Calculate volume of the gases liberated at STP if 1 L of 0.2 molar solution of `CuSO_(4)` is electrolysed by 5.79 A current for 10000 seconds.

Explanation

No. of moles of `e^(-)=(5.79xx10000)/(96500)=(579)/(965)=0.6`
Cathode: `underset(0.2" mole 0.4 mole")(`Cu^(2+)+2e^(-)toCu(s))`
`2H_(2)O(l)+2e^(-)toH_(2)+2OH^(-)`
0.2 mole of `e^(-)to0.1"mole of" H_(2)` at S.T.P
Anode: `2H_(2)O(l)toO_(2)+4H^(+)+4e^(-)`
4 mole of `e^(-)to1` mole of `O_(2)`
0.6 mole of `e^(-)to0.15` mole of `O_(2)`
so total volume `=5.6ltr`