Question 18.PS.5: Electrochemical Cells A simple voltaic cell is assembled wit......

Anode  reaction:      Fe(s) → Fe^{2+}(aq) + 2  e^-
Cathode  reaction:      Cu^{2+}(aq) + 2  e^- → Cu(s)

The electrons flow through the wire from the anode to the cathode. Nitrate ions move from the salt bridge into the anode compartment. Sodium ions move from the salt bridge into the cathode compartment. The completed cell diagram is:

Strategy and Explanation   The net reaction shows that iron is being oxidized from Fe(s) to Fe^{2+}(aq) and that Cu^{2+}(aq) is being reduced to Cu(s). We need to decide at which electrodes these reactions occur. Since Fe metal is being oxidized (increase in oxidation number), the electrode in the Fe(s)/Fe^{2+}(aq) compartment is the anode. The electrode in the Cu(s)/Cu^{2+}(aq) compartment must be the cathode because the Cu^{2+} ions are being reduced to Cu metal (decrease in oxidation number).

The half-reactions are

Fe(s) → Fe^{2+}(aq) + 2  e^-                    (oxidation—anode)

Cu^{2+}(aq) + 2  e^- → Cu(s)                      (reduction—cathode)

Electrons flow from their source (the oxidation of the Fe at the iron electrode which is the anode) through the wire to the electrode where they reduce Cu^{2+} ions to copper metal (the Cu electrode which is the cathode). Because positive Fe^{2+} ions are being produced in the anode compartment, negative NO_3^- ions in the salt bridge move into the anode compartment from the salt bridge to balance the overall charge. Because Cu^{2+} ions are being removed from the cathode compartment, Na^+ ions move into the cathode compartment from the salt bridge to replace the charge of the Cu^{2+} ions reduced to Cu metal. The external part of the electrical circuit is shown in the margin.