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Introduction to Oxidation and Reduction

12.53 mL of 0...

Question

12.53 mL of 0.0509 M $S e O_{2}$ reacted with 25.52 mL 0.1 M $C r S O_{4}$ solution. In the reaction $C r^{2 +}$ was oxidized to $C r^{3 +}$ . To what oxidation state selenium was converted in the reaction? Write the redox change for $S e O_{2}$ .

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Solution

$S e^{4 +} + (a - 4) e \to S e^{[4 - (a - 4)]}$
$n = a - 4$ (where $n$ is valence factor)

or $C r^{2 +} \to C r^{3 +} + e$

$M e q . o f S e O_{2} = M e q . o f C r S O_{4}$
$0.0509 \times n \times 12.53 = 0.1 \times 1 \times 25.52$

$n = 4$
$a - 4 = 4$
or $a = 8$
Thus redox change is: $S e^{4 +} + 4 e \to S e$

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