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Nature of Reactant and Rate of Reaction

It requires 4...

Question

It requires $40.05 m l$ 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+
$Number of equivalents of C e^{4 +} = Number of equivalents of S n^{2 +}$
$40.05 \times 1 \times n-factor = 20 \times 1 \times 2$
$\Rightarrow n-factor of C e^{4 +} = 1$
Therefore oxidation state change will be $C e^{4 +} \to C e^{3 +}$
Hence, the correct answer is option (c).

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