For the reaction :
`4Al+ 3O_(2) + 6H_(2)O+ 4OH^(-) rightarrow4A1(OH)_(4)^(-) E^(@)_(cell) = 2.73V`.
Standard Gibb's free energy change for the reaction is :

Options

A. `-3.16xx10^(3) kJ`
B. `-0.7 6xx10^(3) kJ`
C. `-0.263xx10^(3) kJ`
D. `+0.263xx10^(3) kJ`

Explanation

To find the standard Gibbs free energy change (\( \Delta G^\circ \)) for the given reaction, we can use the relationship between Gibbs free energy and cell potential:

\[ \Delta G^\circ = -nFE^\circ \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- \( E^\circ \) = standard cell potential (given as \( 2.73 \, \text{V} \))

### Step 1: Determine the number of moles of electrons transferred (\( n \))

In the reaction:
\[ 4Al + 3O_2 + 6H_2O + 4OH^- \rightarrow 4Al(OH)_4^- \]

Aluminum (Al) goes from an oxidation state of 0 to +3. Since there are 4 moles of Al, the total change in oxidation state is:
\[ 4 \times 3 = 12 \, \text{electrons} \]
Thus, \( n = 12 \).

### Step 2: Substitute the values into the Gibbs free energy equation

Now we can substitute \( n \), \( F \), and \( E^\circ \) into the Gibbs free energy equation:
\[ \Delta G^\circ = -nFE^\circ = -12 \times 96500 \, \text{C/mol} \times 2.73 \, \text{V} \]

### Step 3: Calculate \( \Delta G^\circ \)

Calculating the above expression:
\[ \Delta G^\circ = -12 \times 96500 \times 2.73 \]
Calculating \( 12 \times 96500 = 1158000 \):
\[ \Delta G^\circ = -1158000 \times 2.73 = -3161610 \, \text{J/mol} \]

### Step 4: Convert Joules to Kilojoules

Since the answer is typically expressed in kilojoules:
\[ \Delta G^\circ = -3161610 \, \text{J/mol} = -3161.61 \, \text{kJ/mol} \]

### Final Answer

Thus, the standard Gibbs free energy change for the reaction is:
\[ \Delta G^\circ \approx -3161.61 \, \text{kJ/mol} \quad (\text{rounded to } -3.16 \times 10^3 \, \text{kJ/mol}) \]