Question

In a fuel cell $H_2$ and $O_2$ react to produce electricity. In the process $H_2$ gas is oxidized at anode and $O_2$ at cathode. If 67.2 litre of $H_2$ at STP reacts in 15 minutes. The average current produced is:

• A. 643.33 A
• B. 642.33 A
• C. 682.33 A
• D. None of the above

Answer 1

The correct option is A 643.33 A

The redox changes in fuel cell are:
$2H_2\left(g\right)+O_2\left(g\right)\to 2H_{2}O\left(l\right)$
Anode : $H_2+2O{H}^{-}\to 2H_{2}O+2e$
Cathode: $O_2+2H_{2}O+4e\to 4O{H}^{-}$
$\therefore$ Mole of $H_2$ reacting $=$ $\frac{67.2}{22.4}$$=$3
$\therefore$ Eq. of $H_2$ used $=$$3\times 2$ $=$6
Now $\frac{m}{E}=\frac{i.t}{96500}$
$\therefore$ $6=\frac{i\times 15\times 60}{96500}$
or $i=643.33$
Also Eq. of $H_2$ = Eq.of Cu formed =6
$M_{Cu}=\frac{6\times 63.5}{2}=190.5g$