Question

In a fuel cell, $H_2$ and $O_2$ react to produce electricity. In the process, $H_2$ gas is oxidized at the anode and $O_2$ gas is reduced at the cathode. If $67.2\text{litres}$ of $H_2$ at $STP$ react in 15 minutes and the entire current is used for electrode deposition of $Cu$ from $Cu\left(II\right)$ solution, how many grams of $Cu$ will be deposited and what is the average current produced?
Anode : $H_2+2O{H}^{-}\to 2H_{2}O+2e^{-}$
Cathode : $O_2+2H_{2}O+4e^{-}\to 4O{H}^{-}$

• A. $\text{190.5 g of Cu}$
• B. $\text{381 g of Cu}$
• C. $\text{643.33 A}$
• D. $\text{321.6 A}$

Answer 1

Answer: (A), (C)

$\text{643.33 A}$

Reaction at anode,
$H_2\left(g\right)\to 2H^{+}+2e^{-}$
67.2 L of $H_2$ ​= $22.42\times 96500\times 67.2$ coulomb
Time = $15\times 60$ seconds
Average current =$\frac{2\times 96500\times 67.2}{22.4\times 15\times 60}=643.3A$
Mass of copper deposited by $\frac{2\times 96500\times 67.2}{22.4}$ coulomb
$=\frac{2\times 96500\times 63.5\times 67.2}{2\times 96500\times 22.4}$
=190.5 g.