Question

The standard emf of the galvanic cell involves $3$ moles of electrons in a redox reaction$0.59\mathrm{V}$. The equilibrium constant in the cell reaction is?

Answer 1

Step 1: Given data:

• The standard emf of a galvanic cell that involves $3$ moles of electrons in its redox reaction is$0.59\mathrm{volt}$.

$$\begin{gathered} {{\mathrm{E}}_{\mathrm{cell}}}^{°}=0.59\mathrm{volt} \\ \mathrm{n}=3 \end{gathered}$$

Step 2: Formula for the equilibrium constant:

• The equilibrium constant for the reaction of the cell is given as follows:

$$\begin{gathered} \ln \mathrm{K}=\frac{\mathrm{nFE}{°}_{\mathrm{cell}}}{\mathrm{RT}} \\ \mathrm{R}=\mathrm{Gas}\mathrm{constant}=8.314\mathrm{J} \\ \mathrm{F}=\mathrm{Faraday}=96500\mathrm{Coulombs}\mathrm{per}\mathrm{mole} \\ \mathrm{T}=\mathrm{Temperature}=298\mathrm{K} \end{gathered}$$

Step 3: Substituting the values:

• Substituting the values in the equilibrium constant equation,

$$\begin{gathered} \ln \mathrm{K}=\frac{\mathrm{nFE}{°}_{\mathrm{cell}}}{\mathrm{RT}} \\ \ln \mathrm{K}=\frac{3\mathrm{x}96500\mathrm{x}0.59}{8.314\mathrm{x}298} \\ \ln \mathrm{K}=68.9 \\ \mathrm{K}~{10}^{30} \end{gathered}$$

• Therefore, the equilibrium constant for the reaction of the cell is ${10}^{30}$.