On the basis of information available from the reaction
`(4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1)`, the minimum emf required of `O_(2)` to carry out of the electrolysis of `Al_(2)O_(3)` is `(F=96,500 C mol^(-1))`

Options

A. `2.14V`
B. `4.28 V`
C. `6.42 V`
D. `8.56 V`

Explanation

To solve the problem, we need to calculate the minimum emf required for the electrolysis of Al2O3 based on the given reaction and the Gibbs free energy change (ΔG). Let's break down the steps:

### Step 1: Write down the reaction and identify the components
The reaction given is:
\[ \frac{4}{3} \text{Al} + \text{O}_2 \rightarrow \frac{2}{3} \text{Al}_2\text{O}_3 \]
The ΔG for this reaction is given as -827 kJ/mol.

### Step 2: Convert ΔG from kJ to J
Since the standard units for Gibbs free energy in the equation we will use is in joules, we convert ΔG:
\[ \Delta G = -827 \text{ kJ/mol} = -827 \times 10^3 \text{ J/mol} = -827000 \text{ J/mol} \]

### Step 3: Determine the number of electrons transferred (n)
From the reaction, we can see that oxygen (O2) is reduced to oxide ions (O2-). To balance the reaction, we find that:
\[ \text{O}_2 + 4e^- \rightarrow 2\text{O}^{2-} \]
Thus, the number of electrons transferred (n) is 4.

### Step 4: Use the relationship between ΔG, n, F, and E
The relationship between Gibbs free energy change (ΔG), the number of moles of electrons transferred (n), Faraday's constant (F), and the cell potential (E) is given by:
\[ \Delta G = -nFE \]
Rearranging this formula to find E:
\[ E = -\frac{\Delta G}{nF} \]

### Step 5: Substitute the known values into the equation
Substituting the values we have:
- ΔG = -827000 J/mol
- n = 4
- F = 96500 C/mol

\[ E = -\frac{-827000}{4 \times 96500} \]
Calculating the denominator:
\[ 4 \times 96500 = 386000 \]
Now substituting back:
\[ E = \frac{827000}{386000} \approx 2.14 \text{ V} \]

### Step 6: Final answer
Thus, the minimum emf required for the electrolysis of Al2O3 is approximately:
\[ \boxed{2.14 \text{ V}} \]