The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a

Exercises · 7.21

Q. 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.

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Concept understanding — Redox Reaction Stoichiometry

Redox Reaction Stoichiometry – From Intuition to Precision

Imagine you're balancing a seesaw. On one side, electrons are being lost; on the other, they're being gained. The seesaw stays level only when the number of electrons lost equals the number gained. That's the core idea behind redox stoichiometry.

The Intuition: Electrons Are the Currency

In any redox reaction, two things happen simultaneously:

Oxidation: a substance loses electrons (its oxidation state increases).
Reduction: a substance gains electrons (its oxidation state decreases).

Think of electrons as money. If one person gives away ₹10, another must receive exactly ₹10. You can't have ₹5 floating in the air. Similarly, the total number of electrons lost in oxidation must equal the total number of electrons gained in reduction.

This simple equality is what makes redox stoichiometry work. It's not about balancing atoms first — it's about balancing electrons first.

The Precise Statement

\text{Total electrons lost (by reducing agent)} = \text{Total electrons gained (by oxidising agent)}

This equality is the foundation of the half-reaction method (also called the ion-electron method) for balancing redox equations.

How It Works in Practice

Let's walk through a classic example: the reaction between permanganate ions ( $\text{MnO}_4^-$ ) and iron(II) ions ( $\text{Fe}^{2+}$ ) in acidic medium.

Step 1: Write the two half-reactions (unbalanced).

Oxidation half: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

Reduction half: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Notice: iron loses 1 electron per atom, while permanganate gains 5 electrons per ion.

Step 2: Balance electrons between the halves.

To make electrons lost = electrons gained, multiply the oxidation half by 5:

$5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-$

Now both halves involve 5 electrons.

Step 3: Add the halves and cancel common terms.

$5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}$

The electrons cancel because they appear on opposite sides. The equation is now balanced in both atoms and charge.

Tip

Always check that the net charge on both sides is equal after balancing. In the example above: left side charge = $5(+2) + (-1) + 8(+1) = +17$ ; right side = $5(+3) + (+2) + 0 = +17$ . Matches perfectly.

Why This Matters for Exams

In Indian board exams (CBSE, ICSE, state boards), redox stoichiometry appears in two main forms:

Balancing equations using the half-reaction method (acidic or basic medium).
Titration calculations where you use the electron equality to find unknown concentrations.

For titrations, the key formula is:

\frac{n_1}{n_2} = \frac{M_1 V_1}{M_2 V_2}

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The key idea is that Mn³⁺ disproportionates — one Mn³⁺ is oxidised to MnO₂ and another is reduced to Mn²⁺ — and the balanced ionic equation is $2\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4\text{H}^+$ .

Disproportionation is a special kind of redox reaction where a single species (here, Mn³⁺) acts as both the oxidising agent and the reducing agent. One part of it gets oxidised (loses electrons, oxidation number increases) and another part gets reduced (gains electrons, oxidation number decreases). The trick is to figure out the two products and then balance atoms and charge.

Let’s work through it.

Identify the oxidation states. In Mn³⁺, manganese is in the +3 oxidation state. In Mn²⁺, it’s +2 — that’s a decrease of 1 electron per ion (reduction). In MnO₂, oxygen is −2 each, so Mn must be +4 — that’s an increase of 1 electron per ion (oxidation). So the disproportionation is:

\text{Mn}^{3+} \rightarrow \text{Mn}^{2+} \quad (\text{reduction, gain of 1 e}^-)

\text{Mn}^{3+} \rightarrow \text{MnO}_2 \quad (\text{oxidation, loss of 1 e}^-)

Balance the electron transfer.

Each Mn³⁺ that becomes Mn²⁺ gains 1 electron. Each Mn³⁺ that becomes MnO₂ loses 1 electron. So the electrons already cancel if we take one of each — but we also need to balance atoms, especially oxygen. That’s where water and H⁺ come in.
Write the half-reactions in acidic medium.

Reduction half:

\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}

Oxidation half: Mn³⁺ to MnO₂. Start with:

\text{Mn}^{3+} \rightarrow \text{MnO}_2

Balance oxygen by adding water:

\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2

Balance hydrogen by adding H⁺:

\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+

Now balance charge: left side has +3, right side has +4 (from 4H⁺). So add 1 electron to the right:

\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+ + e^-

Combine the half-reactions. The reduction half gives 1 electron, the oxidation half gives 1 electron — they cancel directly. Add them:

(\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}) + (\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+ + e^-)

Cancel the electrons: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$\mathrm{X} \mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{YSO}_2(\mathrm{~g}) \rightarrow \mathrm{Mn}^{+2}(\mathrm{aq})+\mathrm{HSO}_4^{-}(\mathrm{aq})$
›Reveal solutionSolution
The key is balancing the half‑reactions in acidic medium: permanganate gains 5 electrons, sulfur dioxide loses 2 electrons. The stoichiometric ratio X:Y (MnO₄⁻ : SO₂) is therefore 2:5, which corresponds to option (C).

Concept and intuition

In any redox reaction, the total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidizing agent. Here, MnO₄⁻ is reduced to Mn²⁺ (Mn changes oxidation state from +7 to +2, a gain of 5 electrons), and SO₂ is oxidized to HSO₄⁻ (S changes from +4 to +6, a loss of 2 electrons). To make electrons equal, we need the smallest whole‑number ratio that balances the electron transfer: 2 MnO₄⁻ (gaining 10 e⁻) for every 5 SO₂ (losing 10 e⁻). That gives X:Y = 2:5.

Step‑by‑step reasoning

Assign oxidation numbers

In MnO₄⁻: O is –2 (×4 = –8), overall charge –1 ⇒ Mn = +7.

In Mn²⁺: Mn = +2.

In SO₂: O is –2 (×2 = –4), molecule neutral ⇒ S = +4.

In HSO₄⁻: O is –2 (×4 = –8), H is +1, overall charge –1 ⇒ S = +6.

Write the half‑reactions

Reduction: MnO₄⁻ → Mn²⁺ Change: Mn from +7 to +2 ⇒ gain of 5 e⁻. In acidic medium: balance O with H₂O, then H⁺:

$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$

Oxidation: SO₂ → HSO₄⁻ Change: S from +4 to +6 ⇒ loss of 2 e⁻. Balance O with H₂O, then H⁺:

$\mathrm{SO_2 + 2H_2O \rightarrow HSO_4^- + 3H^+ + 2e^-}$

Equalize electrons transferred

Reduction half‑reaction involves 5 e⁻; oxidation involves 2 e⁻.

LCM of 5 and 2 is 10. Multiply reduction by 2, oxidation by 5:

$2\mathrm{MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O}$ $5\mathrm{SO_2 + 10H_2O \rightarrow 5HSO_4^- + 15H^+ + 10e^-}$

Add the half‑reactions and simplify

Combine: $2\mathrm{MnO_4^- + 16H^+ + 5SO_2 + 10H_2O \rightarrow 2Mn^{2+} + 8H_2O + 5HSO_4^- + 15H^+} …$
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COMEDK 2026Set 2026-A1 markMCQ
$x$
›Reveal solutionSolution
The key is to balance the redox reaction by equating the total electrons lost by ferrous oxalate (Fe²⁺ and C₂O₄²⁻) to the total electrons gained by dichromate. Solving gives $x = 0.5$ , so the correct option is (D).

We start by recognizing that ferrous oxalate is a mixed reducing agent: it contains both Fe²⁺ (which oxidises to Fe³⁺) and oxalate ion C₂O₄²⁻ (which oxidises to CO₂). In acidic medium, K₂Cr₂O₇ is a strong oxidising agent, with Cr⁶⁺ being reduced to Cr³⁺. The stoichiometry is found by equating the total number of electrons transferred.

Write the half-reactions for the reducing agent (ferrous oxalate)

Fe²⁺ → Fe³⁺ + e⁻ (1 electron lost per Fe²⁺)

C₂O₄²⁻ → 2 CO₂ + 2 e⁻ (2 electrons lost per oxalate ion) Since one mole of ferrous oxalate (FeC₂O₄) contains 1 mole of Fe²⁺ and 1 mole of C₂O₄²⁻, the total electrons lost by 1 mole of ferrous oxalate = $1 + 2 = 3$ moles of electrons.

Write the half-reaction for the oxidising agent (dichromate in acid)

$\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$
This shows that 1 mole of K₂Cr₂O₇ accepts 6 moles of electrons.

Set up the electron balance Let $x$ be the moles of K₂Cr₂O₇ required to oxidise 1 mole of ferrous oxalate.

Electrons gained by $x$ moles of dichromate = $x \times 6$

Electrons lost by 1 mole of ferrous oxalate = $3$ For a complete redox reaction: $6x = 3 \quad \Rightarrow \quad x = \frac{3}{6} = 0.5$ …
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KCET 2026Set D31 markMCQ
$C_2O_4^{2-}$
›Reveal solutionSolution
Balance the two half-reactions (oxalate → $CO_2$ losing 2 electrons; permanganate → $Mn^{2+}$ gaining 5 electrons) and combine them so total electrons lost equal total electrons gained.

Step 1 — Oxidation half-reaction

$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$ : carbon goes from +3 to +4 (two carbons), so each oxalate ion loses 2 electrons.

Step 2 — Reduction half-reaction

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ : manganese goes from +7 to +2, so each permanganate ion gains 5 electrons.

Step 3 — Equalise electrons

To balance electrons, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2, so that $5 \times 2 = 10$ electrons lost equal $2 \times 5 = 10$ electrons gained:

$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$

$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$

Step 4 — Combine and read off coefficients …
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COMEDK 2025Set 2025-E1 markMCQ
$Q. Identify the correct coefficients (a), (b),$
(c) and
(d) in the following equations i) $\quad \mathrm{xMnO}_4^{--}+$ (a) $\mathrm{SO}_3{ }^{2-}+$
(b) $\mathrm{H}^{+} \cdots \longrightarrow \mathrm{xMn}^{2+}+$
(a) $\mathrm{SO}_4{ }^{2-}+$
(c) $\mathrm{H}_2 \mathrm{O}$ ii)
(c) $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}+$
(d) $\mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O} \cdots \rightarrow$
(d) $\mathrm{MnO}_2+$
(b) $\mathrm{SO}_4{ }^{2-}+\mathrm{xOH}^{-}$ (A) $(a)=3$
(b) $=8$ $(c)=6$ $(d)=6$ (B) $(a)=5$
(b) $=6$ $(c)=3$ $(\mathrm{d})=8$ (C)
(a) $=2$
(b) $=8$ $(c)=4$ $(\mathrm{d})=4$ (D) $(a)=2$
(b) $=6$ $(c)=6$ $(d)=5$
›Reveal solutionSolution
The key is to balance each redox half‑reaction separately (acidic for the first, basic for the second) and then combine them. The correct coefficients are (a)=5,

(b)=6,

(c)=3,

(d)=8, which corresponds to option (B).

We have two separate redox equations to balance. The first involves permanganate with sulfite in acidic medium; the second involves thiosulfate with permanganate in basic medium. The letters (a),

(b),

(c),

(d) are placeholders for the coefficients we must find. The trick is that the same letter appears in both equations, so the coefficients must be consistent across both.

Concept & Intuition

Redox balancing works by splitting the reaction into oxidation and reduction halves. In acidic solution we add H⁺ and H₂O; in basic solution we add OH⁻ and H₂O. Once each half is balanced for atoms and charge, we multiply them so electrons cancel. The final coefficients give us the values of (a),

(b),

(c),

(d).

Step‑by‑step balancing

Equation (i):
$x\,\mathrm{MnO}_4^- + (a)\,\mathrm{SO}_3^{2-} + (b)\,\mathrm{H}^+ \longrightarrow x\,\mathrm{Mn}^{2+} + (a)\,\mathrm{SO}_4^{2-} + (c)\,\mathrm{H}_2\mathrm{O}$

Identify half‑reactions

Reduction: $\mathrm{MnO}_4^- \to \mathrm{Mn}^{2+}$ (Mn goes from +7 to +2, gain of 5 e⁻)

Oxidation: $\mathrm{SO}_3^{2-} \to \mathrm{SO}_4^{2-}$ (S goes from +4 to +6, loss of 2 e⁻)

Balance the reduction half (acidic)

$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$
(Check: atoms and charge balanced.)

Balance the oxidation half (acidic)

$\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \to \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ + 2e^-$
(Check: atoms and charge balanced.)

Equalize electrons Reduction gives 5 e⁻, oxidation gives 2 e⁻. LCM = 10. Multiply reduction by 2:

$2\mathrm{MnO}_4^- + 16\mathrm{H}^+ + 10e^- \to 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}$
Multiply oxidation by 5:
$5\mathrm{SO}_3^{2-} + 5\mathrm{H}_2\mathrm{O} \to 5\mathrm{SO}_4^{2-} + 10\mathrm{H}^+ + 10e^-$

Add and cancel Add the two half‑reactions:

$2\mathrm{MnO}_4^- + 5\mathrm{SO}_3^{2-} + (16\mathrm{H}^+ + 5\mathrm{H}_2\mathrm{O}) + 10e^- \to 2\mathrm{Mn}^{2+} + 5\mathrm{SO}_4^{2-} + (8\mathrm{H}_2\mathrm{O} + 10\mathrm{H}^+) + 10e^-$
Cancel 10 e⁻, cancel 10 H⁺ from both sides, and cancel 5 H₂O from left with 5 of the 8 H₂O on right:
$2\mathrm{MnO}_4^- + 5\mathrm{SO}_3^{2-} + 6\mathrm{H}^+ \to 2\mathrm{Mn}^{2+} + 5\mathrm{SO}_4^{2-} + 3\mathrm{H}_2\mathrm{O}$
So from equation (i): $x = 2$ , $(a) = 5$ , $(b) = 6$ , $(c) = 3$ .

Equation (ii):
$(c)\,\mathrm{S}_2\mathrm{O}_3^{2-} + (d)\,\mathrm{MnO}_4^- + \mathrm{H}_2\mathrm{O} \longrightarrow (d)\,\mathrm{MnO}_2 + (b)\,\mathrm{SO}_4^{2-} + x\,\mathrm{OH}^-$
We already know from (i): $b = 6$ , $c = 3$ . So we need to find $d$ and check consistency.

Identify half‑reactions

Reduction: $\mathrm{MnO}_4^- \to \mathrm{MnO}_2$ (Mn from +7 to +4, gain of 3 e⁻)

Oxidation: $\mathrm{S}_2\mathrm{O}_3^{2-} \to \mathrm{SO}_4^{2-}$ (S in thiosulfate: average oxidation state +2; in sulfate +6; each S loses 4 e⁻, but there are 2 S atoms per thiosulfate, so total loss = 8 e⁻ per $\mathrm{S}_2\mathrm{O}_3^{2-}$ )

Balance the reduction half (basic)

In basic medium:

$\mathrm{MnO}_4^- + 2\mathrm{H}_2\mathrm{O} + 3e^- \to \mathrm{MnO}_2 + 4\mathrm{OH}^-$
(Check: atoms and charge balanced.)

Balance the oxidation half (basic) Start with $\mathrm{S}_2\mathrm{O}_3^{2-} \to 2\mathrm{SO}_4^{2-}$ . Balance S: already 2 on each side. Balance O: left 3 O, right 8 O → add 5 H₂O to left:

$\mathrm{S}_2\mathrm{O}_3^{2-} + 5\mathrm{H}_2\mathrm{O} \to 2\mathrm{SO}_4^{2-} + 10\mathrm{H}^+$
But we are in basic medium, so add 10 OH⁻ to both sides to neutralise H⁺:
$\mathrm{S}_2\mathrm{O}_3^{2-} + 5\mathrm{H}_2\mathrm{O} + 10\mathrm{OH}^- \to 2\mathrm{SO}_4^{2-} + 10\mathrm{H}_2\mathrm{O}$
Cancel 5 H₂O from both sides:
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{H}^{+}$
›Reveal solutionSolution
The key is to balance the half-reaction for the oxidation of sulphite to sulphate, which shows that 2 moles of electrons are transferred per mole of sulphite. For 3.0 moles of the oxidised product (sulphate), the total moles of electrons is 6. The correct option is (D).

Concept and Intuition

In any redox reaction, the number of electrons transferred is directly tied to the change in oxidation state of the species being oxidised or reduced. Here, the question asks for the number of moles of electrons involved in producing 3.0 moles of the oxidised product. That means we focus on the oxidation half-reaction: sulphite ion ( $\mathrm{SO}_3^{2-}$ ) is oxidised to sulphate ion ( $\mathrm{SO}_4^{2-}$ ). The dichromate ( $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ ) is the oxidising agent, but we don’t need its full balancing — we only need the electron count per mole of product.

Step-by-step reasoning

Identify the oxidation half-reaction Sulphite ion ( $\mathrm{SO}_3^{2-}$ ) is oxidised to sulphate ion ( $\mathrm{SO}_4^{2-}$ ). In acidic medium, the half-reaction is:

$\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ + 2e^-$
This shows that each mole of sulphite that becomes sulphate releases 2 moles of electrons.

Confirm the oxidation state change

In $\mathrm{SO}_3^{2-}$ , sulfur has oxidation state +4 (since oxygen is -2, total -6, so S = +4).

In $\mathrm{SO}_4^{2-}$ , sulfur has oxidation state +6.

The change is from +4 to +6, which is a loss of 2 electrons per sulfur atom. This matches the half-reaction.

Relate to the question

The “oxidised product” is sulphate ( $\mathrm{SO}_4^{2-}$ ). We are producing 3.0 moles of it.

Since each mole of sulphate formed comes from one mole of sulphite and releases 2 moles of electrons, the total moles of electrons is:
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{w}, \mathrm{x}, \mathrm{y}$
›Reveal solutionSolution
Balancing the redox: the chlorate Cl( $+5$ ) gains 5 electrons while five $\mathrm{Cl^-}$ each lose one, giving $\mathrm{ClO_3^-}+5\mathrm{Cl^-}+6\mathrm{H^+}\to 3\mathrm{H_2O}+3\mathrm{Cl_2}$ , i.e. $w=5,\ x=6,\ y=3,\ z=3$ .

Oxidation states: In $\mathrm{ClO_3^-}$ , Cl is $+5$ ; it is reduced to $0$ in $\mathrm{Cl_2}$ — gains $5\,e^-$ . Each $\mathrm{Cl^-}$ ( $-1$ ) is oxidised to $0$ — loses $1\,e^-$ .

Electron balance: to supply the 5 electrons gained by one chlorate, we need 5 $\mathrm{Cl^-}$ , so $w=5$ .

Atom balance:

Cl: LHS $=1+5=6$ ; RHS $=2z \Rightarrow z=3$ . …
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COMEDK 2021Set 20211 markMCQ
$_4$
›Reveal solutionSolution

For 1 mole of KMnO4: moles of H2O2 = 5/2.

Concept: Redox stoichiometry - H2O2 acts as a REDUCING agent towards acidified KMnO4, decolourising the purple MnO4- to colourless Mn2+.

Balanced equation (acidic medium):

2 MnO4- + 5 H2O2 + 6 H+ -> 2 Mn2+ + 5 O2 + 8 H2O

Equivalently by n-factors: MnO4- (Mn +7 -> +2) gains 5 electrons; H2O2 (O -1 -> 0) loses 2 electrons. For electron balance: …
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Q.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.

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