Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
Derive the relationship between standard cell potential and equilibrium constant of cell reaction.
✓
Answer
For any galvanic cell, the overall cell reaction at equilibrium can be represented as, Reactants $\rightleftharpoons$ Products.
[For example for Daniell cell,
$\left. Zn _{( s )}+ Cu _{( aq )}^{2+} \rightleftharpoons Zn _{( aq )}^{2+}+ Cu _{( s )}\right]$
The equilibrium constant, $K$ for the reversible reaction will be, $K=\frac{[\text { Products }]}{[\text { Reactants }]}$ (E.g. For Daniell cell, $\left.K=\frac{\left[ Zn ^{++}\right]}{\left[ Cu ^{++}\right]}\right)$The equilibrium constant is related to the stan-dard free energy change $\Delta G^0$, as follows,
$\Delta G^0=-R T \ln K$
If $E_{\text {cell }}^0$ is the standard cell potential (or emf) of the galvanic cell, then $\Delta G ^0=-$
$nFE _{\text {cell }}^0$
The equilibrium constant is related to the stan-dard free energy change $\Delta G^0$, as follows,
$\Delta G^0=-R T \ln K$
If $E_{\text {cell }}^0$ is the standard cell potential (or emf) of the galvanic cell, then $\Delta G ^0=-nFE _{\text {cell }}^0$
By comparing above equations,
$\Delta G^0=-n F E_{\text {cell }}^0=-R T \ln K$
$\therefore n F E_{\text {cell }}^0=R T \ln K$
$\therefore E_{\text {cell }}^0=\frac{R T}{n F} \ln K$
OR $E_{\text {cell }}^0=\frac{2.303 R T}{n F} \log _{10} K$
At $25^{\circ} C$,
$\frac{2.303 \times R T}{F}=\frac{2.303 \times 8.314 \times 298}{96500}=0.0592$
$\therefore E_{\text {cell }}^0=\frac{0.0592}{n} \log _{10} K$
This is a relation between equilibrium constant and $E_{\text {cell }}^0$
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