Question

A current of $2.68$ ampere is passed for one hour through an aqueous solution of copper sulphate using copper electrode. Calculate the change in mass of cathode and that of the anode. (At. mass of copper $=63.5$).

Answer 1

The electrode reaction are:
$\underset{1mole}{C{u}^{2+}}+\underset{2\times 96500C}{2e^{-}}\to Cu\left(Cathode\right)$
$Cu\times C{u}^{2+}+2e^{-}\left(Anode\right)$
Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves.
Charge passed through cell $=2.68\times 60\times 60coulomb$
Copper deposited or dissolved $=\frac{63.5}{2\times 96500}\times 2.68\times 60\times 60=3.174g$
Increase in mass of cathode $=$ Decrease in mass of anode $=3.174g$.