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Introduction to Oxidation and Reduction

12.5 ml of 0....

Question

12.5 ml of 0.05 M $S e O_{2}$ reacts with 25 ml of 0.1 M $C r S O_{4}$ which is oxidised to $C r^{3 +}$ . To what oxidation state was the selenium converted by the reaction?

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Solution

Here, $C r S O_{4}$ is oxidised to $C r^{3 +}$ , so n-factor is 1.
In $S e O_{2}$ Oxidation state of $S e$ is +4 lets assume it is reduced to +x oxidation state.
So, n-factor of $S e$ = $4 - x$
equating the meqv. we get
$(4 - x) \times 12.5 \times 0.05$ = $25 \times 0.1$
x=0

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