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Valency and Oxidation State

It requires ...

Question

It requires $40 m L$ of $0.5 M C e^{4 +}$ to titrate $10 m L$ of $1 M S n^{4 +}$ . What is the oxidation state of the cerium in the product?

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Solution

The reduced oxidation state of cerium in the reduced product is $C e^{3 +}$
1) Law of equivalents
$n$ - factor of $s n^{2 +} = 4 - 2 = 2$
$x$ - Change in oxidation state of $C e^{4 +}$
No. of equivalents of $C e^{4 +} =$ No of equivalents of $S n^{2 +}$
$40 \times 0.5 \times x = 10 \times 2 \times 1$
$x = 1$
Therefore oxidation state of $C e^{4 +}$ is rediced by 1 unit i.e. $C e^{3 +}$

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