100ml of 0.1M aq CuSO₄ solution was electrolysed by passing 0.1F electricity, using pt electrodes. Then products deposited / liberated at cathode is /are 

$n_{cu}^{+2}=\frac{m\times v}{1000}$
$\begin{array}{rl} & =\frac{0.1\times 100}{1000}\\ & ={10}^{-2}\text{mole}\\ & {\mathrm{Cu}}^{+2}+2\overline{\mathrm{e}}\to \mathrm{Cu}\end{array}$
1 mole 2F
${10}^{-2}mole-2\times {10}^{-2}F$ required
0.02 F are enough to discharge 10^-2 mole Cu⁺² –ions present in 100 ml
The remaining 0.08F are utilised to discharge the $H^{\oplus }$   ions