Question #285738

The following cell has a potential of +1.2272 V at 25°C: Zn(s)|ZnCl2(aq 0.0005m)||Hg2Cl2(s)|Hg(l)|Pt(s). Determine E for this cell assuming the Debye Hückel equation is satisfied.



1
Expert's answer
2022-01-10T14:01:42-0500



Zn(s)++ Hg2+(aq) \rightleftharpoons\> Zn(aq)2++Hg(l)Zn^{2+}_{(aq)}+Hg_{(l)}


E=E0+RTnFE=E^0\>'+\frac{RT}{nF} In C0CR\frac{C^*_0}{C^*_R}



Where E0E^0\>' (V) is formal potential


C[MolL1]C^*[Mol\>L^{-1}] is the bulk concentration for the considered species


RTnF=8.314×298.152×96485=0.01285\frac{RT}{nF}=\frac{8.314×298.15}{2×96485}=0.01285


The activity for a very dilute solution is the same as the concentration


E=1.2272+0.01285In0.00051\therefore\>E=1.2272+0.01285\> In\> \>\frac{0.0005}{1}


=1.1295V=1.1295V


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