Question
Derive nernst equation for Daniel cell
✓
Answer
→ Concentration of Electrolyte in Electrochemic cell is Unity (1M). It is not true always.
→ Nernst derive equation for calculating E when concentration of electrolyte is not unin this equation is known as nernst equation.
$M ^{ n +}( aq )+ ne ^{-} \rightarrow M ( s )$
→ Electrode potential of above reaction is
$E _{\left( M ^{ n +} \mid M \right)}= E _{\left( M ^{ n +} \mid M \right)}^{\ominus}-\frac{ RT }{ nF } \ln \frac{1}{\left[ M ^{ n +}\right]}$
but concentration of solid M is taken as unity and we have
$E _{\left( M ^{ n +} \mid M \right)}= E _{\left( M ^{ n +} \mid M \right)}^{\ominus}-\frac{ RT }{ nF } \ln \frac{1}{\left[ M ^{ n +}\right]}$
Where, $E ^{\ominus}{ }_{ M ^{ n +} \mid M }=$ Standard Electrode potential
$
\begin{array}{l}
R =\text { gas constant } \\
T =\text { Temperature } \\
F =\text { Faraday }\left(96427 C \cdot mol ^{-1}\right. \text { ) } \\
{\left[ M ^{ n +}\right]=\text { concentration of } M ^{ n +}} \\
n =\text { No. of Electron }
\end{array}
$
→ For Daniell Cell Zn act as anode and Cu act as cathode so,
For cathode :
$E _{\left( Cu ^{2+} \mid Cu \right)}= E _{\left( Cu ^{2+} \mid Cu \right)}^{\ominus}-\frac{ RT }{2 F} \ln \frac{1}{\left[ Cu ^{2+}( aq )\right]}$
For Anode :
$E _{\left( Zn ^{2+} \mid Zn \right)}= E _{\left( Zn ^{2+} \mid Zn \right)}^{\ominus}-\frac{ RT }{2 F} \ln \frac{1}{\left[ Zn ^{2+}( aq )\right]}$
The cell potential,
$
\begin{aligned}
E _{\text {cell }}= & E _{\left( Cu ^{2+} \mid Cu \right)}- E _{\left( Zn ^{2+} \mid Zn \right)} \\
= & E _{\left( Cu ^{2+} \mid Cu \right)}-\frac{ RT }{2 F} \ln \frac{1}{\left[ Cu ^{2+}( aq )\right]} \\
& \quad- E _{\left( Zn ^{2+} \mid Zn \right)}^{\ominus}+\frac{ RT }{2 F} \ln \frac{1}{\left[ Zn ^{2+}( aq )\right]} \\
= & E _{\left( Cu ^{2+} \mid Cu \right)}^{\ominus}- E _{\left( Zn ^{2+} \mid Zn \right)} \\
& -\frac{ RT }{2 F} \ln \left(\ln \frac{1}{\left[ Cu ^{2+}( aq )\right]}-\ln \frac{1}{\left[ Zn ^{2+}( aq )\right]}\right) \\
E _{\text {cell }}= & E _{\text {cell }}^{\ominus}-\frac{ RT }{2 F} \ln \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}
\end{aligned}
$
→ It can be seen that $E _{\text {(cell) }}$ depends on the concentration of both $Cu ^{2+}$ and $Zn ^{2+}$ ions.
It increase with increase in the concentration of $Cu ^{2+}$ ions and decrease in the concentration of $Zn ^{2+}$ ions.
$
\begin{array}{l}
→ R =8.314 J mol ^{-1} k ^{-1} \\
T=298 K \\
F =96487 C \\
ln =2.303 log
\end{array}
$
Substituting this value in above equation.
$
E _{\text {(cell) }}= E _{\text {(cell) }}^{\ominus}-\frac{0.059}{2} \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$
→ Nernst derive equation for calculating E when concentration of electrolyte is not unin this equation is known as nernst equation.
$M ^{ n +}( aq )+ ne ^{-} \rightarrow M ( s )$
→ Electrode potential of above reaction is
$E _{\left( M ^{ n +} \mid M \right)}= E _{\left( M ^{ n +} \mid M \right)}^{\ominus}-\frac{ RT }{ nF } \ln \frac{1}{\left[ M ^{ n +}\right]}$
but concentration of solid M is taken as unity and we have
$E _{\left( M ^{ n +} \mid M \right)}= E _{\left( M ^{ n +} \mid M \right)}^{\ominus}-\frac{ RT }{ nF } \ln \frac{1}{\left[ M ^{ n +}\right]}$
Where, $E ^{\ominus}{ }_{ M ^{ n +} \mid M }=$ Standard Electrode potential
$
\begin{array}{l}
R =\text { gas constant } \\
T =\text { Temperature } \\
F =\text { Faraday }\left(96427 C \cdot mol ^{-1}\right. \text { ) } \\
{\left[ M ^{ n +}\right]=\text { concentration of } M ^{ n +}} \\
n =\text { No. of Electron }
\end{array}
$
→ For Daniell Cell Zn act as anode and Cu act as cathode so,
For cathode :
$E _{\left( Cu ^{2+} \mid Cu \right)}= E _{\left( Cu ^{2+} \mid Cu \right)}^{\ominus}-\frac{ RT }{2 F} \ln \frac{1}{\left[ Cu ^{2+}( aq )\right]}$
For Anode :
$E _{\left( Zn ^{2+} \mid Zn \right)}= E _{\left( Zn ^{2+} \mid Zn \right)}^{\ominus}-\frac{ RT }{2 F} \ln \frac{1}{\left[ Zn ^{2+}( aq )\right]}$
The cell potential,
$
\begin{aligned}
E _{\text {cell }}= & E _{\left( Cu ^{2+} \mid Cu \right)}- E _{\left( Zn ^{2+} \mid Zn \right)} \\
= & E _{\left( Cu ^{2+} \mid Cu \right)}-\frac{ RT }{2 F} \ln \frac{1}{\left[ Cu ^{2+}( aq )\right]} \\
& \quad- E _{\left( Zn ^{2+} \mid Zn \right)}^{\ominus}+\frac{ RT }{2 F} \ln \frac{1}{\left[ Zn ^{2+}( aq )\right]} \\
= & E _{\left( Cu ^{2+} \mid Cu \right)}^{\ominus}- E _{\left( Zn ^{2+} \mid Zn \right)} \\
& -\frac{ RT }{2 F} \ln \left(\ln \frac{1}{\left[ Cu ^{2+}( aq )\right]}-\ln \frac{1}{\left[ Zn ^{2+}( aq )\right]}\right) \\
E _{\text {cell }}= & E _{\text {cell }}^{\ominus}-\frac{ RT }{2 F} \ln \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}
\end{aligned}
$
→ It can be seen that $E _{\text {(cell) }}$ depends on the concentration of both $Cu ^{2+}$ and $Zn ^{2+}$ ions.
It increase with increase in the concentration of $Cu ^{2+}$ ions and decrease in the concentration of $Zn ^{2+}$ ions.
$
\begin{array}{l}
→ R =8.314 J mol ^{-1} k ^{-1} \\
T=298 K \\
F =96487 C \\
ln =2.303 log
\end{array}
$
Substituting this value in above equation.
$
E _{\text {(cell) }}= E _{\text {(cell) }}^{\ominus}-\frac{0.059}{2} \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$
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