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Valency and Oxidation State

It required ...

Question

It required $40.05$ mL of $1 M C e^{4 +}$ to titrate $20$ mL of $1 M S n^{2 +}$ to $S n^{4 +}$ . What is the oxidation state of the cerium in the product?

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Solution

The correct option is C $+ 3$
$C e^{+ 4} + S n^{+ 2} \to S n^{+ 4} + C e^{+ n}$

$40.05$ $20$ Volume required
$1 M$ $1 M$ Molarity

Meq. of $C e^{+ 4} =$ Meq. of $S n^{+ 2}$

$40.05 \times 1 \times (4 - n) = 20 \times 1 \times 2$

$(4 - n) = \frac{20 \times 2}{40.05} ≃ 1$

$∴ n = 3$

Therefore, the oxidation state of $C e$ in the product is $+ 3$ .

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