Question 20.9: Calculating the Free-Energy Change from Electrode Potentials...
Calculating the Free-Energy Change from Electrode Potentials
Using standard electrode potentials, calculate the standard free-energy change at 25^{\circ} \mathrm{C} for the reaction
\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)
PROBLEM STRATEGY
Substitute into \Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}. Use a table of standard electrode potentials to obtain E_{\text {cell }}^{\circ}. The cell reaction equals the sum of the half-reactions after they have been multiplied by factors so that the electrons cancel in the summation. Note that n is the number of moles of electrons involved in each half-reaction.
The half-reactions, corresponding half-cell potentials, and their sums are displayed below.
\begin{array}{rlr} \mathrm{Zn}(s) & \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} & -E^{\circ}=0.76 \mathrm{~V} \\ 2 \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Ag}(s) & E^{\circ}=0.80 \mathrm{~V} \\ \hline \mathrm{Zn}(s) +2 \mathrm{Ag}^{+}(a q) & \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s) & E_{\text {cell }}^{\circ}=1.56 \mathrm{~V} \end{array}
Note that each half-reaction involves the transfer of two moles of electrons; hence, n=2. Also, E_{\text {cell }}^{\circ}=1.56 \mathrm{~V}, and the faraday constant, F, is 9.65 \times 10^{4} \mathrm{C} / \mathrm{mol} \mathrm{e}^{-}. Therefore,
\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}=-2 \mathrm{~mol} \mathrm{e}^{-} \times 96,500 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-} \times 1.56 \mathrm{~V}=-3.01 \times 10^{5} \mathrm{~J}
Recall that (coulombs ) \times( volts )= joules. Thus, the standard free-energy change is \mathbf{- 3 0 1} \mathbf{k J}.