Question 18.PS.6: Determining a Half-Cell Potential The voltaic cell shown in ......
Determining a Half-Cell Potential
The voltaic cell shown in the drawing below generates a potential of E° = 0.36 V under standard conditions at 25 °C. The net cell reaction is
Zn(s) + Cd^{2+}(aq, 1 M) → Zn^{2+}(aq, 1 M) + Cd(s)
The standard half-cell potential for Zn(s)/Zn^{2+}(aq, 1 M) is -0.76 V.
(a) Determine which electrode is the anode and which is the cathode.
(b) Show the direction of electron flow through the circuit outside the cell, and complete the cell diagram.
(c) Calculate the standard potential for the half-cell Cd^{2+}(aq) + 2 e^- → Cd(s).
(a) Zinc is the anode, and cadmium is the cathode.
(b) The completed cell diagram is shown below.
(c) The standard cell potential is -0.40 V.
Strategy and Explanation The electrode where oxidation occurs is the anode.
Because Zn(s) is oxidized to Zn^{2+}(aq), the Zn electrode is the anode. Cadmium(II) ions are reduced at the Cd electrode, so it is the cathode.
The net cell potential and the potential for the Zn(s)/Zn^{2+}(aq, 1 M) half-cell are known, so the value of E° for Cd^{2+}(aq, 1 M) + 2 e^- → Cd(s) can be calculated.
Zn( s ) → Zn^{2+} (aq, 1 M) + 2 e^- E^\circ_{anode} =- 0.76 V (anode)
\underline{Cd^{2+}(aq, 1 M) + 2 e^- → Cd(s) E^\circ_{cathode} =? V (cathode)}
Zn(s) + Cd^{2+}(aq, 1 M) → Zn^{2+}(aq, 1 M) + Cd(s) E^\circ_{cell} = +0.36 V
Using E^\circ_{cell} = E^\circ_{cathode} – E^\circ_{anode}, we can solve for E^\circ_{cathode}.
E^\circ_{cathode} = E^\circ_{cell} + E^\circ_{anode} = 0.36 V + (- 0.76 V) = – 0.40 V
At 25 °C, the value of E° for the Cd^{2+}(aq) + 2 e^- → Cd(s) half-reaction is – 0.40 V.
