
The pressure of \[{{H}_{2}}\] required to make the potential of \[{{H}_{2}}\] electrode zero in pure water at $298K$ is:
A. \[\mathbf{1}{{\mathbf{0}}^{-\mathbf{14}}}\mathbf{atm}\]
B. \[\mathbf{1}{{\mathbf{0}}^{-\mathbf{12}}}\mathbf{atm}\]
C. \[\mathbf{1}{{\mathbf{0}}^{-\mathbf{10}}}\mathbf{atm}\]
D. \[\mathbf{1}{{\mathbf{0}}^{-\mathbf{4}}}\mathbf{atm}\]
Answer
533.7k+ views
Hint:We know that the potential of \[{{H}_{2}}\] electrode to zero, pH2 should be equal to \[{{\left[ {{H}^{+}} \right]}^{2}}\] . 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 \[{{10}^{-14}}atm\] as from the question, we have a below equation:
\[2{{H}^{+}}+\text{ }2{{e}^{-}}\to {{H}_{2}}\]
As pure water has \[\left[ {{H}^{+}} \right]\text{ }=\text{ }\left( -OH \right)\text{ }={{10}^{-7}}\]
According to Nernst equation,
\[E={{E}^{\circ }}-\dfrac{0.0591}{n}\log \dfrac{{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}\]
Considering the $1.0M$hydrogen ion solution with pressurize hydrogen around $1atm$
\[E=0-\dfrac{0.0591}{2}\log \dfrac{1}{{{[1]}^{2}}}\]
\[E=0-\dfrac{0.0591}{2}\log 1\]
$E=0-0$
$\Rightarrow E=0$
Whereas in pure water, hydrogen ion conc. of ${{10}^{-7}}M$ with pressurize hydrogen of around \[P\text{ }atm\]
\[E={{E}^{\circ }}-\dfrac{0.0591}{n}\log \dfrac{{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}\]
Substituting the values and we get the value of $P$ on evaluating;\[0=0-\dfrac{0.0591}{2}\log \dfrac{P}{{{[{{10}^{-7}}]}^{2}}}\]
\[0=\log \dfrac{P}{{{[{{10}^{-7}}]}^{2}}}\]
\[1=\dfrac{P}{{{({{10}^{-7}})}^{2}}}\]
$\Rightarrow P={{10}^{-14}}atm$
Therefore correct answer is Option A i.e. the pressure of \[{{H}_{2}}\] required to make the potential of \[{{H}_{2}}\] electrode zero in pure water at $298K$ is ${{10}^{-14}}atm$.
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 ${{H}_{2}}$
Complete step-by-step answer:According to our question, correct answer is \[{{10}^{-14}}atm\] as from the question, we have a below equation:
\[2{{H}^{+}}+\text{ }2{{e}^{-}}\to {{H}_{2}}\]
As pure water has \[\left[ {{H}^{+}} \right]\text{ }=\text{ }\left( -OH \right)\text{ }={{10}^{-7}}\]
According to Nernst equation,
\[E={{E}^{\circ }}-\dfrac{0.0591}{n}\log \dfrac{{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}\]
Considering the $1.0M$hydrogen ion solution with pressurize hydrogen around $1atm$
\[E=0-\dfrac{0.0591}{2}\log \dfrac{1}{{{[1]}^{2}}}\]
\[E=0-\dfrac{0.0591}{2}\log 1\]
$E=0-0$
$\Rightarrow E=0$
Whereas in pure water, hydrogen ion conc. of ${{10}^{-7}}M$ with pressurize hydrogen of around \[P\text{ }atm\]
\[E={{E}^{\circ }}-\dfrac{0.0591}{n}\log \dfrac{{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}\]
Substituting the values and we get the value of $P$ on evaluating;\[0=0-\dfrac{0.0591}{2}\log \dfrac{P}{{{[{{10}^{-7}}]}^{2}}}\]
\[0=\log \dfrac{P}{{{[{{10}^{-7}}]}^{2}}}\]
\[1=\dfrac{P}{{{({{10}^{-7}})}^{2}}}\]
$\Rightarrow P={{10}^{-14}}atm$
Therefore correct answer is Option A i.e. the pressure of \[{{H}_{2}}\] required to make the potential of \[{{H}_{2}}\] electrode zero in pure water at $298K$ is ${{10}^{-14}}atm$.
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 ${{H}_{2}}$
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