Question
Explain dependence of cell potential on concentration.###Explain Nernst equation for cell potential.
✓
Answer
Consider following general reaction taking place in the galvanic cell.
$a A+b B \rightarrow c C+d D$
The cell voltage is given by,
$E_{\text {cell }}=E_{\text {cell }}^0-\frac{R T}{n F} \ln \frac{[ C ]^{ c } \times[ D ]^{ d }}{[ A ]^{ a } \times[ B ]^{ b }}$
OR
$E_{\text {cell }}=E_{\text {cell }}^0-\frac{2.303 R T}{n F} \log _{10} \frac{[ C ]^c \times[ D ]^d}{[ A ]^a \times[ B ]^b}$
where,
$T \rightarrow$ temperature
$R \rightarrow$ Gas constant
$F \rightarrow$ Faraday
$n \rightarrow$ Number of electrons in the redox cell reaction.
At $25^{\circ} C , \frac{2.303 R T}{F}=\frac{2.303 \times 8.314 \times 298}{96500}=0.0592$
$\therefore E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.0592}{n} \log _{10} \frac{[ C ]^{ c } \times[ D ]^{ d }}{[ A ]^{ a } \times[ B ]^{ 5 }}$
This is Nernst equation for cell potential. It is used to calculate cell potential and electrode potentials.
$a A+b B \rightarrow c C+d D$
The cell voltage is given by,
$E_{\text {cell }}=E_{\text {cell }}^0-\frac{R T}{n F} \ln \frac{[ C ]^{ c } \times[ D ]^{ d }}{[ A ]^{ a } \times[ B ]^{ b }}$
OR
$E_{\text {cell }}=E_{\text {cell }}^0-\frac{2.303 R T}{n F} \log _{10} \frac{[ C ]^c \times[ D ]^d}{[ A ]^a \times[ B ]^b}$
where,
$T \rightarrow$ temperature
$R \rightarrow$ Gas constant
$F \rightarrow$ Faraday
$n \rightarrow$ Number of electrons in the redox cell reaction.
At $25^{\circ} C , \frac{2.303 R T}{F}=\frac{2.303 \times 8.314 \times 298}{96500}=0.0592$
$\therefore E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.0592}{n} \log _{10} \frac{[ C ]^{ c } \times[ D ]^{ d }}{[ A ]^{ a } \times[ B ]^{ 5 }}$
This is Nernst equation for cell potential. It is used to calculate cell potential and electrode potentials.
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