Out of the following half cells, which two half cell combination gives the largest \(E_{cell}\) value?
| \(A^{3-} \longrightarrow A^{2-} + e^-\) | \(E^o = + 1.5 V\) -------(I) |
| \(B^+ + e^- \longrightarrow B\) | \(E^o = - 0.5 V\) -------(II) |
| \(C^{2+} + e^- \longrightarrow C^+\) | \(E^o = + 0.5 V\) -------(III) |
| \(D \longrightarrow D^{2+} + 2e^-\) | \(E^o = - 1.5 V\) -------(IV) |
Answer & explanation
Correct answer: option 2
The correct answer is option 2. I and IV.
Given Half-Cells and Their Standard Potentials:
(I) \(A^{3-} \longrightarrow A^{2-} + e^-\) with \(E^\circ = +1.5 \text{ V}\)
(II) \(B^+ + e^- \longrightarrow B\) with \(E^\circ = -0.5 \text{ V}\)
(III) \(C^{2+} + e^- \longrightarrow C^+\) with \(E^\circ = +0.5 \text{ V}\)
(IV) \(D \longrightarrow D^{2+} + 2e^-\) with \(E^\circ = -1.5 \text{ V}\)
The cell potential (\(E_{\text{cell}}\)) is given by:
\(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\)
Where:
Cathode: The half-cell where reduction occurs (higher reduction potential).
Anode: The half-cell where oxidation occurs (lower reduction potential).
Evaluating Each Combination:
1. Combination I and III:
Reduction (Cathode): Half-cell I (\(A^{3-} \longrightarrow A^{2-} + e^-\)) with \(E^\circ = +1.5 \text{ V}\).
Oxidation (Anode): Half-cell III (\(C^+ \longrightarrow C^{2+} + e^-\)) with \(E^\circ = +0.5 \text{ V}\) (reversed to \(-0.5 \text{ V}\) for oxidation).
\(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 1.5 - (-0.5) = 1.5 + 0.5 = +2.0 \text{ V}\)
2. Combination I and IV:
Reduction (Cathode): Half-cell I (\(A^{3-} \longrightarrow A^{2-} + e^-\)) with \(E^\circ = +1.5 \text{ V}\).
Oxidation (Anode): Half-cell IV (\(D \longrightarrow D^{2+} + 2e^-\)) with \(E^\circ = -1.5 \text{ V}\) (reversed to \(+1.5 \text{ V}\) for oxidation).
\(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 1.5 -(- 1.5) = 3 \text{ V}\)
3. Combination II and IV:
Reduction (Cathode): Half-cell II (\(B^+ + e^- \longrightarrow B\)) with \(E^\circ = -0.5 \text{ V}\).
Oxidation (Anode): Half-cell IV (\(D \longrightarrow D^{2+} + 2e^-\)) with \(E^\circ = -1.5 \text{ V}\) (reversed to \(+1.5 \text{ V}\) for oxidation).
\(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = -0.5 - 1.5 = -2.0 \text{ V}\)
4. Combination III and IV:
Reduction (Cathode): Half-cell III (\(C^{2+} + e^- \longrightarrow C^+\)) with \(E^\circ = +0.5 \text{ V}\).
Oxidation (Anode): Half-cell IV (\(D \longrightarrow D^{2+} + 2e^-\)) with \(E^\circ = -1.5 \text{ V}\) (reversed to \(+1.5 \text{ V}\) for oxidation).
\(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.5 - 1.5 = -1.0 \text{ V}\)
Conclusion: The combination of half-cells that gives the largest \(E_{\text{cell}}\) is combination I and IV, where:
\(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 1.5 - (-1.5) = 1.5 + 1.5 = +3.0 \text{ V}\)
Thus, Combination I and IV indeed provides the highest cell potential, confirming that the correct answer is 2. I and IV.