Concept:In a redox reaction,
oxidation corresponds to an increase in oxidation number (loss of electrons), whereas
reduction corresponds to a decrease in oxidation number (gain of electrons).
Formula / Reaction Analysis:$$10\text{Cl}^- + 16\text{H}^+ + 2\text{MnO}_4^- \longrightarrow 5\text{Cl}_2 + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$$
Solution:- Manganese: In the permanganate ion (\(\text{MnO}_4^-\)), let the oxidation number of \(\text{Mn}\) be \(x\):
$$x + 4(-2) = -1 \implies x = +7$$
In the product \(\text{Mn}^{2+}\), the oxidation number is \(+2\).
Since the oxidation state decreases from \(+7 \to +2\), Manganese is reduced (it gains 5 electrons).
- Chlorine: Reactant is \(\text{Cl}^-\) (oxidation state \(-1\)) and product is \(\text{Cl}_2\) (oxidation state \(0\)).
Since the oxidation state increases from \(-1 \to 0\), Chloride ions are oxidized (loss of electrons).
Why other options are incorrect:- Opt_A: Manganese undergoes a decrease in oxidation number (\(+7 \to +2\)), which is reduction, not oxidation.
- Opt_B: Chlorine in the reactants is present as \(\text{Cl}^-\) (state \(-1\)), not zero. The forward reaction oxidizes \(-1 \to 0\).
- Opt_C: Going from \(-1 \to 0\) is an algebraic increase in oxidation number (loss of \(e^-\)), which defines oxidation, not reduction.
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