The equilibrium constant of a 2 electron redox reaction at 298 K is `3.8xx10^(-3)` The cell potential `E^@` (in V) and the free energy change o `DeltaG^@` (in `kJ "mol"^(–1)` ) for this equilibrium respectively, are

Options

A. `-0.071, - 13.8`
B. `-0.071, 13.8`
C. `0.71, - 13.8`
D. `0.071, - 13.8`

Explanation

To solve the problem, we need to calculate the cell potential \( E^\circ \) and the free energy change \( \Delta G^\circ \) for the given equilibrium constant \( K \) of a two-electron redox reaction at 298 K.

### Step 1: Calculate \( \Delta G^\circ \)

The relationship between the free energy change \( \Delta G^\circ \) and the equilibrium constant \( K \) is given by the equation:

\[ \Delta G^\circ = -2.303 RT \log K \]

Where:
- \( R \) is the universal gas constant \( (8.314 \, \text{J mol}^{-1} \text{K}^{-1}) \)
- \( T \) is the temperature in Kelvin (298 K)
- \( K \) is the equilibrium constant \( (3.8 \times 10^{-3}) \)

Substituting the values into the equation:

\[ \Delta G^\circ = -2.303 \times 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \times 298 \, \text{K} \times \log(3.8 \times 10^{-3}) \]

Calculating \( \log(3.8 \times 10^{-3}) \):

\[ \log(3.8 \times 10^{-3}) \approx -2.421 \]

Now substituting this back into the equation:

\[ \Delta G^\circ = -2.303 \times 8.314 \times 298 \times (-2.421) \]

Calculating this gives:

\[ \Delta G^\circ \approx 13809.38 \, \text{J/mol} \approx 13.8 \, \text{kJ/mol} \]

### Step 2: Calculate \( E^\circ \)

The relationship between \( \Delta G^\circ \) and the cell potential \( E^\circ \) is given by:

\[ \Delta G^\circ = -nFE^\circ \]

Where:
- \( n \) is the number of moles of electrons transferred (2 for this reaction)
- \( F \) is Faraday's constant \( (96500 \, \text{C/mol}) \)

Rearranging the equation to solve for \( E^\circ \):

\[ E^\circ = -\frac{\Delta G^\circ}{nF} \]

Substituting the values:

\[ E^\circ = -\frac{13809.38 \, \text{J/mol}}{2 \times 96500 \, \text{C/mol}} \]

Calculating this gives:

\[ E^\circ \approx -0.071 \, \text{V} \]

### Final Results

- The cell potential \( E^\circ \) is approximately \(-0.071 \, \text{V}\).
- The free energy change \( \Delta G^\circ \) is approximately \( 13.8 \, \text{kJ/mol} \).

### Summary

Thus, the values for the cell potential and free energy change for the given equilibrium constant are:
- \( E^\circ \approx -0.071 \, \text{V} \)
- \( \Delta G^\circ \approx 13.8 \, \text{kJ/mol} \)