Balance the following redox reactions by ion-electron method: (a) MnO4– (aq) + I– (aq) → MnO2 (s) + I2 (s) (in

Exercises · 7.18

Q. 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)

(d) Cr2O7^2– (aq) + SO2

(g) → Cr3+ (aq) + SO4^2– (aq) (in acidic solution)

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Concept understanding — Oxidation Reduction

Let’s start with something you already know from everyday life.

The intuition: what does “oxidation” really mean?

Think of a piece of iron left out in the rain. Over time, it turns into reddish-brown rust. Or think of a slice of apple turning brown when you leave it on the table. Or a fire burning wood to ash. In all these cases, something is combining with oxygen — that’s the original meaning of “oxidation.” The iron combines with oxygen from the air to form iron oxide (rust). The apple’s chemicals react with oxygen in the air. The wood burns because carbon in the wood combines with oxygen.

So the first, simplest idea: oxidation = adding oxygen. And the reverse — taking oxygen away — was called reduction. For example, if you heat iron oxide with carbon, the carbon steals the oxygen away, leaving pure iron. That’s reduction: removing oxygen.

But chemists soon realised this was too narrow. Many reactions that look like oxidation-reduction don’t involve oxygen at all. For instance, when sodium metal reacts with chlorine gas to make table salt, no oxygen is involved — yet the sodium clearly “rusts” in a sense, and the chlorine “steals” something from it.

So the definition had to be broadened.

The precise modern definition: electron transfer

Here’s the clean, exam-ready statement:

Oxidation is the loss of electrons by a substance.

Reduction is the gain of electrons by a substance.

They always happen together — you cannot have one without the other. That’s why we call them redox reactions (short for reduction-oxidation).

Let’s see this with the sodium-chlorine example:

Sodium atom ( $\text{Na}$ ) loses one electron to become $\text{Na}^+$ . That’s oxidation.
Chlorine atom ( $\text{Cl}$ ) gains that electron to become $\text{Cl}^-$ . That’s reduction.

You can write the two halves separately:

\text{Na} \rightarrow \text{Na}^+ + e^- \quad (\text{oxidation})

\text{Cl} + e^- \rightarrow \text{Cl}^- \quad (\text{reduction})

Add them together:

\text{Na} + \text{Cl} \rightarrow \text{Na}^+ + \text{Cl}^-

That’s table salt.

Tip

A handy mnemonic: OIL RIG — Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

How to spot a redox reaction without seeing electrons

You can’t watch electrons move directly. So chemists use oxidation numbers (also called oxidation states) — a bookkeeping system that tracks electrons.

Rules (simplified for first-time learners):

An atom in its elemental form has oxidation number 0.
A monatomic ion has oxidation number equal to its charge (e.g., $\text{Na}^+$ is +1, $\text{Cl}^-$ is -1).
Oxygen is usually -2 (except in peroxides).
Hydrogen is usually +1 (except in metal hydrides).
The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion’s charge.

Then:

Oxidation = increase in oxidation number.
Reduction = decrease in oxidation number.

Example: Rusting of iron.

4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3

Fe starts at 0 (elemental). In $\text{Fe}_2\text{O}_3$ , each Fe is +3. So Fe’s oxidation number goes up from 0 to +3 → oxidation.
O starts at 0 (in $\text{O}_2$ ). In $\text{Fe}_2\text{O}_3$ , each O is -2. So O’s oxidation number goes down from 0 to -2 → reduction.

Watch out

A common mistake: thinking that “reduction” means something becomes smaller or less. It doesn’t — it’s about gaining electrons (or losing oxygen, in the old sense). The name comes from metallurgy: when you “reduce” iron ore to iron, you’re taking away oxygen, so the mass reduces.

One more way to think about it …

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The ion-electron method splits the reaction into separate half-reactions for oxidation and reduction. Each half is balanced for atoms and charge, then combined. For (a) basic medium: $2\text{MnO}_4^- + 3\text{I}^- + 4\text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 3\text{I}_2 + 8\text{OH}^-$ . For (b) acidic: $2\text{MnO}_4^- + 5\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 5\text{HSO}_4^- + \text{H}^+$ . For (c) acidic: $\text{H}_2\text{O}_2 + 2\text{Fe}^{2+} + 2\text{H}^+ \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O}$ . For (d) acidic: $\text{Cr}_2\text{O}_7^{2-} + 3\text{SO}_2 + 2\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-} + \text{H}_2\text{O}$ .

The ion-electron method works because redox reactions are fundamentally two separate processes: one species loses electrons (oxidation) and another gains them (reduction). By treating each half-reaction independently, we can balance atoms and charge stepwise, then recombine so electrons cancel. The medium (acidic or basic) dictates how we handle oxygen and hydrogen — in acid we use $\text{H}_2\text{O}$ and $\text{H}^+$ , in base we use $\text{H}_2\text{O}$ and $\text{OH}^-$ .

Let’s work through each part.

(a) $\text{MnO}_4^- + \text{I}^- \rightarrow \text{MnO}_2 + \text{I}_2$ (basic medium)

1. Identify half-reactions.

Manganese goes from +7 in $\text{MnO}_4^-$ to +4 in $\text{MnO}_2$ — reduction. Iodine goes from -1 in $\text{I}^-$ to 0 in $\text{I}_2$ — oxidation.

2. Write and balance each half for atoms (except H and O).

Reduction: $\text{MnO}_4^- \rightarrow \text{MnO}_2$ (Mn already balanced).

Oxidation: $2\text{I}^- \rightarrow \text{I}_2$ (balance I by coefficient 2).

3. Balance oxygen by adding $\text{H}_2\text{O}$ .

Reduction: left has 4 O, right has 2 O → add $2\text{H}_2\text{O}$ to right: $\text{MnO}_4^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$ .

Oxidation: no oxygen, skip.

4. Balance hydrogen by adding $\text{H}^+$ (then convert to $\text{OH}^-$ for basic medium).

Reduction: right has 4 H from water, left has none → add $4\text{H}^+$ to left: $\text{MnO}_4^- + 4\text{H}^+ \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$ .

Oxidation: no hydrogen.

5. Balance charge by adding electrons.

Reduction: left charge = $-1 + 4(+1) = +3$ , right charge = 0. Add $3e^-$ to left: $\text{MnO}_4^- + 4\text{H}^+ + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$ .

Oxidation: left charge = $2(-1) = -2$ , right charge = 0. Add $2e^-$ to right: $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$ .

6. Equalise electrons.

LCM of 3 and 2 is 6. Multiply reduction by 2, oxidation by 3:

$2\text{MnO}_4^- + 8\text{H}^+ + 6e^- \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O}$

$6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-$

7. Add and cancel electrons.

$2\text{MnO}_4^- + 8\text{H}^+ + 6\text{I}^- \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} + 3\text{I}_2$

8. Convert to basic medium.

Add $8\text{OH}^-$ to both sides to neutralise $\text{H}^+$ :

Left: $8\text{H}^+ + 8\text{OH}^- = 8\text{H}_2\text{O}$ ; right: add $8\text{OH}^-$ .

$2\text{MnO}_4^- + 8\text{H}_2\text{O} + 6\text{I}^- \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} + 3\text{I}_2 + 8\text{OH}^-$

Cancel $4\text{H}_2\text{O}$ from both sides:

$2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6\text{I}^- \rightarrow 2\text{MnO}_2 + 3\text{I}_2 + 8\text{OH}^-$

Watch out

A common mistake is forgetting to convert $\text{H}^+$ to $\text{OH}^-$ in basic medium. Always add the same number of $\text{OH}^-$ to both sides — this turns $\text{H}^+$ into water and leaves $\text{OH}^-$ on the opposite side.

(b) $\text{MnO}_4^- + \text{SO}_2 \rightarrow \text{Mn}^{2+} + \text{HSO}_4^-$ (acidic solution)

1. Half-reactions.

Mn: +7 to +2 — reduction (gain 5 electrons).

S in $\text{SO}_2$ : +4 to +6 in $\text{HSO}_4^-$ — oxidation (loss 2 electrons).

2. Balance atoms (except H, O).

Reduction: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$ (Mn balanced).

Oxidation: $\text{SO}_2 \rightarrow \text{HSO}_4^-$ (S balanced).

3. Balance oxygen with $\text{H}_2\text{O}$ .

Reduction: left 4 O, right 0 → add $4\text{H}_2\text{O}$ to right: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .

Oxidation: left 2 O, right 4 O → add $2\text{H}_2\text{O}$ to left: $\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^-$ .

4. Balance hydrogen with $\text{H}^+$ .

Reduction: right has 8 H, left none → add $8\text{H}^+$ to left: $\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .

Oxidation: left has 4 H (from $2\text{H}_2\text{O}$ ), right has 1 H → add $3\text{H}^+$ to right: $\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+$ .

5. Balance charge with electrons.

Reduction: left charge = $-1 + 8(+1) = +7$ , right = $+2$ . Add $5e^-$ to left: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .

Oxidation: left charge = 0, right charge = $-1 + 3(+1) = +2$ . Add $2e^-$ to right: $\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+ + 2e^-$ .

6. Equalise electrons (LCM of 5 and 2 = 10).

Multiply reduction by 2, oxidation by 5:

$2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

$5\text{SO}_2 + 10\text{H}_2\text{O} \rightarrow 5\text{HSO}_4^- + 15\text{H}^+ + 10e^-$

7. Add and cancel.

$2\text{MnO}_4^- + 16\text{H}^+ + 5\text{SO}_2 + 10\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{HSO}_4^- + 15\text{H}^+$

Cancel $10\text{H}_2\text{O}$ on left with $8\text{H}_2\text{O}$ on right → net $2\text{H}_2\text{O}$ on left. Cancel $16\text{H}^+$ left with $15\text{H}^+$ right → net $1\text{H}^+$ on left.

Final: $2\text{MnO}_4^- + \text{H}^+ + 5\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 5\text{HSO}_4^-$

Tip

In acidic medium, always check that the final equation has no leftover $\text{H}^+$ or $\text{H}_2\text{O}$ that can be cancelled. The net $\text{H}^+$ coefficient may be 1, as here — that’s fine.

(c) $\text{H}_2\text{O}_2 + \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{H}_2\text{O}$ (acidic solution)

1. Half-reactions.

O in $\text{H}_2\text{O}_2$ : -1 to -2 in $\text{H}_2\text{O}$ — reduction (gain 1 electron per O, but careful: each $\text{H}_2\text{O}_2$ has 2 O, so gain 2 electrons total).

Fe: +2 to +3 — oxidation (loss 1 electron).

2. Balance atoms.

Reduction: $\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O}$ (O balanced, H: left 2, right 4 → add $2\text{H}^+$ to left).

So: $\text{H}_2\text{O}_2 + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O}$ .

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$ (balanced).

3. Balance charge with electrons.

Reduction: left charge = $0 + 2(+1) = +2$ , right = 0. Add $2e^-$ to left: $\text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow 2\text{H}_2\text{O}$ .

Oxidation: left charge = +2, right = +3. Add $1e^-$ to right: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ .

4. Equalise electrons (LCM = 2).

Multiply oxidation by 2: $2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2e^-$ .

5. Add.

$\text{H}_2\text{O}_2 + 2\text{H}^+ + 2\text{Fe}^{2+} \rightarrow 2\text{H}_2\text{O} + 2\text{Fe}^{3+}$ …

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Q.Balance the following redox reactions by ion-electron method:

(a) MnO4−+I−→MnO2+I2\text{MnO}_4^- + \text{I}^- \rightarrow \text{MnO}_2 + \text{I}_2MnO4−​+I−→MnO2​+I2​ (basic medium)

(b) MnO4−+SO2→Mn2++HSO4−\text{MnO}_4^- + \text{SO}_2 \rightarrow \text{Mn}^{2+} + \text{HSO}_4^-MnO4−​+SO2​→Mn2++HSO4−​ (acidic solution)

(c) H2O2+Fe2+→Fe3++H2O\text{H}_2\text{O}_2 + \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{H}_2\text{O}H2​O2​+Fe2+→Fe3++H2​O (acidic solution)

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(a) $\text{MnO}_4^- + \text{I}^- \rightarrow \text{MnO}_2 + \text{I}_2$ (basic medium)

(b) $\text{MnO}_4^- + \text{SO}_2 \rightarrow \text{Mn}^{2+} + \text{HSO}_4^-$ (acidic solution)

(c) $\text{H}_2\text{O}_2 + \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{H}_2\text{O}$ (acidic solution)

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