The rusting of iron takes place as follows `:`
`2H^(o+)+2e^(-) +(1)/(2)O_(2)rarr H_(2)O(l)," "E^(c-)=+1.23V`
`Fe^(2+)+2e^(-) rarr Fe(s)," "E^(c-)=-0.44V`
Calculae `DeltaG^(c-)` for the net process.
(a)`-322kJ mol^(1)`
(b)`-161kJ mol^(-1)`
(c)`-152kJ mol^(-1)`
(d)`-76kJ mol^(-1)`

Options

A. `-322kJ mol^(1)`
B. `-161kJ mol^(-1)`
C. `-152kJ mol^(-1)`
D. `-76kJ mol^(-1)`

Explanation

To calculate the Gibbs free energy change (ΔG°) for the net process of rusting of iron, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.

The two half-reactions given are:
1. \( 2H^+ + 2e^- \rightarrow H_2O \) with \( E^\circ = +1.23 \, V \)
2. \( Fe^{2+} + 2e^- \rightarrow Fe(s) \) with \( E^\circ = -0.44 \, V \)

### Step 2: Determine which half-reaction is oxidation and which is reduction.

- The half-reaction with the higher reduction potential (more positive) will occur at the cathode (reduction).
- The half-reaction with the lower reduction potential (more negative) will occur at the anode (oxidation).

Thus:
- **Cathode (Reduction)**: \( 2H^+ + 2e^- \rightarrow H_2O \) (E° = +1.23 V)
- **Anode (Oxidation)**: \( Fe(s) \rightarrow Fe^{2+} + 2e^- \) (E° = +0.44 V)

### Step 3: Calculate the overall cell potential (E°cell).

The overall cell potential is calculated by subtracting the anode potential from the cathode potential:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
\[ E^\circ_{cell} = 1.23 \, V - (-0.44 \, V) = 1.23 \, V + 0.44 \, V = 1.67 \, V \]

### Step 4: Use the Gibbs free energy equation.

The Gibbs free energy change can be calculated using the equation:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
Where:
- \( n \) = number of moles of electrons transferred (which is 2 for both reactions)
- \( F \) = Faraday's constant \( \approx 96500 \, C/mol \)
- \( E^\circ_{cell} \) = overall cell potential calculated above.

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, C/mol \times 1.67 \, V \]
\[ \Delta G^\circ = -2 \times 96500 \times 1.67 = -322,310 \, J/mol \]
Converting to kJ:
\[ \Delta G^\circ = -322.31 \, kJ/mol \approx -322 \, kJ/mol \]

### Conclusion

The calculated value of \( \Delta G^\circ \) for the net process is approximately \(-322 \, kJ/mol\). Therefore, the correct answer is:
**(a) -322 kJ/mol**