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Question: 1\. Calculate the potential of hydrogen electrodes in contact with a solution whose pH is\[10\]. 2...

1. Calculate the potential of hydrogen electrodes in contact with a solution whose pH is1010.
2. Calculate the EMF of the cell in which the following reaction takes place:
Ni(s) + 2Ag+(0.002 M)→Ni2+(0.160 M) + 2Ag(s).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 E0(cell)=1.05 V{E^0}_{(cell)} = 1.05{\text{ }}V

Explanation

Solution

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 2H+(aq)+2e−→H2(g)  2{H^ + }\left( {aq} \right) + 2{e^ - } \to {H_2}\left( g \right)\;. For calculation using pH use the formula [H+]=10−pH[{H^ + }] = {10^{ - pH}}. EMF of the cell can be calculated using Nernst equation:E(cell)=E0(cell)−0.0591nlog⁡[oxidised  species][reduced  species]{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=10pH = 10. We are going to use the formula [H+]=10−pH[{H^ + }] = {10^{ - pH}}
So [H+] = 10−10M\left[ {{H^ + }} \right]{\text{ }} = {\text{ }}{10^{ - 10}}M
The electrode reaction for hydrogen will be 2H+(aq)+2e−→H2(g)  2{H^ + }\left( {aq} \right) + 2{e^ - } \to {H_2}\left( g \right)\; which can also be written as H+ + e− → 12 H2{H^ + }{\text{ }} + {\text{ }}{e^ - }{\text{ }} \to {\text{ }}\dfrac{1}{2}{\text{ }}{H_2}
We are going to use the formula:
⇒EH+∣H2= E0H+∣H2 − RTnFln[H2][H+]2\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}}}
⇒ 0 - 8.314×2982×96500ln1[H+]2\Rightarrow {\text{ }}0{\text{ - }}\dfrac{{8.314 \times 298}}{{2 \times 96500}}ln\dfrac{1}{{{{\left[ {{H^ + }} \right]}^2}}}
= 0.05915log⁡[H+]={\text{ }}0.05915\log \left[ {{H^ + }} \right]
= - 0.05915pH= {\text{ - }}0.05915pH
= - 0.05915×10= {\text{ - }}0.05915 \times 10
⇒EH+∣H2= - 0.59V\Rightarrow{E_{{H^ + }|{H_2}}} = {\text{ - }}0.59V
Therefore the potential of hydrogen electrode is  - 0.59V{\text{ - }}0.59V
To calculate EMF of the cell. We are given with the reaction:
Ni(s) + 2Ag+(0.002 M)→Ni2+(0.160 M) + 2Ag(s).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)= E0(cell) − 0.0591nlog⁡[oxidisedspecies][reducedspecies]{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(s) →Ni2+(0.160 M) + 2e−  2Ag+(0.002 M)+2e−→2Ag(s).  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 NiNi is getting oxidized and AgAg is getting reduced. So the oxidized species will be Ni2+N{i^{2 + }} and the reduced species will be 2Ag+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 22.
⇒E(cell)= E0(cell) − 0.0591nlog⁡[Ni2+][Ag+]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 E0(cell)=1.05 V{E^0}_{(cell)} = 1.05{\text{ }}V, n=2 = 2 (from the two half cells), [Ni2+]=0.160[N{i^{2 + }}] = 0.160 and [Ag+]=0.002\left[ {A{g^ + }} \right] = 0.002 so by substituting all the values in the Nernst equation we get
⇒E(cell)= 1.05 − 0.05912log⁡[0.160][0.002]2\Rightarrow {E_{(cell)}} = {\text{ }}1.05{\text{ }}-{\text{ }}\dfrac{{0.0591}}{2}\log \dfrac{{[0.160]}}{{{{\left[ {0.002} \right]}^2}}}
= 1.05 − 0.02955log⁡[0.160][0.000004]= {\text{ }}1.05{\text{ }}-{\text{ }}0.02955\log \dfrac{{[0.160]}}{{\left[ {0.000004} \right]}}
= 1.05 − 0.02955log⁡4×104= {\text{ }}1.05{\text{ }}-{\text{ }}0.02955\log 4 \times {10^4}
And hence on doing the simplification we have
= 1.05 − 0.02955(log⁡4+log⁡10000)= {\text{ }}1.05{\text{ }}-{\text{ }}{0.02955_{}}(\log 4 + \log 10000)
Again on solving ,we have
= 1.05 − 0.02955(0.6021+4)={\text{ }}1.05{\text{ }}-{\text{ }}{0.02955_{}}(0.6021 + 4)
⇒E(cell)= 0.914V\Rightarrow {E_{(cell)}} ={\text{ }}{0.914_{}}V
Therefore the EMF of the cell is 0.914V{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.