The number of electrons delivered at the cathode during electrolysis by a current of $${\text{1 ampere}}$$  in 60 seconds is ( charge on electron $${\text{ = 1}}{\text{.60}} \times {\text{1}}{{\text{0}}^{ - 19}}C$$     )

Solution :
$$\eqalign{ & {\text{From Faraday's}}\,{\text{first law of electrolysis,}} \cr & \frac{w}{E} = \frac{{it}}{{96500}}\,\,\,...{\text{(i)}} \cr & {\text{Given,}}\,i = 1\,A;\,t = 60\,s \cr & {\text{Putting these values in Eq}}.{\text{(i), we get}} \cr & \frac{w}{E} = \frac{{1 \times 60}}{{96500}}\,\,{\text{or}}\,\,\frac{w}{E} = \frac{6}{{9650}} \cr & = {\text{Number of mole of electrons}} \cr & \therefore \,\,{\text{Number of electrons}} \cr & = \frac{6}{{9650}} \times 6.022 \times {10^{23}} \cr & = 3.75 \times {10^{20}} \cr} $$