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JEE Advance - Chemistry (1986 - No. 4)

Complete and balance the following equation:

$$ClO_3^-$$ + I- + H2SO4 $$\to$$ Cl- + $$HSO_4^-$$
$$I^- + ClO_3^- + H_2SO_4 \to Cl^- + HSO_4^- + I_2 + H_2O$$
$$6I^- + ClO_3^- + 3H_2SO_4 \to Cl^- + 3HSO_4^- + 3I_2 + 3H_2O$$
$$2I^- + ClO_3^- + 2H_2SO_4 \to Cl^- + 2HSO_4^- + I_2 + H_2O$$
$$I^- + ClO_3^- + 2H^+ \to Cl^- + I_2 + H_2O$$
$$6I^- + ClO_3^- + 6H_2SO_4 \to Cl^- + 6HSO_4^- + 3I_2 + 3H_2O$$

Explanation

To complete and balance the given equation, it's important to first understand that this is a redox reaction where chlorate ion ($$ClO_3^-$$) is reduced to chloride ion ($$Cl^-$$), and iodide ion ($$\text{I}^-$$) is oxidized to iodine ($$I_2$$). The presence of sulfuric acid ($$\text{H}_2\text{SO}_4$$) suggests it mainly acts as a source of protons (H+) and the bisulfate ion ($$HSO_4^-$$) is a byproduct of the reaction.

First, let's write down all the half-reactions and balance them:

Oxidation half-reaction (Iodide to Iodine):

$\text{I}^- \to \text{I}_2$

To balance the iodine atoms, we need 2 I- on the left:

$2\text{I}^- \to \text{I}_2$

Reduction half-reaction (Chlorate to Chloride):

$\text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \to \text{Cl}^- + 3\text{H}_2\text{O}$

The iodine half-reaction needs to be balanced in terms of electrons to conclude the overall redox reaction:

$2\text{I}^- \to \text{I}_2 + 2\text{e}^-$

Now, to combine these half-reactions keeping the electron count balanced, we notice that the reduction half-reaction involves 6 electrons, whereas the oxidation half-reaction involves only 2 electrons. Therefore, to balance electrons, we need to multiply the oxidation half-reaction by 3:

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

Now, adding the balanced half-reactions:

$6\text{I}^- + \text{ClO}_3^- + 6\text{H}^+ \to 3\text{I}_2 + \text{Cl}^- + 3\text{H}_2\text{O}$

Since the reaction occurs in acidic medium (indicated by the presence of sulfuric acid), the $6\text{H}^+$ comes from $3\text{H}_2\text{SO}_4$, but because sulfate ions ($SO_4^{2-}$) are not changed in the reduction process, for every $\text{H}_2\text{SO}_4$ that donates two $H^+$, one $HSO_4^-$ is formed:

$6\text{I}^- + \text{ClO}_3^- + 3\text{H}_2\text{SO}_4 \to 3\text{I}_2 + \text{Cl}^- + 3\text{H}_2\text{O} + 3HSO_4^-$

However, we need to ensure the sulfate ions are properly accounted for. Since each $\text{H}_2\text{SO}_4$ provides two hydrogens and we use 3 $\text{H}_2\text{SO}_4$, it's correct as shown. Thus, the complete and balanced equation in acidic solution is:

$6\text{I}^- + \text{ClO}_3^- + 3\text{H}_2\text{SO}_4 \to 3\text{I}_2 + \text{Cl}^- + 3\text{H}_2\text{O} + 3\text{HSO}_4^-$

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