Aluminium displaces hydrogen from acids, but copper does not. A galvanic 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 V, that of aluminium electrode is

Options

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

Explanation

To find the potential of the aluminum electrode in the galvanic cell consisting of `Cu|Cu^(2+)` and `Al|Al^(3+)`, we can follow these steps:

### Step 1: Understand the Cell Configuration
In the galvanic cell, aluminum (Al) acts as the anode and copper (Cu) acts as the cathode. This is because aluminum can displace hydrogen from acids, indicating that it has a higher tendency to oxidize compared to copper.

### Step 2: Write the Standard Cell Reaction
The overall cell reaction can be represented as:
\[ \text{Al} \rightarrow \text{Al}^{3+} + 3e^- \quad (\text{oxidation at anode}) \]
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (\text{reduction at cathode}) \]

### Step 3: Use the Given Information
We know:
- The standard cell emf (E°_cell) = 2.0 V
- The standard reduction potential of the copper electrode (E°_Cu²⁺/Cu) = +0.34 V

### Step 4: Apply the Formula for Cell Potential
The standard cell potential can be calculated using the formula:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
Where:
- \( E°_{cathode} \) is the potential of the copper electrode.
- \( E°_{anode} \) is the potential of the aluminum electrode.

### Step 5: Rearrange the Formula
We can rearrange the formula to find the potential of the aluminum electrode:
\[ E°_{anode} = E°_{cathode} - E°_{cell} \]

### Step 6: Substitute the Values
Substituting the known values into the equation:
\[ E°_{Al^{3+}/Al} = 0.34 \, \text{V} - 2.0 \, \text{V} \]
\[ E°_{Al^{3+}/Al} = 0.34 \, \text{V} - 2.0 \, \text{V} = -1.66 \, \text{V} \]

### Step 7: Conclusion
The potential of the aluminum electrode is:
\[ E°_{Al^{3+}/Al} = -1.66 \, \text{V} \]