Question

Calculate \(\Delta_r G^\circ\) and \(\log K_C\) of the reaction :
\(\mathrm{Fe^{2+}(aq) + Ag^+(aq) \longrightarrow Fe^{3+}(aq) + Ag(s)}\)
Given : \(E^\circ_{Ag^+/Ag} = 0.80\ \mathrm{V},\ E^\circ_{Fe^{3+}/Fe^{2+}} = 0.77\ \mathrm{V}\)
[\(R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}},\ F = 96500\ \mathrm{C\ mol^{-1}}\)]

Hint:

E0cell = 0.80 - 0.77 = 0.03 V with n = 1. Use dG = -nFE and dG = -2.303 RT log K.

Answer 1

Step 1: Write the cell and find its potential:
The cell is $Pt\,|\,Fe^{2+},Fe^{3+}\,||\,Ag^+\,|\,Ag$. The couple with the higher reduction potential, Ag⁺/Ag, is reduced.
\[ E^\circ_{cell} = 0.80 - 0.77 = +0.03\ \mathrm{V} \]

Step 2: Use the equilibrium relation first:
At 298 K, the Nernst equation at equilibrium gives $E^\circ_{cell} = \dfrac{0.0591}{n}\log K_C$, with $n = 1$.
\[ \log K_C = \frac{0.03}{0.0591} = 0.51 \]
The same value comes from $\log K_C = \dfrac{nFE^\circ}{2.303RT}$ using the given R and F: $\dfrac{96500 \times 0.03}{2.303 \times 8.314 \times 298} = 0.507$.

Step 3: Now find the Gibbs energy:
Use $\Delta_r G^\circ = -nFE^\circ$, where $n = 1$ electron is exchanged.
\[ \Delta_r G^\circ = -1 \times 96500 \times 0.03 = -2895\ \mathrm{J\ mol^{-1}} = -2.895\ \mathrm{kJ\ mol^{-1}} \]
As $\Delta_r G^\circ$ is negative and $K_C > 1$ (log K_C is positive), the forward reaction is favoured.

Final Answer:
Gibbs energy change is -2.895 kJ per mole and log K_C is 0.51. \[ \boxed{\Delta_r G^\circ = -2.895\ \mathrm{kJ\ mol^{-1}},\quad \log K_C = 0.51} \]