Aluminium displaces hydrogen from acids but copper does not. A galvainc cell prepared by combining `Cu//Cu^(2+) and Al//Al^(3+)` has an emf of `2.0` V at 298 K. If the potential of copper electrode is `+0.34` then that of the aluminium electrode is:

Options

A. `+1.66`V
B. `-1.66` V
C. `+2.34`V
D. `-2.3`V

Answer

To find the potential of the aluminum electrode in the galvanic cell, we can use the formula for the cell potential (E_cell):

\[ E_{cell} = E_{cathode} - E_{anode} \]

Where:
- \( E_{cell} \) is the overall cell potential,
- \( E_{cathode} \) is the standard electrode potential of the cathode,
- \( E_{anode} \) is the standard electrode potential of the anode.

### Step-by-Step Solution:

1. **Identify the Components**:
- The galvanic cell is made of copper (Cu) and aluminum (Al).
- The copper electrode is the cathode, and the aluminum electrode is the anode.

2. **Given Values**:
- \( E_{cell} = 2.0 \, V \)
- \( E_{cathode} = +0.34 \, V \) (for Cu)

3. **Set Up the Equation**:
- We know that:
\[ E_{cell} = E_{cathode} - E_{anode} \]
- Rearranging gives us:
\[ E_{anode} = E_{cathode} - E_{cell} \]

4. **Substitute the Known Values**:
- Substitute \( E_{cathode} \) and \( E_{cell} \) into the equation:
\[ E_{anode} = 0.34 \, V - 2.0 \, V \]

5. **Calculate \( E_{anode} \)**:
- Performing the calculation:
\[ E_{anode} = 0.34 \, V - 2.0 \, V = -1.66 \, V \]

6. **Conclusion**:
- The potential of the aluminum electrode is \( -1.66 \, V \).

### Final Answer:
The potential of the aluminum electrode is \( -1.66 \, V \).

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