A solution of copper (II) sulphate is electrolysed between copper electrodes by a current of `10.0` amperes passing for one hour. What changes occur at the electrodes and in the solution?

Explanation

According to Faradays first law of electrolysis:
The reaction at cathode : `{:(Cu^(2+) +, 2e^(-) rarr Cu),(63.5,2 xx 96500 C):}`
The quantity of charge passed `= 1 xx t = (10 amp) xx (60 xx 60s) = 36000C`.
`2 xx 96500 C` of charge deposit copper `= 63.5 g`
`36500 C` of charge deposit copper `= ((63.5g))/((2 xx 96500C)) xx (36000C) = 11.84 g`
Thus, `11.84g` of copper will dissolve from the anode and the same amount from the solution will get deposited on the cathode. The concentration of the solution will remain uncharged.