Question

Calculate the \( \Delta G^\circ \) and \( \log K_c \) for the following cell reaction:
\(\text{Fe (s)} + \text{Ag}^+ (aq) \rightleftharpoons \text{Fe}^{2+} (aq) + \text{Ag (s)}\)
Given: \( E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V}, E^\circ_{\text{Ag}^+/ \text{Ag}} = +0.80 \, \text{V} \)
\( 1 F = 96500 \, \text{C mol}^{-1} \)

Hint:

The Nernst equation can help relate the cell potential to equilibrium constants. A more positive standard electrode potential indicates a more spontaneous reaction.

Answer 1

First, calculate the standard cell potential (\( E^\circ_{\text{cell}} \)): \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 - (-0.44) = 1.24 \, \text{V} \] Next, calculate \( \Delta G^\circ \) using the equation: \[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \] Since the number of electrons transferred \( n = 2 \): \[ \Delta G^\circ = -2 \times 96500 \times 1.24 = -239,360 \, \text{J/mol} = -239.36 \, \text{kJ/mol} \] Now, calculate \( \log K_c \) using the relationship: \[ \Delta G^\circ = -RT \ln K_c \] At 25°C,  R = 8.314 , \(\text{J/mol.K}\), T = 298 ,\(\text{K}\) : \[ -239,360 = -(8.314)(298) \ln K_c \] Solving for \( \ln K_c \): \[ \ln K_c = \frac{239,360}{(8.314)(298)} = 96.5 \] Therefore: \[ K_c = e^{96.5} \quad \Rightarrow \quad \log K_c = 96.5 \, \text{(approx.)} \] Thus, \( \Delta G^\circ = -239.36 \, \text{kJ/mol} \) and \( \log K_c = 96.5 \).