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Faraday's First Law

A current of ...

Question

A current of $2.6 a m p$ was passed through $C u S O_{4}$ solution for $6$ minutes and $20$ seconds. The amount of copper deposited is:

A

0.3175 g

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B

0.0031 g

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C

6.35 g

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D

3.175 g

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Solution

The correct option is A $0.3175 g$
$W = \frac{I t E}{96500} = \frac{2.6 \times 380 \times 31.75}{96500} \approx 0.32 g$ .

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