Question

Consider the cell:
${H_2}\left( {Pt} \right)\left| {{H^ + }_a\left\| {{H^ + }_b\left| {{H_2}\left( {Pt} \right)} \right.} \right.} \right.$
The above cell is not working when
1.$a = 1M$, $b = 1.5M$
2.$a = 0.5M$, $b = 0.75M$
3.$a = 0.02M$, $b = 0.01M$
4.$a = 0.4M$, $b = 0.6M$

Answer 1

This question gives the knowledge about Nernst equation. Nernst equation helps in determining the cell potential that is oxidation potential or reduction potential under various conditions. Also helps in determining the cell potential.

Formula used: The cell potential is determined using Nernst equation as follows:
$E = {E^0} - \dfrac{{RT}}{{zF}}\ln Q$
Where ${E^0}$ is standard potential, $E$ is cell potential, $R$ is universal gas constant, $z$ is ion charge, $F$ is Faraday constant, $Q$ is reaction constant and $T$ is temperature in kelvin.

Complete step-by-step answer: Nernst equation helps in determining the cell potential under various non-standard conditions. Also helps in determining the electrode potential. Nernst equation helps in determining the relation between reduction potential or oxidation potential of an electrochemical cell reaction to the temperature, activities and standard electrode potential of the chemical species.
Consider the cell,
${H_2}\left( {Pt} \right)\left| {{H^ + }_a\left\| {{H^ + }_b\left| {{H_2}\left( {Pt} \right)} \right.} \right.} \right.$
Now, separate the oxidation and reduction potentials as follows:
Oxidation: ${H_2} \to 2{H^ + }_a + 2{e^ - }$
Reduction: $2{H^ + }_b + 2{e^ - } \to \,{H_2}$
According to Nernst equation, we have
$E = {E^0} - \dfrac{{RT}}{{zF}}\ln Q$
Where $Q = \dfrac{{{{\left[ a \right]}^n}}}{{{{\left[ b \right]}^n}}}$ and $n$ is the number of moles.
Substitute ${E^0}$ as $0$, $z$ as $2$, $\dfrac{{RT}}{F}$ as $0.0592$, $a$ as $0.02M$ and $b$ as $0.01M$ in the Nernst equation:
$\Rightarrow E = 0 - \dfrac{{0.0592}}{2}\ln \dfrac{{{{\left[ {0.02} \right]}^2}}}{{{{\left[ {0.01} \right]}^2}}}$
On solving the Nernst equation we have,
$\Rightarrow E = 0 - 0.0295 \times 0.6020$
The chemical cell potential is
$\Rightarrow E = - 0.017759$
Now, consider the Gibbs free energy equation
$\Delta G = - nFE$
Where $\Delta G$ is the Gibbs energy, $F$ is Faraday constant, $E$ is cell potential and $n$ is the
number of moles. Substitute $E$ as $- 0.017759$ in Gibbs free equation
$\Rightarrow \Delta G = - nF\left( { - 0.017759} \right)$
On solving we have,
$\Rightarrow \Delta G = + 0.017759nF$
Here, $\Delta G$ is positive which means the reaction is nonspontaneous and the cell is not working.

Therefore, option $\left( 3 \right)$ is the correct option.

Note: Always remember the values of all the constants. And also remember whenever $\Delta G$ is positive the reaction becomes non- spontaneous and the cell stops working and when $\Delta G$ is negative the reaction becomes spontaneous and the cell starts working.