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Catalytic Transfer Hydrogenation

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Question

It requires $40 mL$ of $1 M C e^{4 +}$ to titrate $20 mL$ of $1 M$ $S n^{2 +} to S n^{4 +}$ . What is the magnitude of the oxidation state of cerium in the product?

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Solution

$Mili equivalents of C e^{4 +} = Mili equivalents of S n^{2 +}$
$\Rightarrow 1 \times Valency factor \times 40 = 1 \times 2 \times 20$
$So, valency factor = 1$
$C e^{4 +} is reduced to C e^{3 +}$

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