Out of the following half cells, which two half cell combination gives the largest \(E_{cell}\) value?

I and IV

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).

\(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 1.5 - 1.5 = 0 \text{ V}\)

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.