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Question: In the cell \[Pt(s),{H_2}(g)/1barHCl(aq)/Ag(S)/Pt(s)\] the cell potential is 0.92 when \({10^{ - 6}}...

In the cell Pt(s),H2(g)/1barHCl(aq)/Ag(S)/Pt(s)Pt(s),{H_2}(g)/1barHCl(aq)/Ag(S)/Pt(s) the cell potential is 0.92 when 106{10^{ - 6}} molal HClHCl solution is used. The standard electrode potential of (AgCl/AgClAgCl/AgC{l^ - }) electrode is
Given: 2.303RTF=0.06V\dfrac{{2.303RT}}{F} = 0.06V at 298K
A. 0.20V
B. 0.076V
C. 0.040V
D. 0.94V

Explanation

Solution

We have to use Nernst equation for this question by writing anode and cathode reactions separately. In electrochemistry the Nernst equation is an equation that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature and activities of the chemical species undergoing oxidation and reduction.
Ecell=E0cell2.303RTnFlog[H+]2[Cl]2{E_{cell}} = {E^0}_{cell} - \dfrac{{2.303RT}}{{nF}}\log {[{H^ + }]^2}{[C{l^ - }]^2}
Ecell={E_{cell}} = Reduction potential of the electrochemical cell
E0cell={E^0}_{cell} = Standard electrode potential of the electrochemical cell
R is Universal gas constant
T is Temperature
n is number of electrons gained or lost
F represents Faraday constant (charge per mole: 96500 coulombs/mole)

Complete step by step answer:
First, we will have a look at the reaction occurring at anode and at the cathode as follows:
Reaction at anode: H22H++2e{H_2} \to 2{H^ + } + 2{e^ - } (Here hydrogen is undergoing oxidation as it is losing two electrons)
Reaction at cathode: 2AgCl(g)+2e2Ag(g)+2Cl2AgCl(g) + 2{e^ - } \to 2Ag(g) + 2C{l^ - } (Here chlorine is undergoing reduction as it is gaining two electrons)
Overall reaction: H2(g)+2AgCl(g)2H+(aq)+2Ag(s)+2Cl(aq){H_2}(g) + 2AgCl(g) \to 2{H^ + }(aq) + 2Ag(s) + 2Cl(aq)
From the above equations we can see that each atom of hydrogen and chlorine is losing and gaining one electron per atom hence n=1n = 1
Given that: 2.303RTF=0.06V\dfrac{{2.303RT}}{F} = 0.06Vat 298 K
Therefore,
0.92=E0AgCl/Ag,Cl0.061log[106][106]\Rightarrow 0.92 = {E^0}_{AgCl/Ag,C{l^ - }} - \dfrac{{0.06}}{1}\log [{10^{ - 6}}][{10^{ - 6}}] (Since the given concentration of HClHCl is 106{10^{ - 6}})
0.92=E0AgCl/Ag,Cl0.06(12)\Rightarrow 0.92 = {E^0}_{AgCl/Ag,C{l^ - }} - 0.06( - 12)
E0AgCl/Ag,Cl=0.20V\Rightarrow {E^0}_{AgCl/Ag,C{l^ - }} = 0.20V

So, the correct answer is Option A.

Note: The value of n has to be calculated for one atom that is the number of electrons gained or lost per atom.
The overall reaction is a combination of reduction half and oxidation half reactions.
The Nernst equation provides the relation between the cell potential of an electrochemical cell, the standard cell potential, temperature and the reaction quotient.
Even under nonstandard conditions the cell potentials of electrochemical cells can be determined with the help of Nernst equation.