Question

The value of \(log K_{c}\) for the given cell reaction at 298 K is
\(Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)\).
(Given: \(E^{\circ}_{Cu^{2+}/Cu}=0.34V, E^{\circ}_{Ag^{+}/Ag}=0.80V; \frac{2.303RT}{F}=0.06\))

• A. 45.33
• B. 20.33
• C. 15.33
• D. 30.66

Hint:

When the cell reaction reaches equilibrium, the net potential is zero, allowing us to calculate the equilibrium constant directly from standard electrode potentials.

Answer 1

The Correct Option is C

Concept: At chemical equilibrium, the cell potential (\(E_{cell}\)) becomes zero. The relationship between the standard Gibbs free energy change and the equilibrium constant is described by the Nernst equation: \(E^{\circ}_{cell} = \frac{2.303RT}{nF} \log K_c\).

Step 1: Calculate the standard cell potential (\(E^{\circ}_{cell}\)).
The reaction is: \(Cu \rightarrow Cu^{2+} + 2e^-\) (oxidation, anode) and \(Ag^+ + e^- \rightarrow Ag\) (reduction, cathode). \[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Ag^+/Ag} - E^{\circ}_{Cu^{2+}/Cu} = 0.80V - 0.34V = 0.46V \]

Step 2: Determine the number of electrons transferred (\(n\)).
The balanced equation involves 2 electrons transferred from Copper to Silver (\(Cu \rightarrow Cu^{2+} + 2e^-\) and \(2Ag^+ + 2e^- \rightarrow 2Ag\)). Thus, \(n = 2\).

Step 3: Calculate \(\log K_c\) using the Nernst relationship.
Using the provided constant \(\frac{2.303RT}{F} = 0.06\): \[ 0.46 = \frac{0.06}{n} \log K_c = \frac{0.06}{2} \log K_c = 0.03 \log K_c \] \[ \log K_c = \frac{0.46}{0.03} \approx 15.33 \]