Question 11.1: Description of a Galvanic Cell Describe completely the galva...
Description of a Galvanic Cell
Describe completely the galvanic cell based on the following half-reactions under standard conditions:
Ag^{+} + e^{-} → Ag ξ° = 0.80 V (1)
Fe^{3+} + e^{-} → Fe^{2+} ξ° = 0.77 V (2)
Since a positive ξ° _{cell} value is required, reaction (2) must run in reverse:
Ag^{+} + e^{-} → Ag ξ° (cathode)= 0.80 V
Fe^{2+} → Fe^{3+} + e^{-} – ξ° (anode)= – 0.77 V
Cell reaction: Ag^{+} (aq) + Fe^{2+}(aq) → Fe^{3+} (aq) + Ag (s) ξ° _{cell} = 0.03 V
Since Ag^{+} receives electrons and Fe^{2+} loses electrons in the cell reaction, the electrons flow from the compartment containing Fe^{2+} to the compartment containing Ag^{+}.
Oxidation occurs in the compartment containing Fe^{2+}. Hence this compartment functions as the anode. Reduction occurs in the compartment containing Ag^{+}, so this compartment is the cathode.
The electrode in the Ag/ Ag^{+} compartment is silver metal, and an inert conductor, such as platinum, must be used in the Fe^{2+} / Fe^{3+} compartment. Appropriate counter ions are assumed to be present. The diagram for this cell is shown in Fig. 11.8
