compounds: $\mathbf{+1}$ • Metallic hydrides (binary): $\mathbf{-1}$	$HCl, H_2O, NH_3, CH_4 (+1)$ $LiH, NaH, CaH_2 (-1)$
Rule 5	Halogens	• Fluorine in ALL compounds: $\mathbf{-1}$ • $Cl, Br, I$ as halides: $\mathbf{-1}$ • $Cl, Br, I$ in oxoacids/oxoanions: $\mathbf{+1 \text{ to } +7}$	$HF, SF_6 (-1)$ $NaCl, KBr (-1)$ $KClO_3 (+5), HClO_4 (+7)$
Rule 6	Algebraic sum of oxidation numbers	• Neutral molecule $= \mathbf{0}$ • Polyatomic ion $= \mathbf{\text{Net charge on ion}}$	$CO_2: x + 2(-2) = 0 \implies x = +4$ $CO_3^{2-}: x + 3(-2) = -2 \implies x = +4$ $Cr_2O_7^{2-}: 2x + 7(-2) = -2 \implies x = +6$

Periodic Trends in Maximum Oxidation States

The highest oxidation state of a representative element equals its Group Number (Groups 1 & 2) or Group Number − 10 (Groups 13 to 17):

$$Na (+1) \quad Mg (+2) \quad Al (+3) \quad Si (+4) \quad P (+5) \quad S (+6) \quad Cl (+7)$$

Stock Notation

Proposed by German chemist Alfred Stock: The oxidation state of a metal in a compound is indicated by a Roman numeral enclosed in parentheses following the metal symbol.

7.3

Using Stock notation, represent the following compounds: HAuCl₄, Tl₂O, FeO, Fe₂O₃, CuI, CuO, MnO and MnO₂.

Calculate oxidation number of the metal:
• $HAuCl_4: +1 + x + 4(-1) = 0 \implies x = +3 \implies \mathbf{HAu(III)Cl_4}$
• $Tl_2O: 2x - 2 = 0 \implies x = +1 \implies \mathbf{Tl_2(I)O}$
• $FeO: x - 2 = 0 \implies x = +2 \implies \mathbf{Fe(II)O}$
• $Fe_2O_3: 2x - 6 = 0 \implies x = +3 \implies \mathbf{Fe_2(III)O_3}$
• $CuI: x - 1 = 0 \implies x = +1 \implies \mathbf{Cu(I)I}$
• $CuO: x - 2 = 0 \implies x = +2 \implies \mathbf{Cu(II)O}$
• $MnO: x - 2 = 0 \implies x = +2 \implies \mathbf{Mn(II)O}$
• $MnO_2: x + 2(-2) = 0 \implies x = +4 \implies \mathbf{Mn(IV)O_2}$

Stock formulas: HAu(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O₂.

The Paradox of Fractional Oxidation Numbers

Electrons are indivisible; they can never be shared or transferred in fractions! When an oxidation state appears as a fraction (e.g., $4/3, 16/3, 2.5$), it represents an average oxidation state over atoms residing in structurally distinct chemical environments.

Structural Resolution of Fractional Oxidation States: C₃O₂, Br₃O₈, and S₄O₆²⁻

+2 Central C* = 0 Average = (2+0+2)/3 = 4/3 Tribromooctaoxide (Br₃O₈) O₃Br — Br*O₂ — BrO₃ +6 +4 +6 Two terminal Br = +6 Central Br* = +4 Average = (6+4+6)/3 = 16/3 Tetrathionate (S₄O₆²⁻) [O₃S — S — S — SO₃]²⁻ +5 0 0 +5 Two terminal S = +5 Two central S = 0 Average = (5+0+0+5)/4 = 2.5

7.3.1 Types of Redox Reactions

Type	Characteristics & Governing Rule	Chemical Reactions with Oxidation Numbers
1. Combination	$A + B \rightarrow C$ At least one reactant must be in elemental form ($0$).	$\overset{0}{C}_{(s)} + \overset{0}{O}_{2(g)} \rightarrow \overset{+4}{C}\overset{-2}{O}_{2(g)}$ $3\overset{0}{Mg}_{(s)} + \overset{0}{N}_{2(g)} \rightarrow \overset{+2}{Mg}_3\overset{-3}{N}_{2(s)}$
2. Decomposition	Breakdown of a compound into $\ge 2$ substances, at least one in elemental state. (Not all decompositions are redox! e.g., $CaCO_3 \rightarrow CaO + CO_2$).	$2\overset{+1}{H}_2\overset{-2}{O}_{(l)} \xrightarrow{\text{electrolysis}} 2\overset{0}{H}_{2(g)} + \overset{0}{O}_{2(g)}$ $2\overset{+1}{K}\overset{+5}{Cl}\overset{-2}{O}_{3(s)} \xrightarrow{\Delta} 2\overset{+1}{K}\overset{-1}{Cl}_{(s)} + 3\overset{0}{O}_{2(g)}$
3. Displacement	$X + YZ \rightarrow XZ + Y$ (a) Metal displacement: More electropositive metal displaces a less active metal. (b) Non-metal displacement: Displacing $H_2$ from water or acids; halogen displacement ($F_2 > Cl_2 > Br_2 > I_2$).	$\overset{+2}{Cu}SO_4 + \overset{0}{Zn} \rightarrow \overset{+2}{Zn}SO_4 + \overset{0}{Cu}$ $2\overset{0}{Na} + 2\overset{+1}{H}_2O \rightarrow 2\overset{+1}{Na}OH + \overset{0}{H}_2$ $\overset{0}{Cl}_2 + 2K\overset{-1}{Br} \rightarrow 2K\overset{-1}{Cl} + \overset{0}{Br}_2$
4. Disproportionation	An element in an intermediate oxidation state is simultaneously oxidised and reduced into higher and lower oxidation states. Requires an element exhibiting $\ge 3$ oxidation states!	$2\overset{+1}{H}_2\overset{-1}{O}_2 \rightarrow 2\overset{+1}{H}_2\overset{-2}{O} + \overset{0}{O}_2$ $\overset{0}{P}_4 + 3OH^- + 3H_2O \rightarrow \overset{-3}{P}H_3 + 3H_2\overset{+1}{P}O_2^-$ $\overset{0}{Cl}_2 + 2OH^- \rightarrow \overset{+1}{Cl}O^- + \overset{-1}{Cl}^- + H_2O$

Halogen Displacement & Organic Layer Test (Oxidising Strength: F₂ > Cl₂ > Br₂ > I₂)

Crucial Disproportionation Exceptions

Fluorine never disproportionates: Being the most electronegative element, fluorine can only exhibit $0$ and $-1$ oxidation states. In alkaline medium: $$2F_2 + 2OH^- \rightarrow 2F^- + OF_2 + H_2O$$ Here, oxygen is oxidised ($-2 \rightarrow +2$) and fluorine is reduced ($0 \rightarrow -1$).

Perchlorate ion ($ClO_4^-$) cannot disproportionate: Chlorine is already in its maximum possible oxidation state ($+7$). It can only undergo reduction!

7.5

Which of the following species do not show disproportionation reaction and why? ClO⁻, ClO₂⁻, ClO₃⁻ and ClO₄⁻. Write reactions for those that disproportionate.

• In $ClO_4^-$, chlorine is in its highest oxidation state ($+7$) and cannot be oxidised further. Hence, $ClO_4^-$ cannot disproportionate.
Disproportionation reactions for the others: $$3\overset{+1}{Cl}O^- \rightarrow 2\overset{-1}{Cl}^- + \overset{+5}{Cl}O_3^-$$ $$6\overset{+3}{Cl}O_2^- \rightarrow 4\overset{+5}{Cl}O_3^- + 2\overset{-1}{Cl}^-$$ $$4\overset{+5}{Cl}O_3^- \rightarrow \overset{-1}{Cl}^- + 3\overset{+7}{Cl}O_4^-$$

ClO₄⁻ does not disproportionate (highest oxidation state +7).

7.6

Suggest a scheme of classification for the following redox reactions: (a) N₂(g) + O₂(g) → 2NO(g); (b) 2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g); (c) NaH(s) + H₂O(l) → NaOH(aq) + H₂(g); (d) 2NO₂(g) + 2OH⁻(aq) → NO₂⁻(aq) + NO₃⁻(aq) + H₂O(l).

(a) Combination redox: Two elements in zero state combine to form $NO$.
(b) Decomposition redox: Lead nitrate breaks down into multiple products including elemental $O_2$.
(c) Displacement redox: Hydride ion ($H^-$) reduces water $H^+$ to produce $H_2$ gas.
(d) Disproportionation redox: $NO_2$ ($+4$) simultaneously forms $NO_2^-$ ($+3$) and $NO_3^-$ ($+5$).

(a) Combination; (b) Decomposition; (c) Displacement; (d) Disproportionation.

7.3.2 Systematic Balancing of Redox Reactions

Two universal methods are recognized: Oxidation Number Method and Half-Reaction (Ion-Electron) Method.

Systematic Half-Reaction (Ion-Electron) Balancing Algorithm

Basic Medium Golden Rule: Balance initially using H⁺ and H₂O as in acidic solution. Then add an equal number of OH⁻ ions to BOTH sides for every H⁺ present. Combine (H⁺ + OH⁻ → H₂O) and simplify water molecules!

7.8

Write the net ionic equation for the reaction of potassium dichromate(VI), K₂Cr₂O₇ with sodium sulphite, Na₂SO₃ in an acid solution to give chromium(III) ion and sulphate ion.

Skeletal equation: $Cr_2O_7^{2-} + SO_3^{2-} \rightarrow Cr^{3+} + SO_4^{2-}$
1. Oxidation half: $SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-$
2. Reduction half: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
3. Multiply oxidation half by 3 to equalize electrons ($6e^-$): $$3SO_3^{2-} + 3H_2O \rightarrow 3SO_4^{2-} + 6H^+ + 6e^-$$ 4. Adding the two halves and simplifying $H^+$ and $H_2O$: $$\mathbf{Cr_2O_7^{2-}(aq) + 3SO_3^{2-}(aq) + 8H^+(aq) \rightarrow 2Cr^{3+}(aq) + 3SO_4^{2-}(aq) + 4H_2O(l)}$$

Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O.

7.10

Permanganate(VII) ion, MnO₄⁻ in basic solution oxidises iodide ion, I⁻ to produce molecular iodine (I₂) and manganese(IV) oxide (MnO₂). Write a balanced ionic equation.

1. Oxidation half: $2I^- \rightarrow I_2 + 2e^-$
2. Reduction half: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$
3. Equalize electrons by multiplying oxidation half by 3 and reduction half by 2 ($6e^-$ total): $$6I^- \rightarrow 3I_2 + 6e^-$$ $$2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$$ 4. Adding together: $$\mathbf{2MnO_4^-(aq) + 6I^-(aq) + 4H_2O(l) \rightarrow 2MnO_2(s) + 3I_2(s) + 8OH^-(aq)}$$

2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.

7.3.3 Redox Titrations and Indicators

Similar to acid-base titrations using pH indicators, redox titrations determine solution concentration using redox-sensitive indicators:

Titration Type	Reagents Involved	Indicator Mechanism & End Point
Permanganate Titration ($KMnO_4$)	$MnO_4^-$ with $Fe^{2+}$ or $C_2O_4^{2-}$ in dil. $H_2SO_4$	Self-Indicator: $MnO_4^-$ is deep purple. The very first excess drop beyond equivalence imparts a lasting faint pink colour (detectable at $10^{-6}\text{ M}$).
Dichromate Titration ($K_2Cr_2O_7$)	$Cr_2O_7^{2-}$ with $Fe^{2+}$	External / Redox Indicator: Diphenylamine oxidises immediately past equivalence point producing an intense blue-violet colour.
Iodometric Titration	$2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 + I_2$ Then $I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$	Starch Indicator: Starch forms an intense deep blue complex with free $I_2$. The blue color sharply vanishes to colourless when all $I_2$ is consumed by thiosulphate.

Visual End Point Detection in Redox Titrations (Permanganate, Dichromate & Iodometry)

7.4 Redox Reactions and Electrode Processes (Daniell Cell)

When zinc metal is placed directly into copper sulphate solution, electron transfer is direct and energy is liberated as heat. In a Galvanic (Voltaic) Cell, the two half-reactions are physically separated so that electron transfer occurs through an external metallic circuit, generating useful electrical energy!

The Daniell Cell (Zn-Cu Galvanic Cell) with Salt Bridge

Functions of the Salt Bridge

Completes the electrical circuit by permitting the migration of ions between the two half-cells.
Maintains electrical neutrality in both compartments by supplying counter-ions ($K^+$ to cathode, $Cl^-$ to anode), preventing junction potential buildup.
Eliminates liquid junction potential and prevents mechanical mixing of the two solutions.

7.4.1 Standard Electrode Potentials ($E^\circ$) & Electrochemical Series

By international convention (IUPAC), Standard Electrode Potential ($E^\circ$) refers strictly to Standard Reduction Potential measured at $298\text{ K}$, $1\text{ bar}$ pressure, and $1\text{ M}$ concentration against the Standard Hydrogen Electrode (SHE, assigned $E^\circ = 0.00\text{ V}$).

$$\mathbf{E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ_{\text{right}} - E^\circ_{\text{left}}}$$ $$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$ Criterion for Spontaneity: A redox reaction is thermodynamically feasible if $\mathbf{E^\circ_{\text{cell}} > 0}$ ($\Delta G^\circ < 0$).

Redox Couple (Reduction Half-Reaction)	$E^\circ$ (Volts) at 298 K	Oxidising / Reducing Strength
$F_{2(g)} + 2e^- \rightarrow 2F^-$	$\mathbf{+2.87\text{ V}}$	Strongest Oxidising Agent ($F_2$ has maximum tendency to gain $e^-$)
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$	$\mathbf{+1.51\text{ V}}$	Powerful oxidant in acidic medium
$Cl_{2(g)} + 2e^- \rightarrow 2Cl^-$	$\mathbf{+1.36\text{ V}}$	Strong oxidant
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$	$\mathbf{+1.33\text{ V}}$	Strong oxidant in acidic medium
$Ag^+ + e^- \rightarrow Ag_{(s)}$	$\mathbf{+0.80\text{ V}}$	Moderate oxidant
$Cu^{2+} + 2e^- \rightarrow Cu_{(s)}$	$\mathbf{+0.34\text{ V}}$	Reduced by hydrogen and active metals
$2H^+ + 2e^- \rightarrow H_{2(g)}$	$\mathbf{0.00\text{ V}}$	Standard Reference (SHE)
$Pb^{2+} + 2e^- \rightarrow Pb_{(s)}$	$\mathbf{-0.13\text{ V}}$	Displaces $H_2$ very slowly
$Fe^{2+} + 2e^- \rightarrow Fe_{(s)}$	$\mathbf{-0.44\text{ V}}$	Reduces $H^+$, displaces $H_2$ from acids
$Zn^{2+} + 2e^- \rightarrow Zn_{(s)}$	$\mathbf{-0.76\text{ V}}$	Active reducing metal
$Al^{3+} + 3e^- \rightarrow Al_{(s)}$	$\mathbf{-1.66\text{ V}}$	Very strong reductant
$Mg^{2+} + 2e^- \rightarrow Mg_{(s)}$	$\mathbf{-2.36\text{ V}}$	Displaces $H_2$ from steam/acids
$Na^+ + e^- \rightarrow Na_{(s)}$	$\mathbf{-2.71\text{ V}}$	Displaces $H_2$ violently from cold water
$Li^+ + e^- \rightarrow Li_{(s)}$	$\mathbf{-3.05\text{ V}}$	Strongest Reducing Agent (highest hydration enthalpy)

Standard Hydrogen Electrode (SHE) Reference Datum (E° = 0.00 V at 298 K)

Standard Conditions: p(H₂) = 1 bar, [H⁺] = 1 M, T = 298 K • Platinized Platinum Foil: Catalyzes reversible adsorption • Half Reaction: 2H⁺(aq) + 2e⁻ ⇌ H₂(g) • Arbitrary Assigned Potential: E° = 0.00 V

Redox Reactions Conceptual Quiz (25 MCQs)

Test your understanding of classical and electronic definitions, oxidation number rules, balancing half-reactions, disproportionation criteria, redox titrations, and electrochemical series. Instant explanatory feedback on every answer.

1 Which of the following processes represents oxidation in terms of electronic concept?

Incorrect: Gain of electrons is reduction (RIG).

Correct: Loss of electrons by a chemical entity is oxidation (OIL).

Incorrect: Addition of hydrogen is reduction.

Incorrect: Removal of oxygen is classical reduction.

2 What is the oxidation number of oxygen in KO₂ (potassium superoxide)?

Incorrect: −2 is for normal oxides.

Incorrect: −1 is for peroxides.

Correct: In superoxide ion (O₂⁻), total charge is −1 shared by two oxygen atoms, giving −1/2 per atom.

Incorrect: +2 is for OF₂.

3 The oxidation state of chromium in dichromate ion (Cr₂O₇²⁻) is:

Incorrect: +3 is the reduced form Cr³⁺.

Incorrect: Chromium is not in +5.

Correct: 2x + 7(−2) = −2 ⇒ 2x = +12 ⇒ x = +6.

Incorrect: Group 6 elements cannot exceed +6.

4 Which of the following compounds exhibits fractional oxidation state due to resonance/structural averaging?

Incorrect: Mn is +7.

Correct: In carbon suboxide (O=C=C*=C=O), the two terminal C atoms are +2 and central C* is 0, giving an average of 4/3.

Incorrect: S is +6.

Incorrect: C is +4.

5 The Stock notation for auric chloride (HAuCl₄) is correctly written as:

Incorrect: +1 is aurous.

Incorrect: Gold is not in +2 state here.

Correct: +1 + x + 4(−1) = 0 ⇒ x = +3. Hence HAu(III)Cl₄.

Incorrect: Not +4.

6 Which of the following oxoanions CANNOT undergo a disproportionation reaction?

Incorrect: Cl is in +1 and can disproportionate.

Incorrect: Cl is in +3 and can disproportionate.

Incorrect: Cl is in +5 and can disproportionate.

Correct: In ClO₄⁻, chlorine is in its maximum possible oxidation state (+7) and cannot be oxidised further!

7 Why does fluorine NEVER exhibit disproportionation?

Incorrect: Size is not the primary chemical reason.

Correct: Disproportionation requires an element to be oxidised to a higher state; fluorine can only form 0 or −1.

Incorrect: Reactivity alone does not dictate oxidation states.

Incorrect: Irrelevant.

8 In the reaction 2H₂O₂(aq) → 2H₂O(l) + O₂(g), oxygen undergoes:

Incorrect: It also forms H₂O.

Incorrect: It also forms O₂.

Correct: Oxygen in peroxide (−1) is oxidised to O₂ (0) and reduced to H₂O (−2) simultaneously.

Incorrect: Clearly a redox reaction.

9 In balancing redox equations in basic medium, each H⁺ ion added should be neutralized by adding:

Incorrect: Water alone does not neutralize H⁺.

Correct: Adding equal OH⁻ to both sides converts H⁺ + OH⁻ to H₂O while preserving mass and charge balance.

Incorrect: Irrelevant.

Incorrect: Electrons balance charges, not protons.

10 In the titration of oxalic acid against potassium permanganate, the indicator used is:

Incorrect: Not used in permanganate titrations.

Incorrect: Used in strong acid-weak base titrations.

Incorrect: Used for dichromate titrations.

Correct: MnO₄⁻ is deep purple and acts as its own indicator; the first lasting pink tinge marks the end point.

11 Which indicator is specifically employed in potassium dichromate (K₂Cr₂O₇) titrations?

Incorrect: Starch is used for iodometry.

Correct: Cr₂O₇²⁻ is not a self-indicator; diphenylamine is oxidized at the end point to give an intense blue-violet colour.

Incorrect: Acid-base indicator.

Incorrect: Litmus is for pH testing.

12 In an operating Daniell cell, electrons flow through the external circuit from:

Incorrect: That is the direction of conventional current.

Correct: Oxidation at the zinc anode releases electrons, which flow through the metallic wire to the copper cathode.

Incorrect: Salt bridge carries ions, not electrons.

Incorrect: Incorrect pathway.

13 What is the primary function of the salt bridge in a galvanic cell?

Incorrect: It does not increase cell potential.

Correct: It allows ion migration (K⁺ and Cl⁻) to maintain electrical neutrality without physical solution mixing.

Incorrect: Precipitation would ruin the cell.

Incorrect: Wires carry electrons; salt bridge carries ions.

14 The Standard Hydrogen Electrode (SHE) has been arbitrarily assigned a potential of:

Incorrect: Not 1.00 V.

Correct: By international IUPAC convention, E° of the standard hydrogen electrode (2H⁺ + 2e⁻ ⇌ H₂) is defined as 0.00 V.

Incorrect: Not −1.00 V.

Incorrect: 0.76 V is the magnitude for Zn/Zn²⁺.

15 Which element is the strongest reducing agent in aqueous solution based on standard electrode potentials?

Incorrect: F₂ is the strongest oxidising agent (+2.87 V).

Incorrect: Na has E° = −2.71 V.

Correct: Li has the most negative reduction potential (E° = −3.05 V) owing to its extraordinarily high hydration enthalpy.

Incorrect: Zn has E° = −0.76 V.

16 Which halogen is the strongest oxidising agent?

Correct: F₂ has E° = +2.87 V, making it the most powerful chemical oxidising agent known.

Incorrect: Cl₂ is +1.36 V.

Incorrect: Br₂ is +1.09 V.

Incorrect: I₂ is +0.54 V.

17 In the Layer Test for halides, adding chlorine water to a solution of NaBr and CCl₄ produces a layer of:

Incorrect: Violet layer indicates free I₂ from iodide.

Correct: Cl₂ displaces Br⁻: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂. Free Br₂ dissolves in CCl₄ layer turning it orange-red.

Incorrect: Not blue.

Incorrect: Not green.

18 What is the oxidation number of sulphur in Caro's acid (peroxomonosulphuric acid, H₂SO₅)?

Incorrect: +8 exceeds maximum valence electron limit of sulphur (6).

Correct: Structure has one peroxo linkage (—O—O—): H—O—SO₂(—O—O—H). 2(+1) + x + 3(−2) + 2(−1) = 0 ⇒ x = +6.

Incorrect: Not +4.

Incorrect: Not +2.

19 For a spontaneous cell reaction, the standard cell EMF (E°_cell) and Gibbs free energy (ΔG°) must be:

Incorrect: ΔG° must be negative.

Incorrect: E°_cell must be positive.

Correct: From ΔG° = −nFE°_cell, a positive E°_cell corresponds to a negative ΔG° (thermodynamic spontaneity).

Incorrect: That is the condition at equilibrium.

20 In the reaction: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O, the equivalent weight of K₂Cr₂O₇ is:

Incorrect: n-factor is not 1.

Incorrect: 3 is the change per Cr atom, but there are 2 Cr atoms.

Correct: Total change in oxidation state per Cr₂O₇²⁻ molecule = 2 × (6 − 3) = 6 electrons. Hence Eq. Wt = M / 6.

Incorrect: n-factor is 6.

21 A metal strip of copper placed in a solution of AgNO₃ turns the solution blue because:

Incorrect: Ag precipitation does not colour the solution.

Correct: Cu is higher in activity than Ag (Cu + 2Ag⁺ → Cu²⁺ + 2Ag). Formation of hydrated Cu²⁺ imparts the characteristic blue colour.

Incorrect: Nitrate is a spectator ion.

Incorrect: No hydrolysis occurs.

22 What is the oxidation number of iron in brown ring complex [Fe(H₂O)₅(NO)]SO₄?

Correct: In the brown ring test, NO exists as nitrosonium ion (NO⁺): x + 5(0) + (+1) + (−2) = 0 ⇒ x = +1.

Incorrect: Common misconception; Fe is +1.

Incorrect: Not +3.

Incorrect: Not zero.

23 Which of the following is NOT a redox reaction?

Incorrect: Combination redox.

Correct: Thermal decomposition of CaCO₃ involves no change in oxidation state: Ca(+2), C(+4), O(−2) remain identical in reactants and products.

Incorrect: Disproportionation redox.

Incorrect: Metal displacement redox.

24 The oxidation number of phosphorus in pyrophosphoric acid (H₄P₂O₇) is:

Incorrect: Not +3.

Incorrect: Not +4.

Correct: 4(+1) + 2x + 7(−2) = 0 ⇒ 2x = 10 ⇒ x = +5.

Incorrect: Phosphorus max is +5.

25 What is the n-factor of KMnO₄ in strongly alkaline medium where it converts to manganate (K₂MnO₄)?

Correct: In strong alkali, MnO₄⁻ (+7) is reduced to MnO₄²⁻ (+6). Change in oxidation state = 7 − 6 = 1. Hence n-factor = 1.

Incorrect: 3 is the n-factor in neutral/faintly alkaline medium (forming MnO₂).

Incorrect: 5 is the n-factor in acidic medium (forming Mn²⁺).

Incorrect: Not 7.

Quiz Results

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NCERT Textbook Exercises (7.1 to 7.30 Fully Solved)

Complete step-by-step solutions for all 30 end-of-chapter questions from CBSE / NCERT Class 11 Chemistry Chapter 7 (Redox Reactions).

7.1

Assign oxidation number to the underlined elements in each of the following species: (a) NaH₂PO₄ (b) NaHSO₄ (c) H₄P₂O₇ (d) K₂MnO₄ (e) CaO₂ (f) NaBH₄ (g) H₂S₂O₇ (h) KAl(SO₄)₂·12H₂O

(a) NaH₂PO₄: $+1 + 2(+1) + x + 4(-2) = 0 \implies 3 + x - 8 = 0 \implies \mathbf{x = +5}$
(b) NaHSO₄: $+1 + 1 + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies \mathbf{x = +6}$
(c) H₄P₂O₇: $4(+1) + 2x + 7(-2) = 0 \implies 4 + 2x - 14 = 0 \implies 2x = 10 \implies \mathbf{x = +5}$
(d) K₂MnO₄: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies \mathbf{x = +6}$
(e) CaO₂: Calcium peroxide with peroxo linkage ($O_2^{2-}$): $+2 + 2x = 0 \implies \mathbf{x = -1}$
(f) NaBH₄: Hydrogen is present as hydride ($H^-$) with electronegative boron: $+1 + x + 4(-1) = 0 \implies \mathbf{x = +3}$
(g) H₂S₂O₇: (Oleum/pyrosulphuric acid): $2(+1) + 2x + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies \mathbf{x = +6}$
(h) KAl(SO₄)₂·12H₂O: (Potash alum): In sulphate ion ($SO_4^{2-}$), $x + 4(-2) = -2 \implies \mathbf{x = +6}$

(a) P = +5; (b) S = +6; (c) P = +5; (d) Mn = +6; (e) O = −1; (f) B = +3; (g) S = +6; (h) S = +6.

7.2

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? (a) KI₃ (b) H₂S₄O₆ (c) Fe₃O₄ (d) CH₃CH₂OH (e) CH₃COOH

(a) KI₃: Contains $K^+$ and triiodide ion ($I_3^-$). In $I_3^-$, an $I_2$ molecule is coordinate bonded to an iodide ion ($I^- \rightarrow I-I$). The two iodine atoms of $I_2$ have oxidation number 0 while the iodide ion has −1. The average is $\mathbf{-1/3}$.
(b) H₂S₄O₆: (Tetrathionate ion): Structure is $HO-SO_2-S-S-SO_2-OH$. The two terminal sulphur atoms attached to oxygens have oxidation number +5 each, while the two central sulphur atoms bonded only to sulphur have 0. Average $= \frac{5 + 0 + 0 + 5}{4} = \mathbf{+2.5}$.
(c) Fe₃O₄: Mixed oxide consisting of equimolar $FeO$ and $Fe_2O_3$ ($Fe^{II}O \cdot Fe^{III}_2O_3$). In $FeO$, iron is +2; in $Fe_2O_3$, iron is +3. Average $= \frac{2 + 2(3)}{3} = \mathbf{+8/3}$.
(d) CH₃CH₂OH: Carbon 1 (methyl): $C-H_3$ gives $x + 3(+1) = 0 \implies C_1 = \mathbf{-3}$. Carbon 2 ($-\text{CH}_2\text{OH}$): $x + 2(+1) - 1 = 0 \implies C_2 = \mathbf{-1}$. Average $= \frac{-3 + (-1)}{2} = \mathbf{-2}$.
(e) CH₃COOH: Carbon 1 (methyl): $x + 3(+1) = 0 \implies C_1 = \mathbf{-3}$. Carbon 2 ($-\text{COOH}$): $x + (-1) + (-2) = 0 \implies C_2 = \mathbf{+3}$. Average $= \frac{-3 + (+3)}{2} = \mathbf{0}$.

Fractional or non-intuitive numbers represent averages over chemically distinct atoms within the structure.

7.3

Justify that the following reactions are redox reactions: (a) CuO(s) + H₂(g) → Cu(s) + H₂O(g); (b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g); (c) 4BCl₃(g) + 3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s); (d) 2K(s) + F₂(g) → 2K⁺F⁻(s); (e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g).

(a) $Cu$ decreases from $+2$ (in $CuO$) to $0$ (in $Cu$) $\implies$ Reduction. $H$ increases from $0$ (in $H_2$) to $+1$ (in $H_2O$) $\implies$ Oxidation.
(b) $Fe$ decreases from $+3$ (in $Fe_2O_3$) to $0$ (in $Fe$) $\implies$ Reduction. $C$ increases from $+2$ (in $CO$) to $+4$ (in $CO_2$) $\implies$ Oxidation.
(c) $B$ decreases from $+3$ (in $BCl_3$) to $-3$ (in $B_2H_6$) $\implies$ Reduction. Hydrogen increases from $-1$ (in $LiAlH_4$) to $+1$ (in $B_2H_6$) $\implies$ Oxidation.
(d) $K$ increases from $0$ to $+1$ (in $KF$) $\implies$ Oxidation. $F$ decreases from $0$ (in $F_2$) to $-1$ (in $KF$) $\implies$ Reduction.
(e) $N$ increases from $-3$ (in $NH_3$) to $+2$ (in $NO$) $\implies$ Oxidation. $O$ decreases from $0$ (in $O_2$) to $-2$ (in $NO$ and $H_2O$) $\implies$ Reduction.

Each reaction involves simultaneous increase and decrease in oxidation numbers; hence all are redox changes.

7.4

Fluorine reacts with ice and results in the change: H₂O(s) + F₂(g) → HF(g) + HOF(g). Justify that this reaction is a redox reaction.

Assign oxidation states to all atoms in the reaction: $$\overset{+1}{H}_2\overset{-2}{O}_{(s)} + \overset{0}{F}_{2(g)} \rightarrow \overset{+1}{H}\overset{-1}{F}_{(g)} + \overset{+1}{H}\overset{-2}{O}\overset{+1}{F}_{(g)}$$ • In $F_2$, fluorine is in $0$ oxidation state.
• In $HF$, fluorine has oxidation state −1 (decreased from 0 to −1 $\implies$ Reduction).
• In $HOF$ (hypofluorous acid), oxygen is more electronegative than hydrogen ($+1$) but less electronegative than fluorine. However, standard rules assign $F = -1, H = +1$, so oxygen is $0$, meaning oxygen increased from $-2$ to $0$ ($\implies$ Oxidation). If viewed in terms of fluorine atoms, one F atom is reduced ($0 \rightarrow -1$ in $HF$) and the other is oxidised ($0 \rightarrow +1$ in $HOF$ under classical polarity assignment).
Thus, the process is a redox change.

Reaction involves simultaneous oxidation and reduction; hence it is a redox reaction.

7.5

Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇²⁻ and NO₃⁻. Suggest structure of these compounds. Count for the fallacy.

1. H₂SO₅ (Caro's acid): By conventional formula: $2(+1) + x + 5(-2) = 0 \implies x = +8$. This is a fallacy because sulphur has only 6 valence electrons and cannot exceed $+6$! Structure contains one peroxide bond: $HO-SO_2-O-O-H$. Calculation: $2(+1) + x + 3(-2) + 2(-1) = 0 \implies \mathbf{x = +6}$.
2. Cr₂O₇²⁻: Structure contains two tetrahedral $CrO_4$ sharing a central bridging oxygen ($O_3Cr-O-CrO_3^{2-}$). $2x + 7(-2) = -2 \implies 2x = +12 \implies \mathbf{x = +6}$ (no fallacy, matches group number).
3. NO₃⁻: Structure has nitrogen forming one double bond to $O$, one single bond with formal charge, and one coordinate bond. $x + 3(-2) = -1 \implies \mathbf{x = +5}$ (no fallacy, matches nitrogen maximum).

H₂SO₅: S = +6 (peroxide linkage); Cr₂O₇²⁻: Cr = +6; NO₃⁻: N = +5.

7.6

Write formulas for the following compounds: (a) Mercury(II) chloride (b) Nickel(II) sulphate (c) Tin(IV) oxide (d) Thallium(I) sulphate (e) Iron(III) sulphate (f) Chromium(III) oxide.

(a) Mercury(II) chloride: $\mathbf{HgCl_2}$
(b) Nickel(II) sulphate: $\mathbf{NiSO_4}$
(c) Tin(IV) oxide: $\mathbf{SnO_2}$
(d) Thallium(I) sulphate: $\mathbf{Tl_2SO_4}$
(e) Iron(III) sulphate: $\mathbf{Fe_2(SO_4)_3}$
(f) Chromium(III) oxide: $\mathbf{Cr_2O_3}$

(a) HgCl₂; (b) NiSO₄; (c) SnO₂; (d) Tl₂SO₄; (e) Fe₂(SO₄)₃; (f) Cr₂O₃.

7.7

Suggest a list of substances where carbon can exhibit oxidation states from −4 to +4 and nitrogen from −3 to +5.

Carbon (−4 to +4):
• $\mathbf{-4}$: $CH_4$ (Methane)
• $\mathbf{-3}$: $C_2H_6$ (Ethane)
• $\mathbf{-2}$: $CH_3Cl$ (Chloromethane)
• $\mathbf{-1}$: $C_2H_2$ (Ethyne)
• $\mathbf{0}$: $CH_2Cl_2$ (Dichloromethane), $C_6H_{12}O_6$ (Glucose)
• $\mathbf{+1}$: $C_2H_2Cl_4$
• $\mathbf{+2}$: $CO$ (Carbon monoxide), $CHCl_3$ (Chloroform)
• $\mathbf{+3}$: $C_2O_4^{2-}$ (Oxalate ion)
• $\mathbf{+4}$: $CO_2$ (Carbon dioxide), $CCl_4$

Nitrogen (−3 to +5):
• $\mathbf{-3}$: $NH_3$ (Ammonia)
• $\mathbf{-2}$: $N_2H_4$ (Hydrazine)
• $\mathbf{-1}$: $NH_2OH$ (Hydroxylamine)
• $\mathbf{0}$: $N_2$ (Dinitrogen)
• $\mathbf{+1}$: $N_2O$ (Nitrous oxide)
• $\mathbf{+2}$: $NO$ (Nitric oxide)
• $\mathbf{+3}$: $HNO_2$ (Nitrous acid), $N_2O_3$
• $\mathbf{+4}$: $NO_2$ (Nitrogen dioxide), $N_2O_4$
• $\mathbf{+5}$: $HNO_3$ (Nitric acid), $N_2O_5$

Listed compounds exhibiting all oxidation numbers across the range.

7.8

While sulphur dioxide (SO₂) and hydrogen peroxide (H₂O₂) can act as oxidising as well as reducing agents, ozone (O₃) and nitric acid (HNO₃) act only as oxidants. Why?

An element can act as both oxidant and reductant if it is in an intermediate oxidation state:
• In $SO_2$, sulphur is in +4 state. Since sulphur exhibits oxidation states from $-2$ to $+6$, it can increase to $+6$ (reducing agent) or decrease to $0$ or $-2$ (oxidising agent).
• In $H_2O_2$, oxygen is in −1 state. It can be oxidised to $O_2$ ($0$) or reduced to $H_2O$ ($-2$).
Conversely:
• In $HNO_3$, nitrogen is in its maximum possible oxidation state (+5). It cannot lose any more electrons and can only decrease its oxidation state; thus it acts strictly as an oxidant.
• In $O_3$, oxygen has an oxidation state of 0. Thermodynamically, ozone readily decomposes into nascent oxygen ($O_3 \rightarrow O_2 + [O]$), powerfully accepting electrons to form $-2$ oxides; hence it acts only as an oxidant.

Because N in HNO₃ is in maximum oxidation state (+5) and O₃ spontaneously decomposes to yield nascent oxygen.

7.9

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 them with 12H₂O and separated O₂? Suggest a technique to investigate their reaction paths.

(a) In photosynthesis, isotopic tracer experiments using water labelled with $^{18}O$ ($H_2^{18}O$) prove that all 6 molecules of liberated $O_2$ originate exclusively from water, not from $CO_2$. Therefore, 12 molecules of $H_2O$ are required to yield 6 molecules of $O_2$: $$\mathbf{6CO_2 + 12H_2O \xrightarrow{h\nu} C_6H_{12}O_6 + 6H_2O + 6O_2}$$ (b) In the reaction $O_3 + H_2O_2 \rightarrow H_2O + 2O_2$, labelling $H_2O_2$ with $^{18}O$ ($H_2^{18}O_2$) proves that one $O_2$ molecule originates completely from $H_2O_2$, while the other $O_2$ comes from ozone ($O_3$). Writing it as: $$\mathbf{O_3 + H_2O_2 \rightarrow H_2O + O_2 + O_2}$$ accurately depicts the two distinct mechanistic sources.
Technique: Isotopic Tracer Technique (using stable isotope $^{18}O$ or radioactive isotopes and detection by mass spectrometry).

Isotopic tracer technique with ¹⁸O reveals the precise molecular origins of oxygen in both processes.

7.10

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

In $AgF_2$, silver is present in the rare +2 oxidation state ($Ag^{2+}$ with electronic configuration $[Kr] 4d^9$). The stable electronic configuration for silver is $Ag^+$ ($[Kr] 4d^{10}$, completely filled $d$-subshell).
Because of the immense thermodynamic drive to achieve the stable $d^{10}$ configuration: $$Ag^{2+} + e^- \rightarrow Ag^+ \qquad E^\circ = +1.98\text{ V}$$ $Ag^{2+}$ avidly pulls electrons from any available species, acting as an extraordinarily powerful oxidising agent.

Ag²⁺ has an unstable 4d⁹ configuration and eagerly gains an electron to form stable 4d¹⁰ (Ag⁺).

7.11

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

Illustration 1: Carbon and Oxygen
• Excess reducing agent ($C$): $2C_{(\text{excess})} + O_2 \rightarrow 2\overset{+2}{C}O$ (Carbon in lower $+2$ state)
• Excess oxidising agent ($O_2$): $C + O_{2(\text{excess})} \rightarrow \overset{+4}{C}O_2$ (Carbon in higher $+4$ state)

Illustration 2: Phosphorus and Chlorine
• Excess reducing agent ($P_4$): $P_{4(\text{excess})} + 6Cl_2 \rightarrow 4\overset{+3}{P}Cl_3$ (Phosphorus in lower $+3$ state)
• Excess oxidising agent ($Cl_2$): $P_4 + 10Cl_{2(\text{excess})} \rightarrow 4\overset{+5}{P}Cl_5$ (Phosphorus in higher $+5$ state)

Illustration 3: Sulphur and Fluorine
• Excess reducing agent ($S$): $S_{(\text{excess})} + 2F_2 \rightarrow \overset{+4}{S}F_4$ (Sulphur in lower $+4$ state)
• Excess oxidising agent ($F_2$): $S + 3F_{2(\text{excess})} \rightarrow \overset{+6}{S}F_6$ (Sulphur in higher $+6$ state)

Demonstrated for C/O₂, P₄/Cl₂, and S/F₂.

7.12

Account for observations: (a) In the manufacture of benzoic acid from toluene, alcoholic KMnO₄ is used instead of acidic/aqueous KMnO₄. Write balanced equation. (b) Adding conc. H₂SO₄ to a mixture containing chloride gives pungent HCl gas, but with bromide gives red vapour of Br₂. Why?

(a) Toluene and $KMnO_4$ do not mix well in pure aqueous medium because toluene is non-polar. Alcohol acts as a mutual solvent, homogenizing both reactants for rapid, intimate contact and smooth oxidation. In alkaline/alcoholic medium, $OH^-$ ions facilitate proton abstraction from the benzylic methyl group: $$\mathbf{C_6H_5CH_3 + 2KMnO_4 \rightarrow C_6H_5COOK + 2MnO_2 + KOH + H_2O}$$ Subsequent acidification with $HCl$ yields pure benzoic acid: $C_6H_5COOK + HCl \rightarrow C_6H_5COOH + KCl$.

(b) $HCl$ is a weak reducing agent; conc. $H_2SO_4$ cannot oxidise $Cl^-$ to $Cl_2$. Thus only a non-redox protonation occurs: $NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\uparrow$ (colourless, pungent gas).
In contrast, $HBr$ is a much stronger reducing agent ($E^\circ_{Br_2/Br^-} = +1.09\text{ V}$, while $E^\circ_{Cl_2/Cl^-} = +1.36\text{ V}$). Conc. $H_2SO_4$ oxidises $Br^-$ to elemental bromine ($Br_2$): $$2NaBr + 3H_2SO_4 \rightarrow 2NaHSO_4 + SO_2 + Br_2\uparrow (\text{red vapour}) + 2H_2O$$

(a) Alcohol acts as mutual solvent; (b) HBr is stronger reducing agent than HCl and is easily oxidised by conc. H₂SO₄ to Br₂.

7.13

Identify the substance oxidised, reduced, oxidising agent and reducing agent for each: (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).

(a) $C_6H_6O_2$ (hydroquinone) loses $H$ $\implies$ Oxidised (Reductant). $AgBr$ ($Ag^+$ to $Ag$) $\implies$ Reduced (Oxidant).
(b) $HCHO$ oxidised to $HCOO^-$ $\implies$ Reductant. $[Ag(NH_3)_2]^+$ ($Ag^+$ to $Ag$) $\implies$ Reduced (Oxidant).
(c) $HCHO$ oxidised to $HCOO^-$ $\implies$ Reductant. $Cu^{2+}$ reduced to $Cu_2O$ ($Cu^+$) $\implies$ Oxidant.
(d) $N_2H_4$ ($-2$ to $0$) $\implies$ Oxidised (Reductant). $H_2O_2$ ($-1$ to $-2$) $\implies$ Reduced (Oxidant).
(e) $Pb$ ($0$ to $+2$) $\implies$ Oxidised (Reductant). $PbO_2$ ($+4$ to $+2$) $\implies$ Reduced (Oxidant).

All species identified based on electron shifts and oxidation number changes.

7.14

Consider the reactions: 2S₂O₃²⁻(aq) + I₂(s) → S₄O₆²⁻(aq) + 2I⁻(aq) and S₂O₃²⁻(aq) + 2Br₂(l) + 5H₂O(l) → 2SO₄²⁻(aq) + 4Br⁻(aq) + 10H⁺(aq). Why does the same reductant (thiosulphate) react differently with iodine and bromine?

Bromine ($E^\circ_{Br_2/Br^-} = +1.09\text{ V}$) is a much stronger oxidising agent than iodine ($E^\circ_{I_2/I^-} = +0.54\text{ V}$).
• The mild oxidant $I_2$ can only oxidise sulphur from $+2$ (in $S_2O_3^{2-}$) to $+2.5$ (in tetrathionate, $S_4O_6^{2-}$).
• The much stronger oxidant $Br_2$ oxidises sulphur completely to its maximum oxidation state of +6 (in sulphate, $SO_4^{2-}$).

Bromine is a much stronger oxidising agent than iodine and oxidises sulphur to its highest state (+6).

7.15

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

1. Fluorine as the best oxidant: $F_2$ has the highest standard reduction potential ($+2.87\text{ V}$) and oxidises all other halide ions to free halogens; it even oxidises water to $O_2$: $$2F_2 + 2H_2O \rightarrow 4HF + O_2\uparrow$$ No other halogen can oxidise water in this manner.
2. HI as the best reductant: Down Group 17, $H-X$ bond dissociation energy decreases ($H-F > H-Cl > H-Br > H-I$). $HI$ has the longest and weakest bond, liberating hydrogen most readily. $HI$ easily reduces sulphuric acid to $SO_2$ and $H_2S$, and reduces $Fe^{3+}$ to $Fe^{2+}$: $$2Fe^{3+} + 2HI \rightarrow 2Fe^{2+} + I_2 + 2H^+$$ $HCl$ and $HF$ cannot perform this reduction.

F₂ has highest E° (+2.87 V) displacing all halogens and water; HI has lowest bond dissociation enthalpy.

7.16

Why does the following reaction occur? XeO₆⁴⁻(aq) + 2F⁻(aq) + 6H⁺(aq) → XeO₃(g) + F₂(g) + 3H₂O(l). What conclusion about Na₄XeO₆ can be drawn from this reaction?

Fluorine has the highest reduction potential of all elements ($E^\circ_{F_2/F^-} = +2.87\text{ V}$), meaning $F^-$ is extraordinarily difficult to oxidise.
In this reaction, perxenate ion ($XeO_6^{4-}$, where $Xe$ is in $+8$ state) oxidises $F^-$ to elemental $F_2$ while being reduced to $XeO_3$ ($Xe = +6$).
Conclusion: Since perxenate can oxidise $F^-$, the perxenate ion $XeO_6^{4-}$ (and its salt $Na_4XeO_6$) is an even stronger oxidising agent than fluorine itself ($E^\circ > +2.87\text{ V}$)!

Perxenate (XeO₆⁴⁻) is an even stronger oxidising agent than fluorine (E° > +2.87 V).

7.17

Consider the reactions: H₃PO₂ reduces AgNO₃ to Ag and CuSO₄ to Cu; C₆H₅CHO reduces Tollen's reagent ([Ag(NH₃)₂]⁺) to Ag, but does not react with Fehling's solution (Cu²⁺). What inference do you draw about Ag⁺ and Cu²⁺?

• In reactions with hypophosphorous acid ($H_3PO_2$, a powerful reducing agent), both $Ag^+$ ($E^\circ = +0.80\text{ V}$) and $Cu^{2+}$ ($E^\circ = +0.34\text{ V}$) are reduced to metallic form.
• Benzaldehyde ($C_6H_5CHO$) is a mild reducing agent. It reduces $[Ag(NH_3)_2]^+$ to metallic silver, but fails to reduce $Cu^{2+}$.
Inference: $Ag^+$ has a higher standard reduction potential ($+0.80\text{ V}$) than $Cu^{2+}$ ($+0.34\text{ V}$). Therefore, $Ag^+$ is a stronger oxidising agent than $Cu^{2+}$.

Ag⁺ is a stronger oxidising agent than Cu²⁺ (E° values: +0.80 V vs +0.34 V).

7.18

Balance the following redox reactions by ion-electron method: (a) MnO₄⁻(aq) + I⁻(aq) → MnO₂(s) + I₂(s) (in basic medium); (b) MnO₄⁻(aq) + SO₂(g) → Mn²⁺(aq) + HSO₄⁻(aq) (in acidic solution); (c) H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l) (in acidic solution); (d) Cr₂O₇²⁻ + SO₂(g) → Cr³⁺(aq) + SO₄²⁻(aq) (in acidic solution).

(a) In basic medium:
Oxidation: $2I^- \rightarrow I_2 + 2e^-$ (multiply by 3)
Reduction: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$ (multiply by 2)
$$\mathbf{2MnO_4^-(aq) + 6I^-(aq) + 4H_2O(l) \rightarrow 2MnO_2(s) + 3I_2(s) + 8OH^-(aq)}$$
(b) In acidic solution:
Oxidation: $SO_2 + 2H_2O \rightarrow HSO_4^- + 3H^+ + 2e^-$ (multiply by 5)
Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ (multiply by 2)
Add: $2MnO_4^- + 5SO_2 + 10H_2O + 16H^+ \rightarrow 2Mn^{2+} + 5HSO_4^- + 15H^+ + 8H_2O$
$$\mathbf{2MnO_4^-(aq) + 5SO_2(g) + 2H_2O(l) + H^+(aq) \rightarrow 2Mn^{2+}(aq) + 5HSO_4^-(aq)}$$
(c) In acidic solution:
Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ (multiply by 2)
Reduction: $H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$
$$\mathbf{2Fe^{2+}(aq) + H_2O_2(aq) + 2H^+(aq) \rightarrow 2Fe^{3+}(aq) + 2H_2O(l)}$$
(d) In acidic solution:
Oxidation: $SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^-$ (multiply by 3)
Reduction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
$$\mathbf{Cr_2O_7^{2-}(aq) + 3SO_2(g) + 2H^+(aq) \rightarrow 2Cr^{3+}(aq) + 3SO_4^{2-}(aq) + H_2O(l)}$$

All four balanced ionic equations derived step-by-step.

7.19

Balance the following equations in basic medium and identify oxidant and reductant: (a) P₄(s) + OH⁻(aq) → PH₃(g) + H₂PO₂⁻(aq); (b) N₂H₄(l) + ClO₃⁻(aq) → NO(g) + Cl⁻(g); (c) Cl₂O₇(g) + H₂O₂(aq) → ClO₂⁻(aq) + O₂(g) + H⁺.

(a) P₄ disproportionation:
Reduction: $P_4 + 12H_2O + 12e^- \rightarrow 4PH_3 + 12OH^-$
Oxidation: $P_4 + 8OH^- \rightarrow 4H_2PO_2^- + 4e^-$ (multiply by 3)
Add: $4P_4 + 12H_2O + 24OH^- \rightarrow 4PH_3 + 12H_2PO_2^- + 12OH^-$
Divide by 4: $\mathbf{P_4(s) + 3OH^-(aq) + 3H_2O(l) \rightarrow PH_3(g) + 3H_2PO_2^-(aq)}$
$P_4$ is both oxidant and reductant.

(b) N₂H₄ + ClO₃⁻ in basic medium:
Oxidation: $N_2H_4 + 8OH^- \rightarrow 2NO + 6H_2O + 8e^-$ (multiply by 3)
Reduction: $ClO_3^- + 3H_2O + 6e^- \rightarrow Cl^- + 6OH^-$ (multiply by 4)
$$\mathbf{3N_2H_4(l) + 4ClO_3^-(aq) \rightarrow 6NO(g) + 4Cl^-(aq) + 6H_2O(l)}$$
Oxidant: $ClO_3^-$; Reductant: $N_2H_4$.

(c) Cl₂O₇ + H₂O₂ in basic medium:
Oxidation: $H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-$ (multiply by 4)
Reduction: $Cl_2O_7 + 3H_2O + 8e^- \rightarrow 2ClO_2^- + 6OH^-$
$$\mathbf{Cl_2O_7(g) + 4H_2O_2(aq) + 2OH^-(aq) \rightarrow 2ClO_2^-(aq) + 4O_2(g) + 5H_2O(l)}$$
Oxidant: $Cl_2O_7$; Reductant: $H_2O_2$.

(a) P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻; (b) 3N₂H₄ + 4ClO₃⁻ → 6NO + 4Cl⁻ + 6H₂O; (c) Cl₂O₇ + 4H₂O₂ + 2OH⁻ → 2ClO₂⁻ + 4O₂ + 5H₂O.

7.20

What sort of information can you draw from the following reaction? (CN)₂(g) + 2OH⁻(aq) → CN⁻(aq) + CNO⁻(aq) + H₂O(l).

1. Disproportionation Redox: In cyanogen $(CN)_2$, carbon is in $+3$ state (or $(CN)_2$ as pseudo-halogen $0$). It simultaneously undergoes reduction to cyanide ($CN^-$, carbon $+2$ or $-1$) and oxidation to cyanate ($CNO^-$, carbon $+4$ or $+1$).
2. Pseudohalogen Behaviour: Cyanogen $(CN)_2$ acts analogously to halogens ($Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$). Hence $(CN)_2$ is a pseudohalogen and $CN^-$ is a pseudohalide ion.

Cyanogen behaves as a pseudohalogen undergoing disproportionation in alkaline medium.

7.21

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

Oxidation half: $Mn^{3+} + 2H_2O \rightarrow MnO_2 + 4H^+ + e^-$
Reduction half: $Mn^{3+} + e^- \rightarrow Mn^{2+}$
Adding both half reactions together directly: $$\mathbf{2Mn^{3+}(aq) + 2H_2O(l) \rightarrow Mn^{2+}(aq) + MnO_2(s) + 4H^+(aq)}$$

2Mn³⁺(aq) + 2H₂O(l) → Mn²⁺(aq) + MnO₂(s) + 4H⁺(aq).

7.22

Consider the elements: Cs, Ne, I and F. (a) Identify the element that exhibits only negative oxidation state; (b) Exhibits only positive oxidation state; (c) Exhibits both positive and negative oxidation states; (d) Exhibits neither positive nor negative oxidation state.

(a) Fluorine (F): Most electronegative element; exhibits only −1 in its compounds (and 0 in $F_2$).
(b) Caesium (Cs): Highly electropositive alkali metal; exhibits only +1 in its compounds.
(c) Iodine (I): Exhibits −1 (in $KI$) as well as positive oxidation states: +1, +3, +5, +7 (in $IF_7, HIO_4$).
(d) Neon (Ne): Noble gas with stable $s^2p^6$ configuration; forms no chemical compounds, exhibiting only 0.

(a) F; (b) Cs; (c) I; (d) Ne.

7.23

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

Chlorine oxidises sulphur dioxide to sulphate, while chlorine is reduced to chloride: $$\mathbf{Cl_2(g) + SO_2(g) + 2H_2O(l) \rightarrow 2Cl^-(aq) + SO_4^{2-}(aq) + 4H^+(aq)}$$ (Or molecular form: $Cl_2 + SO_2 + 2H_2O \rightarrow 2HCl + H_2SO_4$).

Cl₂ + SO₂ + 2H₂O → 2Cl⁻ + SO₄²⁻ + 4H⁺.

7.24

Refer to the periodic table: (a) Select possible non-metals that can show disproportionation; (b) Select three metals that can show disproportionation.

(a) Non-metals: Phosphorus ($P_4$), Sulphur ($S_8$), Chlorine ($Cl_2$), Bromine ($Br_2$), Iodine ($I_2$). (All possess variable intermediate oxidation states).
(b) Metals:
• Copper ($Cu^+$): $2Cu^+ \rightarrow Cu^{2+} + Cu$
• Manganese ($Mn^{3+}$ or $Mn^{6+}$): $2Mn^{3+} + 2H_2O \rightarrow MnO_2 + Mn^{2+} + 4H^+$
• Gallium / Thallium ($Tl^+$ / $Ga^+$): $3Ga^+ \rightarrow Ga^{3+} + 2Ga$

(a) Non-metals: P, S, Cl; (b) Metals: Cu, Mn, Ga.

7.25

In Ostwald's process for manufacture of nitric acid, the first step is oxidation of ammonia by oxygen: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). What is the maximum weight of NO obtained from 10.00 g NH₃ and 20.00 g O₂?

Molar masses: $NH_3 = 17.03\text{ g/mol}$, $O_2 = 32.00\text{ g/mol}$, $NO = 30.01\text{ g/mol}$.
Moles of $NH_3 = \frac{10.00}{17.03} = 0.5872\text{ mol}$.
Moles of $O_2 = \frac{20.00}{32.00} = 0.6250\text{ mol}$.
Stoichiometric ratio requires $\frac{5}{4} = 1.25$ moles $O_2$ per mole $NH_3$:
Required $O_2$ for $0.5872\text{ mol } NH_3 = 0.5872 \times 1.25 = \mathbf{0.734\text{ mol}}$.
Since only $0.6250\text{ mol } O_2$ is available, O₂ is the limiting reagent!
From equation: 5 mol $O_2$ produce 4 mol $NO$:
Moles of $NO$ formed $= 0.6250 \times \frac{4}{5} = \mathbf{0.500\text{ mol}}$.
Weight of $NO = 0.500\text{ mol} \times 30.01\text{ g/mol} = \mathbf{15.00\text{ g}}$.

Maximum weight of NO obtained = 15.00 g (O₂ is the limiting reagent).

7.26

Using standard electrode potentials, predict if reaction is feasible: (a) Fe³⁺(aq) and I⁻(aq); (b) Ag⁺(aq) and Cu(s); (c) Fe³⁺(aq) and Cu(s); (d) Ag(s) and Fe³⁺(aq); (e) Br₂(aq) and Fe²⁺(aq).

Reaction is feasible if $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} > 0$:
(a) $2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$: $E^\circ = 0.77 - 0.54 = \mathbf{+0.23\text{ V} > 0} \implies$ Feasible.
(b) $2Ag^+ + Cu \rightarrow 2Ag + Cu^{2+}$: $E^\circ = 0.80 - 0.34 = \mathbf{+0.46\text{ V} > 0} \implies$ Feasible.
(c) $2Fe^{3+} + Cu \rightarrow 2Fe^{2+} + Cu^{2+}$: $E^\circ = 0.77 - 0.34 = \mathbf{+0.43\text{ V} > 0} \implies$ Feasible.
(d) $Ag + Fe^{3+} \rightarrow Ag^+ + Fe^{2+}$: $E^\circ = 0.77 - 0.80 = \mathbf{-0.03\text{ V} < 0} \implies$ Not Feasible.
(e) $Br_2 + 2Fe^{2+} \rightarrow 2Br^- + 2Fe^{3+}$: $E^\circ = 1.09 - 0.77 = \mathbf{+0.32\text{ V} > 0} \implies$ Feasible.

Feasible: (a), (b), (c), (e); Not feasible: (d).

7.27

Predict products of electrolysis: (i) Aqueous AgNO₃ with Ag electrodes; (ii) Aqueous AgNO₃ with Pt electrodes; (iii) Dilute H₂SO₄ with Pt electrodes; (iv) Aqueous CuCl₂ with Pt electrodes.

(i) AgNO₃ with Ag electrodes: Ag anode dissolves ($Ag \rightarrow Ag^+ + e^-$); silver deposits at cathode ($Ag^+ + e^- \rightarrow Ag$).
(ii) AgNO₃ with Pt electrodes: At cathode: $Ag^+ + e^- \rightarrow Ag(s)$ (silver deposits); At anode: water oxidises instead of $NO_3^-$ ($2H_2O \rightarrow O_2\uparrow + 4H^+ + 4e^-$).
(iii) Dilute H₂SO₄ with Pt electrodes: At cathode: $2H^+ + 2e^- \rightarrow H_2\uparrow$; At anode: $2H_2O \rightarrow O_2\uparrow + 4H^+ + 4e^-$ (Electrolysis of water).
(iv) Aqueous CuCl₂ with Pt electrodes: At cathode: $Cu^{2+} + 2e^- \rightarrow Cu(s)$ (copper deposits); At anode: $2Cl^- \rightarrow Cl_2\uparrow + 2e^-$ (chlorine gas evolved due to overvoltage).

(i) Ag at cathode, Ag dissolves at anode; (ii) Ag at cathode, O₂ at anode; (iii) H₂ at cathode, O₂ at anode; (iv) Cu at cathode, Cl₂ at anode.

7.28

Arrange the following metals in the order in which they displace each other from their salt solutions: Al, Cu, Fe, Mg and Zn.

A metal with more negative reduction potential displaces a metal with less negative / positive potential:
$E^\circ: Mg (-2.36\text{ V}) < Al (-1.66\text{ V}) < Zn (-0.76\text{ V}) < Fe (-0.44\text{ V}) < Cu (+0.34\text{ V})$.
Therefore, the displacement ability order is: $$\mathbf{Mg > Al > Zn > Fe > Cu}$$ $Mg$ displaces all; $Cu$ is displaced by all.

Mg > Al > Zn > Fe > Cu.

7.29

Given standard electrode potentials: K⁺/K = −2.93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = −0.74 V. Arrange in increasing order of reducing power.

More negative reduction potential $\implies$ greater tendency to lose electrons $\implies$ stronger reducing power.
Order of $E^\circ$: $Ag^+ (+0.80) > Hg^{2+} (+0.79) > Cr^{3+} (-0.74) > Mg^{2+} (-2.37) > K^+ (-2.93)$.
Increasing reducing power of metals: $$\mathbf{Ag < Hg < Cr < Mg < K}$$

Ag < Hg < Cr < Mg < K.

7.30

Depict the galvanic cell in which Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place. Show: (i) Which electrode is negatively charged? (ii) Carriers of current in the cell. (iii) Individual reaction at each electrode.

Cell Representation: $$\mathbf{Zn(s) \,|\, Zn^{2+}(aq) \,||\, Ag^+(aq) \,|\, Ag(s)}$$ (i) Negatively charged electrode: Zinc electrode (Anode), where oxidation occurs and electrons accumulate.
(ii) Carriers of current:
• In the external metallic circuit: Electrons (flowing from Zn anode to Ag cathode).
• In the internal electrolyte solution: Ions migrating through the salt bridge.
(iii) Individual electrode reactions:
• At Anode (Oxidation): $\mathbf{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}$
• At Cathode (Reduction): $\mathbf{Ag^+(aq) + e^- \rightarrow Ag(s)}$

Cell: Zn|Zn²⁺||Ag⁺|Ag; (i) Zinc anode; (ii) Electrons in wire, ions in solution; (iii) Anode: Zn→Zn²⁺+2e⁻, Cathode: Ag⁺+e⁻→Ag.

Redox Reactions High-Yield Revision Matrix

Quick-reference cheat-sheets summarizing oxidation state assignment, n-factor calculations, electrochemical series trends, balancing algorithms, and key mnemonics for CBSE Board, NEET, and JEE Main.

1. Master Oxidation State & n-Factor Matrix

Substance / Reagent	Reaction Medium	Product Form	Change in Oxidation State	n-Factor (Eq. Wt = M/n)
$KMnO_4$ (Permanganate)	Acidic Medium ($H^+$)	$Mn^{2+}$	$+7 \rightarrow +2$	$\mathbf{n = 5} \implies \text{Eq. Wt} = M/5$
$KMnO_4$ (Permanganate)	Neutral / Faintly Alkaline	$MnO_2$	$+7 \rightarrow +4$	$\mathbf{n = 3} \implies \text{Eq. Wt} = M/3$
$KMnO_4$ (Permanganate)	Strongly Alkaline Medium	$MnO_4^{2-}$ (Manganate)	$+7 \rightarrow +6$	$\mathbf{n = 1} \implies \text{Eq. Wt} = M/1$
$K_2Cr_2O_7$ (Dichromate)	Acidic Medium ($H^+$)	$2Cr^{3+}$	$2 \times (+6 \rightarrow +3)$	$\mathbf{n = 6} \implies \text{Eq. Wt} = M/6$
$FeSO_4$ / $Fe^{2+}$	Any Medium	$Fe^{3+}$	$+2 \rightarrow +3$	$\mathbf{n = 1} \implies \text{Eq. Wt} = M/1$
$H_2C_2O_4$ (Oxalic Acid)	Acidic Medium	$2CO_2$	$2 \times (+3 \rightarrow +4)$	$\mathbf{n = 2} \implies \text{Eq. Wt} = M/2$
$Na_2S_2O_3$ (Iodometry)	Reaction with $I_2$	$S_4O_6^{2-}$ (Tetrathionate)	$2 \times (+2 \rightarrow +2.5)$	$\mathbf{n = 1} \implies \text{Eq. Wt} = M/1$

2. Electrochemical Series Decision Matrix ($E^\circ$ at 298 K)

Standard Potential Range	Oxidising vs Reducing Character	Chemical Behaviour	Archetype Couple
Highly Positive ($E^\circ > +1.5\text{ V}$)	Extreme Oxidising Power	Eagerly gains electrons; oxidises water to $O_2$	$F_2 / F^- (+2.87\text{ V})$ $Co^{3+} / Co^{2+} (+1.81\text{ V})$
Moderately Positive ($0.00 < E^\circ < +1.5\text{ V}$)	Moderate Oxidants	Reduced by active metals; reduced by $H_2$	$Cl_2 / Cl^- (+1.36\text{ V})$ $Ag^+ / Ag (+0.80\text{ V})$ $Cu^{2+} / Cu (+0.34\text{ V})$
Zero ($E^\circ = 0.00\text{ V}$)	IUPAC Reference Datum	Standard Hydrogen Electrode (SHE)	$2H^+ + 2e^- \rightleftharpoons H_2(g)$
Negative ($-1.5\text{ V} < E^\circ < 0.00\text{ V}$)	Active Reducing Metals	Displace $H_2$ gas from dilute mineral acids	$Fe^{2+} / Fe (-0.44\text{ V})$ $Zn^{2+} / Zn (-0.76\text{ V})$
Highly Negative ($E^\circ < -2.0\text{ V}$)	Extreme Reducing Power	Violently displace $H_2$ from cold water; powerful reductants	$Mg^{2+} / Mg (-2.36\text{ V})$ $Na^+ / Na (-2.71\text{ V})$ $Li^+ / Li (-3.05\text{ V})$

Class 11 Chemistry Chapter 7: Redox Reactions Tests

3 Graded Test Levels: Foundation (CBSE Board essentials & definitions), Intermediate (Balancing & cell potentials), and Advanced (NEET / JEE Main competitive multi-concept problems). Select answers to test yourself.

Level 1: Foundation Test (CBSE Board Level)

F1What is the oxidation state of sulphur in H₂SO₄?

+4 is in H₂SO₃.

Correct: 2(+1) + x + 4(−2) = 0 ⇒ x = +6.

Incorrect.

−2 is in sulphides like H₂S.

F2Which element in a compound is ALWAYS assigned an oxidation number of −1?

Hydrogen is usually +1; only −1 in metal hydrides.

Oxygen exhibits −2, −1, −1/2, +1, +2.

Correct: Fluorine is the most electronegative element and has −1 in all its compounds.

Chlorine can show positive oxidation states (+1 to +7).

F3In the Daniell cell, oxidation takes place at the:

Correct: Zn(s) → Zn²⁺ + 2e⁻ occurs at the negatively charged zinc anode (An Ox).

Reduction occurs at the copper cathode.

The salt bridge carries ions, not electrode reactions.

Voltmeter measures potential difference.

Level 2: Intermediate Test (Application & Stoichiometry Level)

I1How many moles of electrons are involved when 1 mole of Cr₂O₇²⁻ is reduced to Cr³⁺ in acidic medium?

3 is the change per chromium atom, but there are 2 Cr atoms.

Correct: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O involves 6 moles of electrons.

5 is for permanganate in acid.

Incorrect.

I2Which of the following is a disproportionation reaction?

Cl is reduced (+5 to −1) and O is oxidised (−2 to 0); two different elements.

Metal displacement.

Correct: Phosphorus (0) is simultaneously reduced to PH₃ (−3) and oxidised to H₂PO₂⁻ (+1).

Combustion redox.

Level 3: Advanced Test (NEET / JEE Main Level)

A1Standard electrode potentials are: Fe³⁺/Fe²⁺ = +0.77 V, I₂/I⁻ = +0.54 V, Cu²⁺/Cu = +0.34 V, Ag⁺/Ag = +0.80 V. Which reaction is NOT feasible under standard conditions?

E° = 0.77 − 0.54 = +0.23 V > 0 (Feasible).

E° = 0.80 − 0.34 = +0.46 V > 0 (Feasible).

E° = 0.77 − 0.34 = +0.43 V > 0 (Feasible).

Correct: E°_cell = E°(Fe³⁺/Fe²⁺) − E°(Ag⁺/Ag) = 0.77 − 0.80 = −0.03 V < 0. Negative cell potential means NOT feasible!

A2In the reaction CrO₅ + H₂SO₄ → Cr₂(SO₄)₃ + H₂O + O₂, the oxidation number of chromium in blue chromium peroxide (CrO₅) is:

Correct: CrO₅ has a butterfly structure with two peroxo linkages (—O—O—) and one oxo (=O) group: x + 4(−1) + 1(−2) = 0 ⇒ x = +6.

Fallacy assuming all oxygens are normal oxides (−2). Chromium has only 6 valence electrons!

+3 is in the product Cr₂(SO₄)₃.

Incorrect.