Question

When 1000 C of charge is passed through CuSO4 solution, x g of copper is deposited. How much charge should be passed through the solution to deposit 5x g of copper?​

Answer 1

Number of charge = 5018 C

Further explanation

Given

1000 C of charge for x grams of copper

Required

Number of charge

Solution

Faraday's Law :

`\tt W=(e.i.t)/(96500)\\\\W=(e.Q)/(96500)`

For 1000 C, W = x grams

`\tt x=(1000.e)/(96500)=0.0104e`

For 5x grams :

`\tt 5x=(e.Q)/(96500)\\\\5* 0.0104e=(e.Q)/(96500)\\\\Q=(96500* 5* 0.0104e)/(e)=5018~C`