The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 × 10–19 C)
$\dfrac{W}{E} = \dfrac{1 × 60}{96500}$
$=\dfrac{6}{9650}$ = no. of mole e–
no. of e– = $\dfrac{6}{9650}$ × 6.02 × 1023
= 3.75 × 1020
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