Question

Given the equilibrium constant:
\(K_C\) of the reaction :
\(\mathrm{Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)}\) is \(10\times10^{15}\), calculate the \(E^{0}_{cell}\) of this reaction at 298 K
\(\left[2.303\,\frac{RT}{F}\ \text{at 298 K} = 0.059\ \mathrm{V}\right]\)

• A. \(0.4736\ \mathrm{mV}\)
• B. \(0.04736\ \mathrm{V}\)
• C. \(0.4736\ \mathrm{V}\)
• D. \(0.04736\ \mathrm{mV}\)

Hint:

Use \(E^0 = \frac{2.303RT}{nF}\log K_C\) with n = 2 and \(K_C = 10^{16}\).

Answer 1

The Correct Option is C

\(K_C = 10\times10^{15} = 10^{16}\) and n = 2 electrons.
\(E^0_{cell} = \frac{2.303RT}{nF}\log K_C = \frac{0.059}{2}\times\log 10^{16} = \frac{0.059}{2}\times16 = 0.472\ \mathrm{V}\).
The nearest option is 0.4736 V (obtained when 0.0592 V is used: \(0.0296\times16 = 0.4736\)).
Answer: (3) 0.4736 V