Question 20.10: Calculating the Cell emf from Free-Energy Change Suppose the...
Calculating the Cell emf from Free-Energy Change
Suppose the reaction of zinc metal and chlorine gas is utilized in a voltaic cell in which zinc ions and chloride ions are formed in aqueous solution.
\mathrm{Zn}(s)+\mathrm{Cl}_{2}(g) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{Zn}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)
Calculate the standard emf for this cell at 25^{\circ} \mathrm{C} from standard free energies of formation (see Appendix C).
PROBLEM STRATEGY
Calculate \Delta G^{\circ} and substitute it and the value of n into the equation \Delta G^{\circ}= -n F E_{\text {cell. }}^{\circ} Solve for E_{\text {cell. }}^{\circ}
Write the equation with \Delta G_{f}^{\circ} ‘s beneath.
\begin{array}{ccccc} \mathrm{Zn}(s)& +\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{Zn}^{2+}(a q)+ & 2 \mathrm{Cl}^{-}(a q) \\ \Delta G_{f}^{\circ}: 0 & 0 & -147.0 & 2 \times(-131.3) \mathrm{kJ} \end{array}
Hence,
\begin{aligned} \Delta G^{\circ} & \left.=\sum n \Delta G_{f}^{\circ} \text { (products }\right)-\sum m \Delta G_{f}^{\circ}(\text { reactants }) \\ & =[-147.0+2 \times(-131.3)] \mathrm{kJ} \\ & =-410 \mathrm{~kJ}=-4.10 \times 10^{5} \mathrm{~J} \end{aligned}
You obtain n by splitting the reaction into half-reactions.
\begin{aligned} & \mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} \\ & \mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q) \end{aligned}
Each half-reaction involves the transfer of two moles of electrons, so n=2. Now you substitute into
\begin{aligned} \Delta G^{\circ} & =-n F E_{\text {cell }}^{\circ} \\ -4.10 \times 10^{5} \mathrm{~J} & =-2 \mathrm{~mol} \mathrm{e}^{-} \times 96,500 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-} \times 10^{4} \mathrm{C} \times E_{\text {cell }}^{\circ} \end{aligned}
Solving for E_{\text {cell }}^{\circ}, you get \boldsymbol{E}_{\text {cell }}^{\circ}=\mathbf{2 . 1 2} \mathrm{V}.