The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follow :
`(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2), Delta_(r)G= +960 kJ mol^(-1)`
The potential difference needed for the electrolytic reduction of aluminium oxide `(Al_(2)O_(3))` at `500^(@)C` is

Options

A. `5.0 V`
B. `4.5 V`
C. `3.0 V`
D. `2.5 V`

Explanation

To find the potential difference needed for the electrolytic reduction of aluminum oxide (Al₂O₃) at 500°C, we can use the relationship between Gibbs free energy change (ΔG) and the cell potential (E) in electrochemistry. The equation that relates these quantities is:

\[ \Delta G = -nFE \]

Where:
- \(\Delta G\) is the Gibbs free energy change (in joules),
- \(n\) is the number of moles of electrons transferred,
- \(F\) is Faraday's constant (approximately \(96485 \, C/mol\)),
- \(E\) is the cell potential (in volts).

### Step 1: Convert ΔG from kJ to J
Given that \(\Delta_rG = +960 \, kJ/mol\), we need to convert this to joules.

\[ \Delta_rG = 960 \, kJ/mol \times 1000 \, J/kJ = 960000 \, J/mol \]

### Step 2: Determine the number of moles of electrons (n)
From the decomposition reaction:

\[ \frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2 \]

We can see that for every 2 moles of Al₂O₃ decomposed, 4 moles of Al are produced, which involves the transfer of 6 moles of electrons (since each Al atom requires 3 electrons for reduction).

Thus, for 1 mole of Al₂O₃, the number of moles of electrons transferred is:

\[ n = 6 \, \text{(for 2 moles of Al from 1 mole of Al₂O₃)} \]

### Step 3: Rearrange the equation to solve for E
Now we can rearrange the equation to find the cell potential \(E\):

\[ E = -\frac{\Delta G}{nF} \]

### Step 4: Substitute the values into the equation
Substituting the values we have:

\[ E = -\frac{960000 \, J/mol}{6 \times 96485 \, C/mol} \]

Calculating the denominator:

\[ 6 \times 96485 = 578910 \, C/mol \]

Now substituting back into the equation:

\[ E = -\frac{960000}{578910} \approx -1.65 \, V \]

Since we are looking for the potential difference needed for the electrolytic process, we take the absolute value:

\[ E \approx 1.65 \, V \]

### Conclusion
The potential difference needed for the electrolytic reduction of aluminum oxide (Al₂O₃) at 500°C is approximately **1.65 V**.

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