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Question: The standard reduction potential at \[{25^o}C\] of the reaction \(2{H_2}O + 2{e^ - } \to {H_2} + 2O{...

The standard reduction potential at 25oC{25^o}C of the reaction 2H2O+2e−→H2+2OH−2{H_2}O + 2{e^ - } \to {H_2} + 2O{H^ - } is −0.8277 volt - 0.8277{\text{ }}volt. Calculate the equilibrium constant for the reaction, 2H2O⇌H3O++OH−2{H_2}O \rightleftharpoons {H_3}{O^ + } + O{H^ - } at 25oC{25^o}C.

Explanation

Solution

For calculating the equilibrium constant for the given reaction, it is necessary to find out the relation between KW{K_W} and KK for the reaction and then to apply this relation in finding the electrode potential of a particular cell.

Complete step by step answer:
Let us consider a hydrogen electrode as:
2H++2e−→H22{H^ + } + 2{e^ - } \to {H_2} → for this reaction the reduction potential is zero. Thus, ERP(H)o=0E_{RP(H)}^o = 0
In the question, the given electrode is as follows:
2H2O+2e−→H2+2OH−,ERPo=−0.8277V2{H_2}O + 2{e^ - } \to {H_2} + 2O{H^ - },E_{RP}^o = - 0.8277V
Since, oxidation potential of hydrogen is less than that of water, the individual electrode reactions are as follows:
Anode: H2+2OH−→2H2O+2e−,EOPo=+0.8277V{H_2} + 2O{H^ - } \to 2{H_2}O + 2{e^ - },E_{OP}^o = + 0.8277V
Cathode: 2H++2e−→H2,EOPo=02{H^ + } + 2{e^ - } \to {H_2},E_{OP}^o = 0
Thus, the net reaction is: 2H++2OH−⇌2H2O2{H^ + } + 2O{H^ - } \rightleftharpoons 2{H_2}O
Thus, K=[H2O]2[H+]2[OH−]2K = \dfrac{{{{[{H_2}O]}^2}}}{{{{[{H^ + }]}^2}{{[O{H^ - }]}^2}}}
For the reaction, 2H2O⇌[H3O+][OH−] KW=[H3O+][OH−] K=[1KW]2....(i)  2{H_2}O \rightleftharpoons [{H_3}{O^ + }][O{H^ - }] \\\ {K_W} = [{H_3}{O^ + }][O{H^ - }] \\\ K = {[\dfrac{1}{{{K_W}}}]^2}....(i) \\\
Also, Ecell=0.8277+0.0592log⁡[H+]2[OH−]2[H2O]2 Ecell=0.8277+0.0592log⁡1K....(ii)  {E_{cell}} = 0.8277 + \dfrac{{0.059}}{2}\log \dfrac{{{{[{H^ + }]}^2}{{[O{H^ - }]}^2}}}{{{{[{H_2}O]}^2}}} \\\ {E_{cell}} = 0.8277 + \dfrac{{0.059}}{2}\log \dfrac{1}{K}....(ii) \\\
Comparing equations (i) and (ii), we have:
Ecell=0.8277+0.0592log⁡[KW]2{E_{cell}} = 0.8277 + \dfrac{{0.059}}{2}\log {[{K_W}]^2}
We know that, at equilibrium, Ecell=0{E_{cell}} = 0 .
So, \-0.8277=0.059log⁡KW =log⁡KW=0.82770.059 =KW=9.35×10−15  \- 0.8277 = 0.059\log {K_W} \\\ = \log {K_W} = \dfrac{{0.8277}}{{0.059}} \\\ = {K_W} = 9.35 \times {10^{ - 15}} \\\
Thus, the value of equilibrium constant for the desired equation is 9.35×10−159.35 \times {10^{ - 15}} .

Note:
There is an inter-relation between equilibrium constant for water, the cell potential (both reduction and oxidation) and the application of Nernst equation. In order to formulate the value of equilibrium constant, the numerical values are substituted in the Nernst equation.