A solution of copper (II) sulphate us electrolysed between copper electrodes by a current of 10 amperes exactly for 1 hour. What changes occur at the electrodes and in the solution ?

Explanation

Quantity of charge passed (Q) `=Ixxt=(10 amp)xx(60xx60s)`=3600 As =3600 C
Gram equivalent mass of copper `=("Gram atomic mass")/("Valency")=(63.5)/(2)=31.75`
96500 C of charge deposit copper =31.75 g
36000 C of charge deposit copper `=((31.75g))xx((36000 C))/((96500 C))=11.84 g`
Thus, 11.84 g of copper will dissolve from the anode and the same amount of copper will be deposited on the cathode. The concentration of the solution will remain unchanged.