Question

Given the equilibrium constant (K_c) of the reaction:

Cu(s) + 2Ag⁺ (aq) →  Cu²⁺ (aq) + 2Ag(s) is 10 × 10¹⁵, calculate the \(E{^\circ _{cell}}\) of this reaction at 298 K.

\(\left[ {2.303\frac{{RT}}{F}at\;298\;K = 0.059\;V} \right]\)

1. 0.4736 V
2. 0.4736 mV
3. 0.4736 Mv
4. 0.04736 V

Answer 1

Correct Answer - Option 1 : 0.4736 V

Calculation:

According to Nernst equation,

\({E_{{\rm{cell\;}}}} = {\rm{E}}{^\circ _{cell}} - \frac{{2.303RT}}{{nF}}{\rm{log\;}}Q\)

Given, \(\frac{{2.303RT}}{F} = 0.059{\rm{\;V}}\) 

\(\therefore {E_{{\rm{cell}}}} = {\rm{E}}{^\circ _{cell}} - \frac{{0.059}}{n}{\rm{log\;}}Q\)

At equilibrium, E_cell = 0

\({\rm{E}}{^\circ _{cell}} = \frac{{0.059}}{n}{\rm{log\;}}{K_C}\)

For the given reaction, n = 2

Also,

K_C = 10 × 10¹⁵

\(\therefore {\rm{E}}{^\circ _{cell}} = \frac{{0.059}}{2}{\rm{log}}\left( {10 \times {{10}^{15}}} \right)\)

= 0.472 V = 0.473 V