Question

A solution of$CuS{O}_4$ is electrolyzed for $10$ minutes with a current of $1.61$ ampere. The mass of copper $\left(\text{in g}\right)$ deposited at the cathode is (sum the answer upto two decimal places)
(Given : Molar mass of $Cu=63.5\text{g/ mol}$)

Answer 1

Charge (Q) = Current (i) $\times$ time (t)
$Q=1.61\times 10\times 60$
$\Rightarrow Q=1.61\times 600=966\text{C}$
Reaction occuring at the cathode is
$C{u}^{+2}+2e^{-}\to Cu$
Therefore, we require $2F=2\times 96500\text{C}$ charge to deposit $1$ mol or $63.5\text{g}$ of Cu.
For $966\text{C}$, the mass of $Cu$ deposited
$=\frac{63.5}{2\times 96500}\times 966\approx 0.32\text{g}$