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

Step 1: Recall the link at equilibrium.
At equilibrium the cell voltage is zero and we use $E^\circ_{cell}=\frac{2.303RT}{nF}\log K_c$.
Step 2: Find the standard cell voltage.
Copper is oxidised (anode) and silver is reduced (cathode). \[ E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.80-0.34=0.46\ \text{V} \]
Step 3: Count electrons.
Copper gives 2 electrons and two silver ions take one each, so $n=2$.
Step 4: Put values in the formula.
Using $\frac{2.303RT}{F}=0.06$: \[ 0.46=\frac{0.06}{2}\log K_c=0.03\log K_c \]
Step 5: Solve for log Kc.
\[ \log K_c=\frac{0.46}{0.03}\approx15.33 \]
Step 6: State the answer.
\[ \boxed{\log K_c\approx15.33} \]