question_answer

The standard electrode potentials at 298 K are given below: \[E{{{}^\circ }_{Z{{n}^{2+}}/Zn}}=-0.76\,V,\]\[E{{{}^\circ }_{F{{e}^{2+}}/Fe}}=-0.44\,V,\]\[E{{{}^\circ }_{{{H}^{+}}/{{H}_{2}}}}=0.0\,V\]  and\[{{E}^{{}^\circ }}_{C{{u}^{2+}}/Cu}=+0.34V.\] Which of the two electrodes should be combined to form a cell having highest EMF? Identify the cathode and the anode and write the cell reaction. Also mention the direction of flow of electrons in the external as well as the internal circuit.

Answer:

                To have maximum EMF, the anode should have the minimum and cathode should have the maximum \[E{}^\circ \]. Therefore, \[Z{{n}^{2+}}/Zn\] couple should be made as the anode while \[C{{u}^{2+}}/Cu\] couple should be made as the cathode. The cell reaction is: \[Zn(s)+C{{u}^{2+}}(aq)\xrightarrow{\,}Z{{n}^{2+}}(aq)+Cu(s)\] \[EMF=E{{{}^\circ }_{C{{u}^{2+}}/Cu}}-E{{{}^\circ }_{Z{{n}^{2+}}/Zn}}\]\[=\,+\,0.34\,-(-0.76)=+\,1.1\,V\]. The direction of flow of electrons is from Zn to Cu in the external circuit and from Cu to Zn in the internal circuit. The direction of flow of current is, however, in the reverse direction, i.e., from Cu to Zn in the external circuit and from Zn to Cu in the internal circuit.