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The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = $1.60 \times 10^{-19}$ C)
A
$3.75 \times 10^{20}$
B
$7.48 \times 10^{23}$
C
$6 \times 10^{23}$
D
$6 \times 10^{20}$
Explanation
n = It/e.
Detailed Solution
Q = ne, i.e. it = ne
$n = \frac{1\times60}{1.6\times10^{-19}} = 3.75\times10^{20}$ electrons
$n = \frac{1\times60}{1.6\times10^{-19}} = 3.75\times10^{20}$ electrons