Question

The reaction \[\ce{Zn_{(s)} + Co^{2+} -> Co_{(s)} + Zn^{2+}}\] occurs in a cell. Write the electrode reactions and compute the standard emf of the cell. 

Given that \[\ce{E^{\circ}_{{Zn/{Zn}^{2+}}}}\] = +0.76 V and \[\ce{E^{\circ}_{{Co/{Co}^{2+}}}}\] = 0.2 V.

Answer 1

Zinc electrodes have a higher oxidation potential (+0.76 V) than cobalt electrodes (+0.28 V); hence, they operate as anodes. Thus, the cell corresponding to the given reaction can be represented as:

\[\ce{Zn_{(s)} | Zn{^{2+}_{(aq)}} || Co{^{2+}_{(aq)}} | Co_{(s)}}\]

The electrode reactions are as follows.

At anode: \[\ce{Zn_{(s)} -> Zn{^{2+}_{(aq)}} + 2e-}\]

At cathode: \[\ce{Co{^{2+}_{(aq)}} + 2e- -> Co_{(s)}}\]

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

\[\ce{E^{\circ}_{{Co{^{2+}}/{Co}}} = -E^{\circ}_{{Co/{Co}^{2+}}}}\] = −0.28 V

The emf of this cell is given by

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{R}} - E{^{\circ}_{L}}}\]

= \[\ce{E^{\circ}_{{Co^{2+}/{Co}}} - E^{\circ}_{{Zn^{2+}/{Zn}}}}\]

= −0.28 − (−0.76)

= 0.48 V