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Introduction to Oxidation and Reduction

In basic medi...

Question

In basic medium, $C r O_{4}^{2 -}$ oxidize $S_{2} O_{3}^{2 -}$ to form $S O_{4}^{2 -}$ and itself changes to $C r (O H)_{4}^{-} .$ How many mL of 0.154 M $C r O_{4}^{2 -}$ are required to react with 40 mL of 0.246 M $S_{2} O_{3}^{2 -}$ ?

A

200 mL

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B

156.4 mL

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C

170.4 mL

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D

190.4 mL

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Solution

The correct option is C 170.4 mL
$Milliequivalent of S_{2} O_{3}^{2 -} = Milliequivalent of C r O_{4}^{2 -}$
40 $\times$ 0.246 $\times$ 8 = V $\times$ 0.154 $\times$ 3
$∴ V = \frac{40 \times 0.246 \times 8}{0.154 \times 3} = 170.4$ mL
Hence, (c) is the correct answer.

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Introduction to Oxidation and Reduction

Standard XIII Chemistry

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