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.

Options

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

Explanation

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

### Step 1: Identify the half-reactions and their standard reduction potentials
The rusting of iron can be represented by two half-reactions:
1. Reduction half-reaction:
\[ 2H^+ + 2e^- + \frac{1}{2}O_2 \rightarrow H_2O \quad E^\circ = +1.23 \, V \]
2. Oxidation half-reaction:
\[ Fe^{2+} + 2e^- \rightarrow Fe(s) \quad E^\circ = -0.44 \, V \]

### Step 2: Calculate the standard cell potential (E°cell)
To find the standard cell potential for the overall reaction, we need to adjust the oxidation potential to a reduction potential:
- The oxidation potential of the iron reaction is the negative of the given reduction potential:
\[ E^\circ_{oxidation} = +0.44 \, V \]
- Now, we can calculate the overall cell potential:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 1.23 \, V - (-0.44 \, V) = 1.23 \, V + 0.44 \, V = 1.67 \, V \]

### Step 3: Calculate ΔG° using the formula
The Gibbs free energy change can be calculated using the formula:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
Where:
- \( n \) = number of moles of electrons transferred (in this case, \( n = 2 \))
- \( F \) = Faraday's constant \( \approx 96500 \, C/mol \)
- \( E^\circ_{cell} \) = standard cell potential calculated above

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, C/mol \times 1.67 \, V \]

### Step 4: Perform the calculation
Calculating this gives:
\[ \Delta G^\circ = -2 \times 96500 \times 1.67 = -322,390 \, J/mol \]
Converting to kilojoules:
\[ \Delta G^\circ = -322.39 \, kJ/mol \]

### Final Answer
Thus, the Gibbs free energy change for the rusting of iron is:
\[ \Delta G^\circ \approx -322.39 \, kJ/mol \]

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