500 ml, 1.2M KI is completely react with 0.2M, 500 ml $\text{KMnO}_4$ solution in basic medium. $\text{I}^-$ is oxidised to $\text{I}_2$. The liberated $\text{I}_2$ react with 0.1 M $\text{Na}_2\text{S}_2\text{O}_3$ solution. Then find volume (in L) of $\text{Na}_2\text{S}_2\text{O}_3$ solution required to completely react with liberated $\text{I}_2$.
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For redox chains: Equivalents of A = Equivalents of B = Equivalents of C. Be careful with n-factors; for $\text{KMnO}_4$, Acidic=5, Neutral/Weakly Basic=3, Strongly Basic=1. Product $\text{I}_2$ usually implies condition is not strongly alkaline (where iodate forms), fitting n=3.
Step 1: Reaction of $\text{KMnO}_4$ with $\text{KI}$.
In basic medium, $\text{KMnO}_4$ is generally reduced to $\text{MnO}_2$. Change in oxidation state of Mn: $+7 \to +4$.
n-factor for $\text{KMnO}_4 = 3$.
Moles of $\text{KMnO}_4 = M \times V = 0.2 \times 0.5 = 0.1 \text{ mol}$.
Equivalents of $\text{KMnO}_4 = 0.1 \times 3 = 0.3 \text{ Eq}$.
The problem states KI "completely reacts", implying KI is in excess or sufficient. However, usually the oxidant determines the yield of $\text{I}_2$.
According to the principle of equivalence: Equivalents of Oxidant = Equivalents of Reductant ($\text{I}^-$) oxidized = Equivalents of $\text{I}_2$ formed.
So, Equivalents of $\text{I}_2$ liberated = 0.3 Eq.
Step 2: Titration of $\text{I}_2$ with $\text{Na}_2\text{S}_2\text{O}_3$.
Reaction: $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \to 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$.
n-factor for Hypo ($\text{S}_2\text{O}_3^{2-}$) is 1 (avg oxidation state change $+2 \to +2.5$, change per molecule is 1 electron).
Equivalents of $\text{Na}_2\text{S}_2\text{O}_3$ required = Equivalents of $\text{I}_2$.
$M_{\text{hypo}} \times V_{\text{hypo}} \times n_{\text{hypo}} = 0.3$.
$0.1 \times V \times 1 = 0.3$.
$V = 3 \text{ L}$.
