Question

If $1$ A of current is passed through $CuS{O}_4$ solution for $10$ sec, the number of copper atoms deposit at the cathode will be about:

• A. $1.6\times {10}^{20}$
• B. $8\times {10}^{19}$
• C. $3.1\times {10}^{19}$
• D. $6.2\times {10}^{19}$

Answer 1

The correct option is C $3.1\times {10}^{19}$

No of copper ions $=\frac{\text{Charge}}{\text{Charge on one ion}}$

Number of $Cu$ ions liberated, $n=\frac{I\times t}{e\times \text{valence electron}}$

$n=\frac{1\times 10}{1.6\times {10}^{-19}\times 2}$

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

$\therefore n\approx 3.1\times {10}^{19}$.

Hence, option $C$ is correct.