The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Standard Electrode (Half-Cell) Potentials.

[latex]\begin{array}{rll}\text{anode (oxidation):}&\text{Fe}\left(s\right)\longrightarrow {\text{Fe}}^{2+}\left(aq\right)+{\text{2e}}^{-}&{E}_{{\text{Fe}}^{2+}\text{/Fe}}^{\circ }=-\text{0.447 V}\\ \text{cathode (reduction):}&2\times \left({\text{Ag}}^{\text{+}}\left(aq\right)+{\text{e}}^{-}\longrightarrow \text{Ag}\left(s\right)\right)&{E}_{{\text{Ag}}^{\text{+}}\text{/Ag}}^{\circ }=\text{0.7996 V}\end{array}[/latex]

[latex]{E}_{\text{cell}}^{\circ }={E}_{\text{cathode}}^{\circ }-{E}_{\text{anode}}^{\circ }=\qquad{E}_{{\text{Ag}}^{\text{+}}\text{/Ag}}^{\circ }-{E}_{{\text{Fe}}^{2+}\text{/Fe}}^{\circ }=\text{+1.247 V}[/latex]

Remember that the cell potential for the cathode is not multiplied by two when determining the standard cell potential. With n = 2, the equilibrium constant is then

[latex]\begin{array}{rll}{E}_{\text{cell}}^{\circ }&=&\frac{\text{0.0592 V}}{n}\log{K}\\K&=&{10}^{n\times {E}_{\text{cell}}^{\circ }\text{/}\text{0.0592 V}}\\{}&=&{10}^{2\times \text{1.247 V/0.0592 V}}\\{}&=&{10}^{42.128}\\{}&=&1.3\times {10}^{42}\end{array}[/latex]

The two equilibrium constants differ slightly due to rounding in the constants 0.0257 V and 0.0592 V. The standard free energy is then

[latex]\Delta {G}^{\circ }=-nF{E}_{\text{cell}}^{\circ }[/latex]

[latex]\Delta G^{\circ }=-2\times\text{96,485}\frac{\text{J}}{\text{V}\cdot \text{mol}}\times \text{1.247 V}=-240.6\frac{\text{kJ}}{\text{mol}}[/latex]

The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The K value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products.

Spontaneous; n = 2; [latex]{E}_{\text{cell}}^{\circ }=\text{+0.291 V}[/latex]; [latex]\Delta G^{\circ }=-56.2\frac{\text{kJ}}{\text{mol}}[/latex]; K = 6.8 [latex]\times [/latex] 10⁹.

There are two ways to solve the problem. If the thermodynamic information in Standard Thermodynamic Properties for Selected Substances were available, you could calculate the free energy change. If the free energy change is negative, the process is spontaneous. The other approach, which we will use, requires information like that given in Standard Electrode (Half-Cell) Potentials. Using those data, the cell potential can be determined. If the cell potential is positive, the process is spontaneous. Collecting information from Standard Electrode (Half-Cell) Potentials and the problem,

[latex]\begin{array}{rll}\text{Anode (oxidation):}&\text{Co}\left(s\right)\longrightarrow {\text{Co}}^{2+}\left(aq\right)+{\text{2e}}^{-}&{E}_{{\text{Co}}^{2+}\text{/Co}}^{\circ }=-\text{0.28 V}\\ \text{Cathode (reduction):}&{\text{Fe}}^{2+}\left(aq\right)+{\text{2e}}^{-}\longrightarrow \text{Fe}\left(s\right)&{E}_{{\text{Fe}}^{2+}\text{/Fe}}^{\circ }=-\text{0.447 V}\end{array}[/latex]

[latex]{E}_{\text{cell}}^{\circ }={E}_{\text{cathode}}^{\circ }-{E}_{\text{anode}}^{\circ }=\qquad{-}\text{0.447 V}-\left(-\text{0.28 V}\right)=-\text{0.17 V}[/latex]

Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions yields:

[latex]\begin{array}{rll}Q &=&\frac{\left[\text{Co}^{2+}\right]}{\left[\text{Fe}^{2+}\right]}=\dfrac{0.15M}{1.94M}=0.077\\{E}_{\text{cell}}&=&{E}_{\text{cell}}^{\circ }-\frac{0.0592 V}{n}\log{Q}\\{E}_{\text{cell}}&=&-0.1\text{7 V}-\frac{0.0592 V}{2}\log{0.077}\\{E}_{\text{cell}}&=&-\text{0.17 V}+0.033 V=-0.014 V\end{array}[/latex]

The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.

n = 6; Q = 1440; E_cell = +1.97 V, spontaneous

From the information given:

[latex]\begin{array}{rll}\text{Anode:}&\text{Zn}\left(s\right)\longrightarrow {\text{Zn}}^{2+}\left(aq\text{, 0.10}M\right)+{\text{2e}}^{-}&{E}_{\text{anode}}^{\circ }=-\text{0.7618 V}\\ \text{Cathode:}&{\text{Zn}}^{2+}\left(aq\text{, 0.50}M\right)+{\text{2e}}^{-}\longrightarrow \text{Zn}\left(s\right)&{E}_{\text{cathode}}^{\circ }=-\text{0.7618 V}\\ \\ \text{Overall:}&{\text{Zn}}^{2+}\left(aq\text{, 0.50}M\right)\longrightarrow {\text{Zn}}^{2+}\left(aq\text{, 0.10}M\right)&{E}_{\text{cell}}^{\circ }=\text{0.000 V}\end{array}[/latex]

Substituting into the Nernst equation,

[latex]{E}_{\text{cell}}=\text{0.000 V}-\dfrac{\text{0.0592 V}}{2}\log\dfrac{0.10}{0.50}=+0.021 V[/latex]

The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater (E_cathode > E_anode).