Question

A current of 1.6 A is passed through solution of ${CuSO}_4$. How many ${Cu}^{++}$ ions are liberated in one minute? (Electronic charge $=1.6\times {10}^{-19}C)$

• A. $3\times {10}^{20}$
• B. $3\times {10}^{10}$
• C. $6\times {10}^{20}$
• D. $6\times {10}^{10}$

Answer 1

Answer: (A)

$3\times {10}^{20}$

Charge per second$=current=1.6A$

No.of electrons per second ($n$) $=\frac{I}{charge}={10}^{19}\frac{e^{-}}{sec}$

But each copper atom requires two electrons

So, the total copper atoms liberated in one second $\left(N\right)=\frac{n}{2}=5\times {10}^{18}$

Atoms liberated in one minute $=N\times 60=3\times {10}^{20}$