Single Choice

During electrolysis of $$H_2SO_4 (aq)$$ with high charge density, $$H_2S_2O_8$$ formed as by product. In such electrolysis $$22.4\,L \, H_2 (g)$$ and $$8.4\,L \, O_2(g) $$ liberated at $$1$$ atm and $$273\,K$$ at electrode. The moles of $$H_2S_2O_8$$ formed is:

A$$0.25$$
Correct Answer
B$$0.50$$
C$$0.75$$
D$$1.00$$

Solution

$$x_{H_{2}} = x_{O_{2}} + x_{H_2S_2O_8} $$

$$ 2 \times \dfrac{22.4}{22.4} = \dfrac{4 \times 8.4}{22.4} + 2 \times n $$

$$ 2 = 1.5 + 2n \Rightarrow n = 0.25 $$

Hence option (A) is correct.

SIMILAR QUESTIONS

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