$H_{2} O_{2}$ can be prepared by successive reactions:
$2 N H_{4} H S O_{4} \to H_{2} + (N H_{4})_{2} S_{2} O_{8}, (N H_{4})_{2} S_{2} O_{8} + 2 H_{2} O \to 2 N H_{4} H S_{4} + H_{2} O_{2}$
The first reaction is an electrolytic reaction and second is steam distillation. The amount of current in ampere would have to be used in first reaction to produce enough intermediate to yield 200 g pure $H_{2} O_{2}$ per hr is: Assume current efficiency 50%.

315

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125

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248

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630

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Solution

The correct option is D 630
200 g of $H_{2} O_{2}$ corresponds to $\frac{200}{34} = 5.88$ moles.
1 mole of $H_{2} O_{2}$ corresponds to 2 moles of electrons.
Hence, 5.88 moles of $H_{2} O_{2}$ will correspond to $5.88 \times 2 = 11.765$ moles of electrons.
The current to be used is $\frac{11.765 \times 96500}{3600} = 315.33$ A.
However the current efficiency is only 50%.
Hence, the current to be used is $2 \times 315.33 = 630.7$ A.

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Similar questions

2 N H_{4} H S O_{4} \to H_{2} + (N H_{4})_{2} S_{2} O_{8}

(N H_{4})_{2} S_{2} O_{8} + 2 H_{2} O \to 2 N H_{4} H S O_{4} + H_{2} O_{2}

H_{2} O_{2}

N H_{4} H S O_{4} \to N H_{4} S O_{4}^{-} + H^{+}

2 N H_{4} S O_{4}^{-} \to (N H_{4})_{2} S_{2} O_{8} + 2 e^{-}

2 H^{+} + 2 e^{-} \to H_{2}

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(N H_{4})_{2} S_{2} O_{8} + 2 H_{2} O \to 2 N H_{4} H S O_{4} + H_{2} O_{2}

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2 H_{2} O ⟶ H_{2} + H_{2} O_{2}

H_{2} S + H_{2} O_{2} \to S + 2 H_{2} O

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