Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by

Exercises · 7.23

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

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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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Chlorine oxidises sulphur dioxide to sulphuric acid while itself being reduced to chloride ions — the balanced equation is $\ce{Cl2 + SO2 + 2H2O -> H2SO4 + 2HCl}$ .

This is a classic redox reaction that happens in aqueous solution. The key idea is that chlorine is a strong oxidising agent, and sulphur dioxide is a reducing agent. When they meet in water, chlorine pulls electrons from sulphur dioxide, turning itself into harmless chloride ions and converting the sulphur dioxide into sulphuric acid.

Let’s break this down step by step.

Identify the oxidation and reduction half-reactions.

Chlorine ( $\ce{Cl2}$ ) is being reduced — each chlorine atom gains one electron to become chloride ion ( $\ce{Cl-}$ ).

Sulphur dioxide ( $\ce{SO2}$ ) is being oxidised — the sulphur atom goes from the +4 oxidation state (in $\ce{SO2}$ ) to the +6 state (in $\ce{SO4^{2-}}$ or $\ce{H2SO4}$ ).
Write the half-reactions in acidic medium (since water is present).

Reduction: $\ce{Cl2 + 2e- -> 2Cl-}$

Oxidation: $\ce{SO2 + 2H2O -> SO4^{2-} + 4H+ + 2e-}$

Notice that the oxidation half-reaction already balances electrons — both involve 2 electrons.
Combine the half-reactions.

Add them directly because the electron count matches:

$\ce{Cl2 + SO2 + 2H2O -> 2Cl- + SO4^{2-} + 4H+}$
Write the products in molecular form.

In water, the $\ce{H+}$ and $\ce{SO4^{2-}}$ combine to form sulphuric acid ( $\ce{H2SO4}$ ), and the $\ce{Cl-}$ ions pair with $\ce{H+}$ to form hydrochloric acid ( $\ce{HCl}$ ). …

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

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