Question
Derive nernst equation for Daniel cell
✓
Answer
$\rightarrow $ Concentration of Electrolyte in Electrochemic cell is Unity $(1M)$. It is not true always.
$\rightarrow$ Nernst derive equation for calculating $E$ when concentration of electrolyte is not unin this equation is known as nernst equation.
$\ce{M ^{ n +}( aq )+ ne ^{-} \rightarrow M ( s )}$
$\rightarrow$ 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
$ 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 }
$
$\rightarrow$ 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,
$ 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)$
$E _{\text {cell }}= E _{\text {cell }}^{\ominus}-\frac{ RT }{2 F} \ln \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$
$\rightarrow$ 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.
$ \rightarrow R =8.314 J \ \text{mol} ^{-1} k ^{-1}$
$T=298 K$
$F =96487 C$
$ln =2.303 \log $
Substituting this value in above equation.
$E _{\text {(cell) }}= E _{\text {(cell) }}^{\ominus}-\frac{0.059}{2} \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$
$\rightarrow$ Nernst derive equation for calculating $E$ when concentration of electrolyte is not unin this equation is known as nernst equation.
$\ce{M ^{ n +}( aq )+ ne ^{-} \rightarrow M ( s )}$
$\rightarrow$ 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
$ 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 }
$
$\rightarrow$ 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,
$ 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)$
$E _{\text {cell }}= E _{\text {cell }}^{\ominus}-\frac{ RT }{2 F} \ln \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$
$\rightarrow$ 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.
$ \rightarrow R =8.314 J \ \text{mol} ^{-1} k ^{-1}$
$T=298 K$
$F =96487 C$
$ln =2.303 \log $
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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