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)` of `O_(2)`, the minimum emf required to carry out of the electrolysis of `Al_(2)O_(3)` is `(F=96,500 C mol^(-1))`

Options

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

Explanation

To find the minimum emf required to carry out the electrolysis of Al2O3 based on the given reaction and ΔG, we can follow these steps:

### Step 1: Understand the Reaction
The reaction provided is:
\[ \frac{4}{3} \text{Al} + \text{O}_2 \rightarrow \frac{2}{3} \text{Al}_2\text{O}_3 \]
This indicates that aluminum (Al) reacts with oxygen (O2) to form aluminum oxide (Al2O3).

### Step 2: Identify ΔG
We are given:
\[ \Delta G = -827 \text{ kJ/mol} \]
This value indicates that the reaction is spontaneous under standard conditions.

### Step 3: Convert ΔG to Joules
Since we need ΔG in Joules for our calculations, we convert:
\[ \Delta G = -827 \text{ kJ/mol} = -827 \times 10^3 \text{ J/mol} = -827000 \text{ J/mol} \]

### Step 4: Determine the Number of Electrons (n)
From the reaction, we can see that aluminum (Al) is oxidized. Each Al atom loses 3 electrons. Since we have \(\frac{4}{3}\) moles of Al:
\[ n = \frac{4}{3} \text{ moles of Al} \times 3 \text{ electrons/Al} = 4 \text{ moles of electrons} \]

### Step 5: Use the Relationship Between ΔG, n, F, and E
The relationship is given by:
\[ \Delta G = -nFE \]
Where:
- \(E\) is the emf (electromotive force) we want to find,
- \(F\) is Faraday's constant, \(F = 96500 \text{ C/mol}\).

### Step 6: Rearrange the Equation to Solve for E
Rearranging the equation gives:
\[ E = -\frac{\Delta G}{nF} \]

### Step 7: Substitute the Values
Substituting the known values:
\[ E = -\frac{-827000 \text{ J/mol}}{4 \text{ mol} \times 96500 \text{ C/mol}} \]

### Step 8: Calculate E
Calculating the value:
\[ E = \frac{827000}{4 \times 96500} \]
\[ E = \frac{827000}{386000} \]
\[ E \approx 2.14 \text{ V} \]

### Conclusion
The minimum emf required to carry out the electrolysis of Al2O3 is approximately **2.14 V**.

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