Question

Use standard entropies from Appendix D to predict whether the standard potential of the $\mathit{C}\mathit{u}\mathbf{\left|}\mathit{C}{\mathit{u}}^{\mathbf{2}\mathbf{+}}\mathbf{\right|}\mathbf{\left|}\mathit{A}{\mathit{g}}^{\mathbf{+}}\mathbf{\right|}\mathit{A}\mathit{g}$cell, will increase or decrease if the temperature is raised above ${\mathbf{25}}^{\mathbf{o}}\mathit{C}$.

Short answer

The reduction potential is decreased if the temperature is raised above ${\text{25}}^{\text{o}}\text{C}$.

Step-by-step solution

Step (1):- To find overall reaction by adding the two half-reaction

Given cell $Cu\left|C{u}^{2+}\right|\left|A{g}^{+}\right|Ag$

Since at anode oxidation occurs the anode half-reaction is as follows:-

$Cu\left(s\right)\to C{u}^{2+}\left(aq\right)+2e^{-}$

Since at cathode reduction occurs, the cathode half-reaction is as follows:-

$2A{g}^{+}\left(aq\right)+2e^{-}\to 2Ag\left(s\right)$

The overall reaction is obtained by adding the two half-reaction.

${\text{Cu + 2Ag}}^{\text{+}}\to {\text{Cu}}^{\text{2 +}}\text{+ 2Ag}$

Step (2):- To find the entropy change for the reduction.

We have to calculate the entropy change accompanying the overall reactants.

$\begin{array}{rcl}\Delta S& =& {S^0}_{f\left(product\right)}-{S^0}_{f\left(reac\tan ts\right)}\\ \Delta S& =& \left[{S^0}_f\left(C{u}^{2+}\left(aq\right)\right)+2{S^0}_f\left(Ag\left(s\right)\right)\right]-\left[{S^0}_f\left(Cu\left(s\right)\right)+2{S^0}_f\left(A{g}^{+}\left(aq\right)\right)\right]\\ \Delta S& =& \left[\left(99.6J/K\right)+2\left(42.55J/K\right)\right]-\left[(33.15J/K)+2(72.68J/K)\right]\\ \Delta S& =& -99.6J/K+85.10J/K-33.15J/K-145.36J/K\\ \Delta S& =& -193.01J/K\end{array}$

The change in entropy is negative, that is entropy decreases.

Step (3):- Predicting the increase or decrease in reduction potential

The standard Gibbs free energy is the measure of the spontaneity of a chemical reaction. Negative values of Gibbs free energy predicts that the reaction is spontaneous while positive values of Gibbs free energy predict that the reaction is non- spontaneous.

$\begin{array}{rcl}\Delta G^0& =& \Delta H^0-T\Delta S ......1\\ \Delta G^0& =& -nF{E}^0 ......2\\ & & \end{array}$

Both equation 1 and 2 gives an expression for the standard Gibbs free energy. Equating equation 1 and 2 , we get an expression relating the standard reduction potential and the entropy change accompanying the redox reaction.

Thus the standard reduction potential is given by :

$E^0=\frac{T\Delta S-\Delta H^0}{nF}$

Since the entropy change is negative, with increase in temperature the $T∆S$term dominates. This will result in negative values of ${\text{E}}^{\text{o}}$ and consequently positive values of $∆{\text{G}}^{\text{o}}$. Hence, reduction potential will decrease, if the temperature rises above $25°\text{C}$.