Question

In a hydrogen oxygen fuel cell, electricity is produced. In this process $H_2\left(\text{g}\right)$ is oxidised at anode and $O_2\left(\text{g}\right)$ reduced at cathode.
Given : Cathode: $O_2\left(g\right)+2H_{2}O\left(l\right)+4e^{-}\to 4O{H}^{-}\left(aq\right)$
Anode : $H_2\left(g\right)+2O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)+2e^{-}$
$4.48\text{L}$ of $H_2$ at $1\text{atm}$ and $273\text{K}$ is oxidised in $9650\text{sec}$.
The current produced is (in amp):

• A. 1
• B. 4
• C. 2
• D. 8

Answer 1

The correct option is B 4

According to Faraday’s law:
$m=Z\times i\times t$ …….. (1)
Where, m is mass of reactant reacted or product formed
Z $\Rightarrow$ electrochemical equivalent
i$\Rightarrow$ current
t $\Rightarrow$ time
Also, we know that:
$No.ofmoles=\frac{Vol.inlitresatSTP}{22.4}=\frac{mass}{Molarmass}$
$\therefore$
$mass=\frac{Vol.inlitresatSTP}{22.4}\times Molarmass$ ………… (2)
From eq(2) substituting for mass of $H_2\left(g\right)$ in eq(1):
$\frac{Vol.of{H}_2\left(g\right)}{22.4}\times Molarmassof{H}_2\left(g\right)=Z\times i\times t$
By substituting the values we get,
$\frac{4.48}{22.4}\times 2=\frac{2}{2\times 96500}\times i\times 9650$
On solving we get:
$i=4A$