Write balanced chemical equation for the following reactions: (i) Permanganate ion (MnO4^-) reacts with sulphu

NCERT Exemplar · Q21

Q. Write balanced chemical equation for the following reactions:

(i) Permanganate ion (MnO4^-) reacts with sulphur dioxide gas in acidic medium to produce Mn^2+ and hydrogensulphate ion. (Balance by ion electron method)

(ii) Reaction of liquid hydrazine (N2H4) with chlorate ion (ClO3^-) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)

(iii) Dichlorine heptaoxide (Cl2O7) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO2^-) and oxygen gas. (Balance by ion electron method)

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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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Three redox reactions balanced by the ion-electron and oxidation-number methods: (i) permanganate oxidises $\text{SO}_2$ to $\text{HSO}_4^-$ in acid, (ii) hydrazine reduces chlorate to chloride in base, and (iii) dichlorine heptaoxide oxidises peroxide to oxygen in acid.

Redox reactions involve the transfer of electrons between species. Balancing them requires accounting for both mass and charge. Two systematic methods exist: the ion-electron (half-reaction) method splits the reaction into oxidation and reduction half-reactions, balances each separately, then recombines them; the oxidation-number method tracks changes in oxidation states and uses coefficients to equalise electron loss and gain. Both guarantee a balanced equation, but the ion-electron method is often cleaner for ionic species in solution.

(i) Permanganate and sulphur dioxide in acidic medium (Ion-electron method)

The permanganate ion $\text{MnO}_4^-$ is a powerful oxidiser. In acidic solution it is reduced to $\text{Mn}^{2+}$ , while $\text{SO}_2$ (which can be written as $\text{H}_2\text{SO}_3$ in aqueous form) is oxidised to the hydrogensulphate ion $\text{HSO}_4^-$ .

Step 1: Write the skeletal half-reactions.

Reduction:

\text{MnO}_4^- \to \text{Mn}^{2+}

Oxidation:

\text{SO}_2 \to \text{HSO}_4^-

Step 2: Balance atoms other than O and H.

Both Mn and S are already balanced.

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

Reduction:

\text{MnO}_4^- \to \text{Mn}^{2+} + 4\,\text{H}_2\text{O}

Oxidation (the left side has 2 O in $\text{SO}_2$ , the right has 4 O in $\text{HSO}_4^-$ , so add 2 $\text{H}_2\text{O}$ on the left):

\text{SO}_2 + 2\,\text{H}_2\text{O} \to \text{HSO}_4^-

Step 4: Balance hydrogen by adding $\text{H}^+$ (acidic medium).

Reduction (right side has 8 H, so add 8 $\text{H}^+$ on the left):

\text{MnO}_4^- + 8\,\text{H}^+ \to \text{Mn}^{2+} + 4\,\text{H}_2\text{O}

Oxidation (left side has 4 H, right has 1 H, so add 3 $\text{H}^+$ on the right):

\text{SO}_2 + 2\,\text{H}_2\text{O} \to \text{HSO}_4^- + 3\,\text{H}^+

Step 5: Balance charge by adding electrons.

Reduction (left charge: $-1 + 8 = +7$ ; right charge: $+2$ ; add 5 $e^-$ to the left):

\text{MnO}_4^- + 8\,\text{H}^+ + 5\,e^- \to \text{Mn}^{2+} + 4\,\text{H}_2\text{O}

Oxidation (left charge: 0; right charge: $-1 + 3 = +2$ ; add 2 $e^-$ to the right):

\text{SO}_2 + 2\,\text{H}_2\text{O} \to \text{HSO}_4^- + 3\,\text{H}^+ + 2\,e^-

Step 6: Equalise electrons and add the half-reactions.

Multiply the reduction half-reaction by 2 and the oxidation by 5:

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

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

Add and cancel $10\,e^-$ , $10\,\text{H}_2\text{O}$ from the right with part of the left, and $15\,\text{H}^+$ from the right with part of the left:

2\,\text{MnO}_4^- + 5\,\text{SO}_2 + 2\,\text{H}_2\text{O} + \text{H}^+ \to 2\,\text{Mn}^{2+} + 5\,\text{HSO}_4^-

The balanced ionic equation is:

\boxed{2\,\text{MnO}_4^- + 5\,\text{SO}_2 + 2\,\text{H}_2\text{O} + \text{H}^+ \to 2\,\text{Mn}^{2+} + 5\,\text{HSO}_4^-}

In fully molecular form (adding spectator ions):

2\,\text{KMnO}_4 + 5\,\text{SO}_2 + 2\,\text{H}_2\text{O} \to 2\,\text{MnSO}_4 + \text{K}_2\text{SO}_4 + 2\,\text{H}_2\text{SO}_4

(ii) Hydrazine and chlorate in basic medium (Oxidation-number method)

Hydrazine $\text{N}_2\text{H}_4$ is oxidised to nitric oxide $\text{NO}$ , and chlorate $\text{ClO}_3^-$ is reduced to chloride $\text{Cl}^-$ .

Step 1: Assign oxidation numbers.

In $\text{N}_2\text{H}_4$ : each N is $-2$ .

In $\text{NO}$ : N is $+2$ .

Change per N atom: $-2 \to +2$ , an increase of 4. For two N atoms in $\text{N}_2\text{H}_4$ , total increase = $2 \times 4 = 8$ electrons lost.

In $\text{ClO}_3^-$ : Cl is $+5$ .

In $\text{Cl}^-$ : Cl is $-1$ .

Change: $+5 \to -1$ , a decrease of 6 electrons gained per Cl.

Step 2: Equalise electron transfer.

LCM of 8 and 6 is 24.

Multiply $\text{N}_2\text{H}_4$ by 3 (so $3 \times 8 = 24$ electrons lost).

Multiply $\text{ClO}_3^-$ by 4 (so $4 \times 6 = 24$ electrons gained).

Skeletal equation:

3\,\text{N}_2\text{H}_4 + 4\,\text{ClO}_3^- \to 6\,\text{NO} + 4\,\text{Cl}^-

(Note: 3 molecules of $\text{N}_2\text{H}_4$ give $3 \times 2 = 6$ N atoms, hence 6 $\text{NO}$ .)

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

Left: $4 \times 3 = 12$ O in chlorate.

Right: $6$ O in NO.

Deficit on right: $12 - 6 = 6$ O, so add $6\,\text{H}_2\text{O}$ on the right:

3\,\text{N}_2\text{H}_4 + 4\,\text{ClO}_3^- \to 6\,\text{NO} + 4\,\text{Cl}^- + 6\,\text{H}_2\text{O}

Step 4: Balance hydrogen with $\text{OH}^-$ (basic medium).

Left: $3 \times 4 = 12$ H in hydrazine.

Right: $6 \times 2 = 12$ H in water.

Hydrogen is already balanced. However, we must check charge.

Step 5: Balance charge.

Left charge: $4 \times (-1) = -4$ .

Right charge: $4 \times (-1) = -4$ .

Charge is balanced.

The equation is fully balanced:

\boxed{3\,\text{N}_2\text{H}_4(l) + 4\,\text{ClO}_3^-(aq) \to 6\,\text{NO}(g) + 4\,\text{Cl}^-(g) + 6\,\text{H}_2\text{O}(l)}

Watch out

In basic medium, if hydrogen is not balanced after adding water, add $\text{OH}^-$ to the side deficient in H and an equal number of $\text{H}_2\text{O}$ to the other side (or use $\text{H}_2\text{O}$ and $\text{OH}^-$ to balance H and O together). Here, the equation balanced naturally.

(iii) Dichlorine heptaoxide and hydrogen peroxide in acidic medium (Ion-electron method) …

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Q.Write balanced chemical equation for the following reactions:

Concept understanding — Oxidation Reduction

The intuition: what does “oxidation” really mean?

The precise modern definition: electron transfer

How to spot a redox reaction without seeing electrons

One more way to think about it …

Concept: Balancing Redox Equations

(i) Permanganate + sulphur dioxide (acidic, ion–electron method)

(i) Permanganate and sulphur dioxide in acidic medium (Ion-electron method)

(ii) Hydrazine and chlorate in basic medium (Oxidation-number method)

(iii) Dichlorine heptaoxide and hydrogen peroxide in acidic medium (Ion-electron method) …

Q.Write balanced chemical equation for the following reactions:

(i) Permanganate and sulphur dioxide in acidic medium (Ion-electron method)

(ii) Hydrazine and chlorate in basic medium (Oxidation-number method)

(iii) Dichlorine heptaoxide and hydrogen peroxide in acidic medium (Ion-electron method) …

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