Half cell reactions for some electrodes are given below
I. `A + e^(-) rarr A^(-) , E^(@) = 0.96V`
II. `B^(-) + e^(-) rarr B^(2-) , E^(@) = -0.12V`
III. `C^(+) + e^(-) rarr C, E^(@) =+0.18V`
IV. `D^(2+) + 2e^(-) rarr D, E^(@) = -1.12V`
Largest potential will be generated in which cell?

Options

A. `A|A^(+)||B^(+)|B`
B. `D|D^(2+)||A|A^(-)`
C. `B^(-)|B^(2-)||C^(+)|C`
D. `D| D^(2+)||C^(+) |C`

Answer

To determine which cell will generate the largest potential, we need to analyze the given half-cell reactions and their standard reduction potentials (E° values).

### Step-by-Step Solution:

1. **Identify the Half-Cell Reactions and Their Potentials:**
- Reaction I: \( A + e^- \rightarrow A^- \), \( E^\circ = 0.96 \, V \)
- Reaction II: \( B^- + e^- \rightarrow B^{2-} \), \( E^\circ = -0.12 \, V \)
- Reaction III: \( C^+ + e^- \rightarrow C \), \( E^\circ = 0.18 \, V \)
- Reaction IV: \( D^{2+} + 2e^- \rightarrow D \), \( E^\circ = -1.12 \, V \)

2. **Determine the Highest Reduction Potential:**
- The highest reduction potential is for reaction I (A) with \( E^\circ = 0.96 \, V \).

3. **Identify the Lowest Reduction Potential for Oxidation:**
- The lowest reduction potential is for reaction IV (D) with \( E^\circ = -1.12 \, V \). This will act as the oxidizing agent.

4. **Calculate the Cell Potential (E°cell):**
- The cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]
- Here, \( E^\circ_{\text{reduction}} = 0.96 \, V \) (for A) and \( E^\circ_{\text{oxidation}} = -1.12 \, V \) (for D).
- Therefore:
\[ E^\circ_{\text{cell}} = 0.96 - (-1.12) = 0.96 + 1.12 = 2.08 \, V \]

5. **Conclusion:**
- The largest potential will be generated in the cell formed by combining electrode A and electrode D.

### Final Answer:
The largest potential will be generated in the cell with A and D, yielding a cell potential of 2.08 V.

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