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Question: The pressure of \[{{H}_{2}}\] required to make the potential of \[{{H}_{2}}\] electrode zero in pure...

The pressure of H2{{H}_{2}} required to make the potential of H2{{H}_{2}} electrode zero in pure water at 298K298K is:
A. 1014atm\mathbf{1}{{\mathbf{0}}^{-\mathbf{14}}}\mathbf{atm}
B. 1012atm\mathbf{1}{{\mathbf{0}}^{-\mathbf{12}}}\mathbf{atm}
C. 1010atm\mathbf{1}{{\mathbf{0}}^{-\mathbf{10}}}\mathbf{atm}
D. 104atm\mathbf{1}{{\mathbf{0}}^{-\mathbf{4}}}\mathbf{atm}

Explanation

Solution

We know that the potential of H2{{H}_{2}} electrode to zero, pH2 should be equal to [H+]2{{\left[ {{H}^{+}} \right]}^{2}} . In hypothetical material science, it is the energy of two charges isolated by endlessness separation. In electrochemistry, this is picked to be the capability of the hydrogen anode since that is something that can be made reproducibly and hydrogen was the accepted 'zero' for various different amounts.

Complete step-by-step answer: According to our question, correct answer is 1014atm{{10}^{-14}}atm as from the question, we have a below equation:
2H++ 2eH22{{H}^{+}}+\text{ }2{{e}^{-}}\to {{H}_{2}}
As pure water has [H+] = (OH) =107\left[ {{H}^{+}} \right]\text{ }=\text{ }\left( -OH \right)\text{ }={{10}^{-7}}
According to Nernst equation,
E=E0.0591nlogPH2[H+]2E={{E}^{\circ }}-\dfrac{0.0591}{n}\log \dfrac{{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}
Considering the 1.0M1.0Mhydrogen ion solution with pressurize hydrogen around 1atm1atm
E=00.05912log1[1]2E=0-\dfrac{0.0591}{2}\log \dfrac{1}{{{[1]}^{2}}}
E=00.05912log1E=0-\dfrac{0.0591}{2}\log 1
E=00E=0-0
E=0\Rightarrow E=0
Whereas in pure water, hydrogen ion conc. of 107M{{10}^{-7}}M with pressurize hydrogen of around P atmP\text{ }atm
E=E0.0591nlogPH2[H+]2E={{E}^{\circ }}-\dfrac{0.0591}{n}\log \dfrac{{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}
Substituting the values and we get the value of PP on evaluating;0=00.05912logP[107]20=0-\dfrac{0.0591}{2}\log \dfrac{P}{{{[{{10}^{-7}}]}^{2}}}
0=logP[107]20=\log \dfrac{P}{{{[{{10}^{-7}}]}^{2}}}
1=P(107)21=\dfrac{P}{{{({{10}^{-7}})}^{2}}}
P=1014atm\Rightarrow P={{10}^{-14}}atm

Therefore correct answer is Option A i.e. the pressure of H2{{H}_{2}} required to make the potential of H2{{H}_{2}} electrode zero in pure water at 298K298K is 1014atm{{10}^{-14}}atm.

Note: Note that the electrochemistry, the Nernst condition is a condition that relates the decrease capability of an electrochemical response (half-cell or full cell response) to the standard anode potential, temperature, and exercises (frequently approximated by groupings) of the synthetic species going through decrease and oxidation. Nernst equation is used to calculate pressure of H2{{H}_{2}}