Question

In a hydrogen oxygen fuel cell, electricity is produced. In this process $H_2\left(g\right)$ is oxidised at anode and $O_2\left(g\right)$ reduced at cathode.

Given: Cathode $O_2\left(g\right)+2H_{2}O\left(l\right)+4e^{-}\to 4O{H}^{-}\left(aq\right)$

Anode $H_2\left(g\right)+2O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)+2e^{-}$

4.48 litre $H_2$ at 1 atm and 273K oxidised in 9650 sec.

The current produced is (in amp):

• A. 1A
• B. 2A
• C. 4A
• D. 8A

Answer 1

The correct option is C 4A

$2\times \left(\frac{4.48\times 1}{22.4}\right)=\frac{2}{2}\times \frac{I\times 9650}{96,500}$

$I=4$ amp.