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: Split the reaction into two half reactions:
Ag⁺ gains an electron, so silver is the cathode. Fe²⁺ loses an electron, so the \(\mathrm{Fe^{3+}/Fe^{2+}}\) half cell is the anode.
\[ \mathrm{Fe^{2+} \to Fe^{3+} + e^-\quad (anode)} \]
\[ \mathrm{Ag^+ + e^- \to Ag\quad (cathode)} \]
One electron is transferred, so \(n = 1\).

Step 2: Find the standard cell potential:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - 0.77 = 0.03\ \mathrm{V} \]

Step 3: Calculate the standard Gibbs energy:
Use \(\Delta_r G^\circ = -nFE^\circ_{cell}\).
\[ \Delta_r G^\circ = -(1)(96500\ \mathrm{C\ mol^{-1}})(0.03\ \mathrm{V}) = -2895\ \mathrm{J\ mol^{-1}} \]
That is \(-2.895\ \mathrm{kJ\ mol^{-1}}\). The negative sign shows the reaction is spontaneous.

Step 4: Calculate log K_C:
Use \(\Delta_r G^\circ = -2.303\,RT\log K_C\) with \(T = 298\ \mathrm{K}\).
\[ \log K_C = \frac{2895}{2.303 \times 8.314 \times 298} = \frac{2895}{5705.8} = 0.507 \]

Final Answer:
The standard Gibbs energy change is -2895 J per mole and log K_C is about 0.51. \[ \boxed{\Delta_r G^\circ = -2895\ \mathrm{J\ mol^{-1}},\quad \log K_C \approx 0.51} \]