Question

Write Nernst equation for daniel cell at equilibrium condition and derive relationship between${{\mathrm{E}}^{°}}_{\mathrm{cell}}$ and equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$.

Answer 1

Step 1: Writing equation for equilibrium condition:

• Nernst equation for Daniel cell at equilibrium condition,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}°–\frac{0.592}{\mathrm{n}}\mathrm{logK} \\ {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}°–\frac{0.592}{2}\log \frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]} \end{gathered}$$

Step 2: Stating the relationship:

• Relationship between ${{\mathrm{E}}^{°}}_{\mathrm{cell}}$ and equilibrium constant $\left({\mathrm{K}}_{\mathrm{c}}\right)$

$$\begin{gathered} \mathrm{\Delta }\mathrm{G}°=–{\mathrm{nFE}}_{\mathrm{cell}}°—\left[1\right] \\ \mathrm{\Delta }\mathrm{G}°=-\mathrm{RTlnK}—\left[2\right] \end{gathered}$$

• From the equation$\left[1\right]\&\left[2\right]$, we get:

$$\begin{gathered} -{\mathrm{nFE}}_{\mathrm{Cell}}^{°}=-2.303\mathrm{RT}\log \mathrm{K} \\ {\mathrm{E}}_{\mathrm{Cell}}^{°}=\frac{2.303\mathrm{RT}\log \mathrm{K}}{\mathrm{nF}} \end{gathered}$$

Step 3: Substituting the values:

$$\begin{gathered} \mathrm{At}{25}^{°}\mathrm{C} \\ {\mathrm{E}}_{\mathrm{cell}}^{°}=\frac{2.303\mathrm{RT}}{\mathrm{nF}}\log \mathrm{K} \\ \mathrm{R}(\mathrm{gas}\mathrm{constant})=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1} \\ \mathrm{F}\left(\mathrm{Faraday}\right)=96487\mathrm{coulomb} \\ \mathrm{Hence},{\mathrm{E}}_{\mathrm{cell}}^{°}=\frac{0.0592}{\mathrm{n}}\log \mathrm{K} \end{gathered}$$