Question

The reduction potential of a hydrogen half-cell will be negative if:
[A] $p\left( {{H}_{2}} \right)=1atm\text{ and }\left[ {{H}^{+}} \right]=2M$
[B] $p\left( {{H}_{2}} \right)=1atm\text{ and }\left[ {{H}^{+}} \right]=1M$
[C] $p\left( {{H}_{2}} \right)=2atm\text{ and }\left[ {{H}^{+}} \right]=1M$
[D] $p\left( {{H}_{2}} \right)=2atm\text{ and }\left[ {{H}^{+}} \right]=2M$

Answer 1

To solve this you can consider using the Nernst equation. Firstly write down the reduction half reaction for hydrogen. Then put the pressure and concentration terms in the equation. Remember that for reduction potential to be negative $\dfrac{\left[ reduction \right]}{\left[ oxidation \right]}$ in the Nernst equation should be positive.

Complete step by step solution:
Firstly, let us discuss what reduction potential is.
The reduction potential is basically the tendency of a species to be reduced.
We know that the Nernst equation is-
${{E}_{cell}}={{E}^{\circ }}-\dfrac{RT}{nF}\ln \dfrac{\left[ reduction \right]}{\left[ oxidation \right]}$
We can also write the Nernst equation as-
${{E}_{cell}}={{E}^{\circ }}-\dfrac{0.059}{n}\log \dfrac{[{{M}_{red}}]}{[{{M}_{ox}}]}$
Here, we have to find what will make the reduction potential of the hydrogen half-cell reaction negative.
So, firstly let us write down the reduction half-cell reaction for hydrogen-
$2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}(g)$
Now, from the Nernst equation we can write down that-
${{E}_{{{{H}^{+}}}/{{{H}_{2}}}\;}}={{E}^{\circ }}_{{{{H}^{+}}}/{{{H}_{2}}}\;}-\dfrac{0.059}{2}\log \dfrac{{{p}_{{{H}_{2}}}}}{{{\left[ {{H}^{+}} \right]}^{2}}}$
Now, we need the reduction potential of the cell to be negative. All the terms except the ones including hydrogen concentration are constant. So, ${{p}_{{{H}_{2}}}}$ and ${{\left[ {{H}^{+}} \right]}^{2}}$ will affect the reduction potential.
Now, for it to be negative we need the term $\dfrac{0.059}{2}\log \dfrac{{{p}_{{{H}_{2}}}}}{{{\left[ {{H}^{+}} \right]}^{2}}}$ to be negative and thus is possible when $\dfrac{{{p}_{{{H}_{2}}}}}{{{\left[ {{H}^{+}} \right]}^{2}}}$ is positive. This means that ${{p}_{{{H}_{2}}}}$ must be greater than $\left[ {{H}^{+}} \right]$.
So, $\left[ {{H}^{+}} \right]$ will be 1 M and ${{p}_{{{H}_{2}}}}$ will be 2 atm.

Therefore, the correct answer is option [C] $p\left( {{H}_{2}} \right)=2atm\text{ and }\left[ {{H}^{+}} \right]=1M$

Note: We should remember that the Nernst’s equation is only valid for a chemical reaction at equilibrium. We consider a hydrogen electrode as the standard electrode and it is called the SHE i.e. standard hydrogen electrode. We measure it at standard temperature and pressure conditions. The electrode potential of this electrode is referred as ${{E}^{\circ }}$ and is zero at any temperature. It consists of a platinum electrode immersed in a solution of hydrogen ion of concentration 1M.