The rusting of iron takes place as `2H^++2e+1/2O_2rarrH_2O(l),E^@=+1.23V`
`Fe^(2+)+2e rarrFe(s),E^@=-0.44V` Thus, `DeltaG^@` for the net process is

Options

A. `-322 "kJ/mol"`
B. `-161"kJ/mol"`
C. `-1522"kJ/mol"`
D. `-76"kJ/mol"`

Answer

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

### Step 1: Identify the half-reactions and their standard electrode potentials (E°)
1. **Reduction half-reaction**:
\[ 2H^+ + 2e^- + \frac{1}{2}O_2 \rightarrow H_2O(l), \quad E° = +1.23 \, V \]
2. **Oxidation half-reaction**:
\[ Fe^{2+} + 2e^- \rightarrow Fe(s), \quad E° = -0.44 \, V \]

### Step 2: Determine the oxidation potential of iron
- The oxidation potential is the negative of the reduction potential for the oxidation half-reaction:
\[ E°_{oxidation} = -(-0.44) = +0.44 \, V \]

### Step 3: Calculate the total standard electrode potential (E°) for the overall reaction
- The overall cell potential (E°) is given by:
\[ E°_{total} = E°_{cathode} + E°_{anode} \]
\[ E°_{total} = 1.23 \, V + 0.44 \, V = 1.67 \, V \]

### Step 4: Calculate ΔG° using the relation ΔG° = -nFE°
- Where:
- n = number of moles of electrons transferred (2 in this case)
- F = Faraday's constant (approximately 96500 C/mol)
- E° = total standard electrode potential (1.67 V)

Using the formula:
\[ \Delta G° = -nFE° \]
Substituting the values:
\[ \Delta G° = -2 \times 96500 \, C/mol \times 1.67 \, V \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ \Delta G° = -2 \times 96500 \times 1.67 = -322310 \, J/mol \]

### Step 6: Convert to kilojoules per mole
To convert from joules to kilojoules, divide by 1000:
\[ \Delta G° = -322.31 \, kJ/mol \]

### Step 7: Round to appropriate significant figures
Rounding to three significant figures gives:
\[ \Delta G° \approx -322 \, kJ/mol \]

### Final Answer
Thus, the standard Gibbs free energy change for the rusting of iron is:
\[ \Delta G° = -322 \, kJ/mol \]

---