Question

Consider the following galvanic cell at 25°C:

$\mathit{P}\mathit{t}\left|C{r}^{2+}(0.3 M),C{r}^{3+}\left(2M\right)\right|\left|C{o}^{2+}(0.20 M)\right|\mathit{C}\mathit{o}$
The overall reaction and equilibrium constant value are
$\mathbf{2}\mathit{C}{\mathit{r}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{o}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{C}\mathit{o}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathit{K}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{79}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}$

Calculate the cell potential E for this galvanic cell and$\mathit{\Delta }\mathit{G}$
for the cell reaction at these conditions.

Short answer

The value of $\Delta G\&E_{cell}$ are 65.47kJ &$-0.3393 V$ respectively.

Step-by-step solution

Half-cell reaction at anode and cathode

The half-cell reaction at anode and cathode can be denoted as:

$$\begin{gathered} C{r}^{+2}\left(aq\right)\to C{r}^{+3}\left(aq\right)+e^{-} \\ C{o}^{+2}\left(aq\right)+2e^{-}\to Co\left(s\right) \end{gathered}$$

Calculation of overall cell potential

The overall cell potential can be calculated using,

${E^o}_{cell}={E^o}_{cathode}-{E^o}_{anode}$

On substituting the values,

$$\begin{gathered} {E^o}_{cell}=-0.5 V-(-0.23 V) \\ {E^o}_{cell}=-0.27 V \end{gathered}$$

Expression of equilibrium constant

The expression of the equilibrium constant for chemical equation can be denoted as:

$$\begin{gathered} K=\frac{{\left[C{r}^{+3}\right]}^2}{\left[C{o}^{2+}\right]{\left[C{r}^{+2}\right]}^2} \\ K=\frac{{\left(2\right)}^2}{(0.2){(0.3)}^2} \\ K=222.22 \end{gathered}$$

Expression of Nernst equation

The Nernst equation can be denoted as:

$$\begin{gathered} E_{cell}={E^o}_{cell}-\frac{0.0591}{n}\log Q \\ {E}_{cell}=-0.27-\frac{0.0591}{2}\log (222.22) \\ {E}_{cell}=-0.27-0.069 \\ {E}_{cell}=-0.3393 V \end{gathered}$$

Finding Gibbs free energy change

The Gibbs free energy change can be expressed as:

$$\begin{gathered} \Delta G^o=-nF{E^o}_{cell} \\ \Delta G^o=-(2 mol)(96485 C/mol)(-0.3393 V)\left(\frac{{10}^{-3 }kJ}{1 J}\right) \\ \Delta G^o=65.47 kJ \end{gathered}$$