Question 25.3: Write the equations for the electrode half reactions and the......
Write the equations for the electrode half reactions and the net cell reaction for the electrochemical cell
Cd(s)|CdSO_{4}(aq)|Hg_{2}SO_{4}(s)|Hg(l)This cell does not have a salt bridge because it has only one electrolyte solution CdSO_{4}(aq) (all the other species are either liquids or solids).
By our convention, oxidation is assumed to take place at the lefthand electrode in the cell diagram. The oxidation of Cd(s) yields Cd^{2+}(aq), so we write
Cd(s) → Cd^{2+}(aq) + 2\ e^− (oxidation)
Because oxidation occurs at the left-hand electrode, reduction must occur at the right-hand electrode. The only element besides cadmium that appears in two different oxidation states in the cell is mercury, which has an oxidation state of 0 in Hg(l) and +1 in water-insoluble Hg_{2}SO_{4}(s). For the reduction at the mercury electrode, therefore, we write
Hg_{2}SO_{4}(s) → 2\ Hg(l) (not balanced)
The balanced equation for the electrode half reaction is
Hg_{2}SO_{4}(s) + 2\ e^− → 2\ Hg(l ) + SO_{4}^{2−}(aq) (reduction)
The sum of the equations for the oxidation and reduction half reactions gives the equation for the net cell reaction:
Cd(s) + Hg_{2}SO_{4}(s) → 2\ Hg(l) + Cd^{2+}(aq) + SO_{4}^{2−}(aq)The cell described here is similar to the Weston standard cell (Figure 25.9), which is widely used as a source of accurately known voltage. The voltage of the cell is 1.018 V and is essentially independent of temperature, a desirable feature for a voltage reference. As we shall show in Section 25-4, in most cases the voltage of a cell changes significantly with the temperature.
