The number of electrons lost or gained during electrolysis of $2gC{l}^{-}$ ions are x, 1 g $Z{n}^{2+}$ ions are y, then x + y =  $\_\times {10}^{22}$
 

(a) Eq. of Cl- used  $=\frac{2}{35.5}$ for   $2C{\mathcal{l}}^{-}\to C{\mathcal{l}}_2+2e$
 1 eq. of an element = 1 faraday charge = $6.23\times {10}^{23}$ electrons
$\therefore \frac{2}{35.5}eq.ofC{\mathcal{l}}^{-}=\frac{6.023\times {10}^{23}\times 2}{35.5}=3.39\times {10}^{22}$ electrons lost
(b)  Similarly, calculate for  $Z{n}^{2+}+2e\to Zn$
Electrons gained  $=1.85\times {10}^{22}$ electrons