Question

Apply the Nernst equation to a Daniell cell.

Answer 1

The overall redox reaction in a Daniell cell is \[\ce{Zn_{(s)} + Cu{^{2+}_{(aq)}} -> Zn{^{2+}_{(aq)}} + Cu_{(s)}}\]

Standard electrode potentials are 

\[\ce{E^{\circ}_{{Zn^{2+}/{Zn}}}}\] = −0.76 V

\[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = +0.34 V

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]

= 0.34 − (−0.76) 

= 1.10 V

The general form of the Nernst equation is

\[\ce{E = E{^{\circ}_{cell}} - \frac{0.0591}{n} log Q}\]

For the Daniell cell, the reaction quotient \[\ce{(Q) = \frac{[Zn^{2+}]}{[Cu^{2+}]}}\]

and n = 2

∴ \[\ce{E = 1.10 - \frac{0.0591}{2} log \frac{[Zn^{2+}]}{[Cu^{2+}]}}\]