Question

1. Calculate the potential of hydrogen electrodes in contact with a solution whose pH is\[10\].
2. Calculate the EMF of the cell in which the following reaction takes place:
\[Ni\left( s \right){\text{ }} + {\text{ }}2A{g^ + }\left( {0.002{\text{ }}M} \right) \to N{i^{2 + }}\left( {0.160{\text{ }}M} \right){\text{ }} + {\text{ }}2Ag\left( s \right).\]Given that \[{E^0}_{(cell)} = 1.05{\text{ }}V\]

Answer 1

Hint: In hydrogen electrodes the electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The hydrogen electrode reaction is based on \[2{H^ + }\left( {aq} \right) + 2{e^ - } \to {H_2}\left( g \right)\;\]. For calculation using pH use the formula $[{H^ + }] = {10^{ - pH}}$. EMF of the cell can be calculated using Nernst equation:\[{E_{(cell)}} = {E^0}_{(cell)}-\dfrac{{0.0591}}{n}\log \dfrac{{\left[ {oxidised\;species} \right]}}{{\left[ {reduced\;species} \right]}}\] where n is the number of electrons taking part in the reaction.

Complete answer: For hydrogen electrodes, given that \[pH = 10\]. We are going to use the formula $[{H^ + }] = {10^{ - pH}}$
So \[\left[ {{H^ + }} \right]{\text{ }} = {\text{ }}{10^{ - 10}}M\]
The electrode reaction for hydrogen will be \[2{H^ + }\left( {aq} \right) + 2{e^ - } \to {H_2}\left( g \right)\;\] which can also be written as \[{H^ + }{\text{ }} + {\text{ }}{e^ - }{\text{ }} \to {\text{ }}\dfrac{1}{2}{\text{ }}{H_2}\]
We are going to use the formula:
\[\Rightarrow {E_{{H^ + }|{H_2}}} = {\text{ }}{E^0}_{{H^ + }|{H_2}}{\text{ }}-{\text{ }}\dfrac{{RT}}{{nF}}ln\dfrac{{\left[ {{H_2}} \right]}}{{{{\left[ {{H^ + }} \right]}^2}}}\]
\[ \Rightarrow {\text{ }}0{\text{ - }}\dfrac{{8.314 \times 298}}{{2 \times 96500}}ln\dfrac{1}{{{{\left[ {{H^ + }} \right]}^2}}}\]
\[ ={\text{ }}0.05915\log \left[ {{H^ + }} \right]\]
\[ = {\text{ - }}0.05915pH\]
\[ = {\text{ - }}0.05915 \times 10\]
\[ \Rightarrow{E_{{H^ + }|{H_2}}} = {\text{ - }}0.59V\]
Therefore the potential of hydrogen electrode is \[{\text{ - }}0.59V\]
To calculate EMF of the cell. We are given with the reaction:
\[Ni\left( s \right){\text{ }} + {\text{ }}2A{g^ + }\left( {0.002{\text{ }}M} \right) \to N{i^{2 + }}\left( {0.160{\text{ }}M} \right){\text{ }} + {\text{ }}2Ag\left( s \right).\] We are going to use Nernst equation here, which is given as:
\[{E_{(cell)}} = {\text{ }}{E^0}_{(cell)}{\text{ }}-{\text{ }}\dfrac{{0.0591}}{n}\log \dfrac{{\left[ {oxidise{d_{}}species} \right]}}{{\left[ {reduce{d_{}}species} \right]}}\]
For this we should know the two half cells and which among them is oxidized and which is reduced.
The two half-cell reactions are:
\[
  Ni\left( s \right){\text{ }} \to N{i^{2 + }}\left( {0.160{\text{ }}M} \right){\text{ }} + {\text{ 2}}{{\text{e}}^ - } \\
  {\text{ }}2A{g^ + }\left( {0.002{\text{ }}M} \right) + 2{e^ - } \to 2Ag\left( s \right). \\
 \]
In this \[Ni\] is getting oxidized and \[Ag\] is getting reduced. So the oxidized species will be \[N{i^{2 + }}\] and the reduced species will be \[2A{g^ + }\]. From the two half -cell we also get the value of n (total number of electrons used up in the reaction) which is \[2\].
\[\Rightarrow {E_{(cell)}} = {\text{ }}{E^0}_{(cell)}{\text{ }}-{\text{ }}\dfrac{{0.0591}}{n}\log \dfrac{{[N{i^{2 + }}]}}{{{{\left[ {A{g^ + }} \right]}^2}}}\]
We are given that \[{E^0}_{(cell)} = 1.05{\text{ }}V\], n$ = 2$ (from the two half cells), \[[N{i^{2 + }}] = 0.160\] and \[\left[ {A{g^ + }} \right] = 0.002\] so by substituting all the values in the Nernst equation we get
\[\Rightarrow {E_{(cell)}} = {\text{ }}1.05{\text{ }}-{\text{ }}\dfrac{{0.0591}}{2}\log \dfrac{{[0.160]}}{{{{\left[ {0.002} \right]}^2}}}\]
\[ = {\text{ }}1.05{\text{ }}-{\text{ }}0.02955\log \dfrac{{[0.160]}}{{\left[ {0.000004} \right]}}\]
\[ = {\text{ }}1.05{\text{ }}-{\text{ }}0.02955\log 4 \times {10^4}\]
And hence on doing the simplification we have
\[ = {\text{ }}1.05{\text{ }}-{\text{ }}{0.02955_{}}(\log 4 + \log 10000)\]
Again on solving ,we have
\[ ={\text{ }}1.05{\text{ }}-{\text{ }}{0.02955_{}}(0.6021 + 4)\]
\[ \Rightarrow {E_{(cell)}} ={\text{ }}{0.914_{}}V\]
Therefore the EMF of the cell is \[{0.914_{}}V\].

Note:
Writing the two half-cell reactions, oxidation half-cell reaction and reduction half-cell reaction is important to understand which species is undergoing oxidation and which is undergoing reduction. The oxidizing agent oxidizes the other and itself gets reduced and vice-versa.