Question

Question:Consider a galvanic cell based on the following theoretical

half-reactions:

What is the value of $\mathbf{∆}{\mathbf{G}}^{\mathbf{\circ }}$and $\mathbf{K}$for this cell?

Short answer

Value of $∆G^{\circ }=-3.1\times {10}^5$

Value of K will be $6.6\times {10}^{54}$

Step-by-step solution

Analyzing the given data

The two half equations can be written as

$$\begin{gathered} M^{4+}+4e^{-}\to M\left(E^{\circ }=0.66V\right) \\ {N}^{3+}+3e^{-}\to N\left(E^{\circ }=0.39V\right) \end{gathered}$$

The equation can be written as,

$$\begin{gathered} 3{\mathrm{M}}^{+4}+4\mathrm{N}\to 3\mathrm{M}+4{\mathrm{N}}^{+3} \\ {\mathrm{E}}_{\mathrm{cell}}^{\circ }={\mathrm{E}}_{\mathrm{cathode}}^{\circ }-{\mathrm{E}}_{\mathrm{anode}}^{\circ } \\ {\mathrm{E}}_{\mathrm{cell}}^{\circ }=0.66\mathrm{V}-0.39\mathrm{V} \\ {\mathrm{E}}_{\mathrm{cell}}^{\circ }=0.27\mathrm{V} \end{gathered}$$

Determining the value of Gibbs free energy (ΔG0)

$$\begin{gathered} ∆G^{\circ }=-nF{E}_{cell}^{\circ } \\ ∆G^{\circ }=-12\mathrm{mol}\times 96485\mathrm{C}/\mathrm{mol}\times 0.27\mathrm{J}/\mathrm{C} \\ ∆G^{\circ }=-312.6\mathrm{kJ} \end{gathered}$$

Hence, change in Gibb’s free energy will be $\mathbf{}\mathbf{-}\mathbf{312}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{kJ}\mathbf{.}$

Determining the value of k

$$\begin{gathered} \mathrm{E}={\mathrm{E}}^{\circ }-\frac{0.0591}{\mathrm{n}}\mathrm{logK} \\ 0=0.27\mathrm{V}-\frac{0.0591}{12}\mathrm{logK} \\ \mathrm{K}=6.6\times {10}^{54} \end{gathered}$$