Question

In a fuel cell $H_2$ and $O_2$ react to produce electricity. In the process $H_2$ gas is oxidised at the anode and $O_2$ is reduced at the cathode. If $67.2$ litre of $H_2$ at $NTP$ reacts in $15$ minute, what is the average current produced? If the entire current is used for electro-deposition of $Cu$ from $C{u}^{2+}$, how many $g$ of $Cu$ are deposited?

Answer 1

Reaction at anode of fuel cell,
$\underset{\underset{22.4L}{1mole}}{H_2\left(g\right)}\to 2H^{+}+\underset{\underset{2F}{2mole}}{2e^{-}}$
$67.2L$ of $H_{2}correspond=\frac{2\times 96500}{22.4}\times 67.2coulomb$
$Time=15\times 60second$
Average current $=\frac{2\times 96500\times 67.2}{22.4\times 15\times 60}=643.3amp$
Mass of copper deposited by $\frac{2\times 96500}{22.4}\times 67.2coulomb$
$=\frac{63.5}{2\times 96500}\times \frac{2\times 96500\times 67.2}{22.4}$
$=190.5g$.