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Valency and Oxidation State

For a titrati...

Question

For a titration of $100 {c m}^{3}$ of $0.1$ $M$ ${S n}^{2 +}$ to ${S n}^{4 +}$ , $50$ ${c m}^{3}$ of $0.40$ $M$ ${C e}^{4 +}$ solution was required. The oxidation state of cerium in the reduction product is :

+ 1

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+ 2

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+ 3

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0

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Solution

The correct option is C $+ 3$
According to given conditions, we have
${S n}^{2 +} ⟶ {S n}^{4 +} + 2 e^{-}; i n c r e a s e i n O . N = 2$
${C e}^{4 +} + n e^{-} ⟶ {C e}^{(4 - n)}; d e c r e a s e i n O . N = n$
On balancing, we get
$n {S n}^{2 +} + 2 {C e}^{4 +} ⟶ n {S n}^{4 +} + 2 {C e}^{(4 - n)}$
Given,
Millimoles of ${S n}^{2 +} = 100 \times 0.1 = 10$
Millimoles of ${C e}^{4 +} = 50 \times 0.4 = 20$
At equivalent point, number of equivalents of both reactant and product are same.
Therefore, $\frac{n}{2} = \frac{10}{20}$
or, $n = 1$
Thus, oxidation state of cerium in the reduced product $({C e}^{(4 - 1)} = {C e}^{3 +}) = + 3$ .

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