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: ${H_2} + 2\mathop {OH}\limits^ - \to 2{H_2}O + 2{e^ - }$
Cathode reaction: ${O_2} + 2{H_2}O + 2{e^ - } \to 4\mathop O\limits^ - H$

Answer 1

Hint: We can calculate the average current using the product of time, equivalent of hydrogen used divided by Faraday’s constant. We can calculate the equivalent of hydrogen used using the moles of hydrogen and number of electrons transferred. We can calculate the moles of hydrogen reacting using the volume of hydrogen to the molar volume.

Complete step by step answer:
Given data contains,
Volume of hydrogen is $67.2L$.
Time taken for the reaction is $15\min $.
The amounts of gas are compared at a set of standard conditions of temperature and pressure, abbreviated as STP. At STP, one mole of any gas has the same volume \[{\text{22}}{\text{.4L}}\] called the standard molar volume.
The volumes can be converted to moles by,
$L \times \dfrac{{\,mol}}{{22.4L}}$
$\Rightarrow 67.2L \times \dfrac{{mol}}{{22.4L}} = 3mol$
The number of moles hydrogen reacting is $3mol$.
The reactions taking place in hydrogen fuel cell is,
Anode reaction: ${H_2} + 2\mathop {OH}\limits^ - \to 2{H_2}O + 2{e^ - }$
Cathode reaction: ${O_2} + 2{H_2}O + 2{e^ - } \to 4\mathop O\limits^ - H$
Overall reaction: $2{H_2} + {O_2} \to 2{H_2}O$
Let us now calculate the equivalent of hydrogen used.
Equivalent is the number of moles reaction multiplied by the number of electrons transferred.
The number of electrons transferred in this reaction is two and we have calculated the number of moles of hydrogen reacting is $3mol$.
Therefore, the equivalent of hydrogen used is $3 \times 2 = 6$.
The number of equivalents of hydrogen used is six.
From this, we can calculate the current as follows,
$\dfrac{W}{{{E_w}}} = \dfrac{{I \times t}}{{96500}}$
Here, $\dfrac{W}{{{E_w}}}$ is the equivalent
I is the current
t is time
Faraday's constant is $96500$
Let us substitute the known values in the equation to get the value of average current.
$\dfrac{W}{{{E_w}}} = \dfrac{{I \times t}}{{96500}}$
$ \Rightarrow $$6 = \dfrac{{I \times 15 \times 60}}{{96500}}$
$ \Rightarrow $$I = 643.333A$

$\therefore $ The average current is $643.33A$.

Note:
We also know The primary advantage of using a hydrogen fuel cell is that the by-products obtained from reacting hydrogen and oxygen in a fuel cell are water and heat, so hydrogen fuel cells do not liberate greenhouse gases or any other air pollutants as engines. We can use hydrogen fuel cells to power any device which uses electricity. Therefore, we can use hydrogen fuel cells to power vehicles which run off electricity.