Question

In a fuel cell, hydrogen and oxygen react to produce electricity. In the process, hydrogen gas is oxidized at anode and oxygen at cathode. If 67.2 L of $H_2$ at STP reacts in 15 min, the average current (in A) produced is :

Anode reaction : $H_2+2\stackrel{\ominus }{O}H\to 2H_{2}O+2e^{-}$

Cathode reaction : $O_2+2H_{2}O+2e^{-}\to 4\stackrel{\ominus }{O}H$

(write your answer to nearest integer)

Answer 1

Moles of $H_2$ reacting = $\frac{67.2}{22.4}$ = 3.

$\therefore \text{equivalent of}{H}_2\text{used}=3\times 2=6$

Now, $\frac{W}{E_w}=\frac{It}{96500}$

$\therefore 6=\frac{I\times 15\times 60}{96500}$

$\therefore I=643.33A$

Also, equivalent of $H_2$ = equivalent of Cu formed $\therefore$ Equivalent of Cu deposited = 6

$\therefore W_{Cu}=6\times =\frac{63.5}{2}=190.5g$

Thus, weight of Cu deposited = 190.5 g.