To determine which cell generates the largest potential from the given half-cell reactions, we will calculate the standard cell potential (E°cell) for each combination of half-reactions. The cell potential can be calculated using the formula:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
### Step 1: Identify the half-reactions and their potentials
We have the following half-cell reactions and their standard reduction potentials:
1. \( A + e^- \rightarrow A^- \), \( E° = 0.96 \, V \)
2. \( B^- + e^- \rightarrow B^{2-} \), \( E° = -0.12 \, V \)
3. \( C^+ + e^- \rightarrow C \), \( E° = 0.18 \, V \)
4. \( D^{2+} + 2e^- \rightarrow D \), \( E° = -1.12 \, V \)
### Step 2: Calculate E°cell for each combination
#### Combination 1: A and B
- **Anode**: \( B^- + e^- \rightarrow B^{2-} \) (oxidation, reverse the sign)
- **Cathode**: \( A + e^- \rightarrow A^- \)
\[
E°_{cell} = E°_{A} - E°_{B} = 0.96 - (-0.12) = 0.96 + 0.12 = 1.08 \, V
\]
#### Combination 2: A and D
- **Anode**: \( D^{2+} + 2e^- \rightarrow D \) (oxidation, reverse the sign)
- **Cathode**: \( A + e^- \rightarrow A^- \)
\[
E°_{cell} = E°_{A} - E°_{D} = 0.96 - (-1.12) = 0.96 + 1.12 = 2.08 \, V
\]
#### Combination 3: C and B
- **Anode**: \( B^- + e^- \rightarrow B^{2-} \) (oxidation, reverse the sign)
- **Cathode**: \( C^+ + e^- \rightarrow C \)
\[
E°_{cell} = E°_{C} - E°_{B} = 0.18 - (-0.12) = 0.18 + 0.12 = 0.30 \, V
\]
#### Combination 4: C and D
- **Anode**: \( D^{2+} + 2e^- \rightarrow D \) (oxidation, reverse the sign)
- **Cathode**: \( C^+ + e^- \rightarrow C \)
\[
E°_{cell} = E°_{C} - E°_{D} = 0.18 - (-1.12) = 0.18 + 1.12 = 1.30 \, V
\]
### Step 3: Compare the potentials
Now we compare the calculated cell potentials:
- Combination 1 (A and B): \( 1.08 \, V \)
- Combination 2 (A and D): \( 2.08 \, V \)
- Combination 3 (C and B): \( 0.30 \, V \)
- Combination 4 (C and D): \( 1.30 \, V \)
### Conclusion
The largest potential is generated in the cell with half-reactions A and D, yielding a cell potential of \( 2.08 \, V \).
