0. 2.5 Faradays of electricity is passed through a solution of   ${\mathrm{CuSO}}_4$  .The number of gram equivalents

 of copper deposited at cathode are : 

$\begin{array}{cc}{\mathrm{CuSO}}_4\rightleftharpoons {\mathrm{Cu}}^{2+}+{\mathrm{SO}}_4^{2-}& \\ {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}⟶\mathrm{Cu}& \\ 2\mathrm{F}& 1\mathrm{g}.\mathrm{eq}\end{array}$

2F electricity deposits  $\mathrm{Cu}=1\mathrm{g}$ . equivalent 

$2.5\mathrm{F}\text{ electricity deposits }\mathrm{Cu}=\frac{1\mathrm{g}\cdot \text{ equivalent }\times 2.5}{2}$

                                            $=1.25 \mathrm{g}$