Deduce from the following `E^(c-)` values of half cells, what combination of two half cells would results in a cell with the largest potential?
`{:(I. , A+e(-) rarr A^(c-),,,,E^(c-)=-0.24V),(II., B^(c-)+e^(-)rarr B^(2-),,,,E^(c-)=+1.25V),(III.,C^(c-)+2e^(-) rarr C^(3-),,,,E^(c-)=-1.25V),(IV. , D+2e^(c-)rarr D^(2-),,,, E^(c-)=+0.68V):}`

Options

A. `II,IV`
B. `II,III`
C. `III,IV`
D. `I,II`

Explanation

To determine the combination of two half-cells that results in the largest cell potential, we can follow these steps:

### Step 1: Identify the half-cell potentials
We have the following half-cell reactions and their corresponding standard reduction potentials (E°):

1. \( A^+ + e^- \rightarrow A \), \( E° = -0.24 \, V \)
2. \( B^{2+} + 2e^- \rightarrow B \), \( E° = +1.25 \, V \)
3. \( C^{3+} + 3e^- \rightarrow C \), \( E° = -1.25 \, V \)
4. \( D^{2+} + 2e^- \rightarrow D \), \( E° = +0.68 \, V \)

### Step 2: Determine which half-cell will act as the cathode and which as the anode
In an electrochemical cell, the half-cell with the higher reduction potential will act as the cathode (reduction occurs), and the half-cell with the lower reduction potential will act as the anode (oxidation occurs).

### Step 3: Calculate the cell potential for possible combinations
We will calculate the cell potential (E°cell) for the combinations of half-cells:

1. **Combination of B and A**:
- Cathode: \( B^{2+} + 2e^- \rightarrow B \) (E° = +1.25 V)
- Anode: \( A^+ + e^- \rightarrow A \) (E° = -0.24 V)
- \( E°_{cell} = E°_{cathode} - E°_{anode} = 1.25 - (-0.24) = 1.25 + 0.24 = 1.49 \, V \)

2. **Combination of B and C**:
- Cathode: \( B^{2+} + 2e^- \rightarrow B \) (E° = +1.25 V)
- Anode: \( C^{3+} + 3e^- \rightarrow C \) (E° = -1.25 V)
- \( E°_{cell} = E°_{cathode} - E°_{anode} = 1.25 - (-1.25) = 1.25 + 1.25 = 2.50 \, V \)

3. **Combination of B and D**:
- Cathode: \( B^{2+} + 2e^- \rightarrow B \) (E° = +1.25 V)
- Anode: \( D^{2+} + 2e^- \rightarrow D \) (E° = +0.68 V)
- \( E°_{cell} = E°_{cathode} - E°_{anode} = 1.25 - 0.68 = 0.57 \, V \)

4. **Combination of D and A**:
- Cathode: \( D^{2+} + 2e^- \rightarrow D \) (E° = +0.68 V)
- Anode: \( A^+ + e^- \rightarrow A \) (E° = -0.24 V)
- \( E°_{cell} = E°_{cathode} - E°_{anode} = 0.68 - (-0.24) = 0.68 + 0.24 = 0.92 \, V \)

5. **Combination of D and C**:
- Cathode: \( D^{2+} + 2e^- \rightarrow D \) (E° = +0.68 V)
- Anode: \( C^{3+} + 3e^- \rightarrow C \) (E° = -1.25 V)
- \( E°_{cell} = E°_{cathode} - E°_{anode} = 0.68 - (-1.25) = 0.68 + 1.25 = 1.93 \, V \)

### Step 4: Identify the combination with the largest potential
From the calculated potentials, the combination of half-cells B and C yields the largest potential of **2.50 V**.

### Final Answer
The combination of half-cells that results in the largest potential is **B and C**.

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