Question

If 1A current is passed through $CuS{O}_4$ solution for 10 s, the number of copper atoms deposited at the cathode will approximately be

• A. $3.1\times {10}^{19}$
• B. $6.2\times {10}^{19}$
• C. $9.3\times {10}^{19}$
• D. $1.6\times {10}^{19}$

Answer 1

Answer: (A)

$3.1\times {10}^{19}$

The current passing through electrolyte is
$I=\frac{2ne}{t}$

where, $n$ is number of atoms deposited at the cathode, $e$ is charge on electron and $t$ is the time.

$n=\frac{It}{2e}$

$n=\frac{\left(1\right)\left(10\right)}{2(1.6\times {10}^{-19})}$

$n=\frac{10}{3.2\times {10}^{-19}}$

$n=3.125\times {10}^{19}$