During electrolysis of 1 litre of aq. `CuSO_4` , the `P^(H)` of the solution is changed from 5.0 to 4.3 . Then the number of faradays consumed in the process is `(x xx 10^(-4))` , here , `(x + y) = ?`

Answer

9

Explanation

`(H_2O + CuSO_4) overset("electrolysis/2F")to Cu+ (1)/(2)O_2 uarr + (2H^(oplus) + SO_4^(-2))`
` 2F overset("passed ") to 2` moles of `H^(oplus)`
`Q to n xx 4 xx 10^(-5) , n = (Q xx 2)/(2F) = 4 xx 10^(-5) , Q = 4 xx 10^(-5) xx F`
` Q = x xx 10^(-4) xx F , (x + y) = (4+5) = 9`