`DeltaG` for the reaction `:`
`(4)/(3) Al+O_(2)rarr (2)/(3)Al_(2)O_(3)`
is `-772 kJ mol^(-1)` of `O_(2)`.
Calculate the minimum `EMF` in volts required to carry out an electrolysis of `Al_(2)O_(3)`

Explanation

To calculate 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 given is:
\[ \frac{4}{3} \text{Al} + \text{O}_2 \rightarrow \frac{2}{3} \text{Al}_2\text{O}_3 \]
This indicates that aluminum is being oxidized to form aluminum oxide.

### Step 2: Identify the Change in Gibbs Free Energy (ΔG)
We are given:
\[ \Delta G = -772 \text{ kJ/mol of } \text{O}_2 \]
We need to convert this value into joules:
\[ \Delta G = -772 \times 10^3 \text{ J/mol of } \text{O}_2 = -772000 \text{ J/mol} \]

### Step 3: Determine the Number of Electrons Transferred (n)
From the reaction, we can see that:
- 4/3 moles of Al are oxidized.
- Each mole of Al loses 3 electrons to form Al³⁺.

Thus, for 4/3 moles of Al:
\[ n = 3 \times \frac{4}{3} = 4 \text{ moles of electrons} \]

### Step 4: Use the Relationship Between ΔG and EMF
The relationship between ΔG and EMF (E) is given by the equation:
\[ \Delta G = -nFE \]
Where:
- F is Faraday's constant, approximately \(96500 \text{ C/mol}\).

### Step 5: Rearrange the Equation to Solve for EMF (E)
Rearranging the equation gives:
\[ E = -\frac{\Delta G}{nF} \]

### Step 6: Substitute the Values into the Equation
Substituting the values we have:
\[ E = -\frac{-772000 \text{ J/mol}}{4 \times 96500 \text{ C/mol}} \]

### Step 7: Calculate the EMF
Calculating the denominator:
\[ 4 \times 96500 = 386000 \text{ C/mol} \]
Now substituting back:
\[ E = \frac{772000}{386000} \approx 2 \text{ volts} \]

### Final Answer
The minimum EMF required to carry out the electrolysis of Al₂O₃ is:
\[ \boxed{2 \text{ volts}} \]