`DeltaG^(c-)` or the reaction is `,`
`4Al+3O_(2)+6H_(2)O+4overset(c-)(O)H rarr 4 Al(OH)_(4)^(c-)`
`E^(c-)._(cell)=2.73V`
`Delta_(f)G^(c-)._((overset(c)(O)H))=-157k J mol^(-1)`
`Delta_(f)G^(c-)._((overset(c-)(H2O))=-237k J mol^(-1)`

(a)`-3.16xx10^(3)kJ mol^(-1)`
(b)`-0.79xx10^(3)kJmol^(-1)`
(c)`-0.263xx10^(3)k J mol^(-1)`
(d)`+0.263xx10^(3)k J mol^(-1)`

Options

A. `-3.16xx10^(3)kJ mol^(-1)`
B. `-0.79xx10^(3)kJmol^(-1)`
C. `-0.263xx10^(3)k J mol^(-1)`
D. `+0.263xx10^(3)k J mol^(-1)`

Explanation

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the given electrochemical reaction using the relationship between Gibbs free energy and cell potential.

### Step-by-step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ 4Al + 3O_2 + 6H_2O + 4OH^- \rightarrow 4Al(OH)_4^- \]

2. **Determine the Number of Electrons Transferred (n)**:
- At the cathode, the reduction reaction involves oxygen and water gaining electrons:
\[ 3O_2 + 6H_2O + 12e^- \rightarrow 4OH^- \]
- At the anode, aluminum is oxidized:
\[ 4Al + 4OH^- \rightarrow 4Al(OH)_4^- + 12e^- \]
- Therefore, the total number of electrons (n) transferred in the overall reaction is 12.

3. **Use the Formula for Gibbs Free Energy**:
The relationship between Gibbs free energy change and cell potential is given by:
\[ \Delta G° = -nFE° \]
where:
- \( n = 12 \) (number of electrons)
- \( F = 96500 \, C/mol \) (Faraday's constant)
- \( E°_{cell} = 2.73 \, V \) (given cell potential)

4. **Substitute the Values into the Formula**:
\[ \Delta G° = -12 \times 96500 \times 2.73 \]

5. **Calculate ΔG°**:
- First, calculate \( 12 \times 96500 \):
\[ 12 \times 96500 = 1158000 \, C/mol \]
- Now multiply by the cell potential:
\[ 1158000 \times 2.73 = 3164340 \, J/mol \]
- Convert Joules to kilojoules:
\[ \Delta G° = -3164.34 \, kJ/mol \approx -3.16 \times 10^3 \, kJ/mol \]

6. **Final Answer**:
The value of ΔG° for the reaction is:
\[ \Delta G° \approx -3.16 \times 10^3 \, kJ/mol \]
Therefore, the correct option is (a) \(-3.16 \times 10^3 \, kJ/mol\).