Gujarat BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry4 Marks
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]}$
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