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. `-322kJ mol^(-10`
B. `-161kJmol^(-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: Write down the half-reactions and their standard electrode potentials.
The rusting of iron can be represented by the following half-reactions:

1. Reduction of hydrogen ions to water:
\[ 2H^+ + 2e^- \rightarrow H_2O \quad E^\circ = +1.23 \, V \]

2. Reduction of iron ions to solid iron:
\[ Fe^{2+} + 2e^- \rightarrow Fe(s) \quad E^\circ = -0.44 \, V \]

### Step 2: Determine the net reaction.
In the rusting process, iron is oxidized, and hydrogen ions are reduced. The overall reaction can be obtained by reversing the second half-reaction (iron reduction) and adding it to the first half-reaction (water formation):

\[ Fe(s) \rightarrow Fe^{2+} + 2e^- \quad E^\circ = +0.44 \, V \, (\text{reversed}) \]
\[ 2H^+ + 2e^- \rightarrow H_2O \quad E^\circ = +1.23 \, V \]

Adding these reactions gives:
\[ Fe(s) + 2H^+ \rightarrow Fe^{2+} + H_2O \]

### Step 3: Calculate the standard cell potential (E°cell).
The standard cell potential for the overall reaction can be calculated as:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Where:
- \(E^\circ_{cathode} = +1.23 \, V\) (reduction of H⁺)
- \(E^\circ_{anode} = -0.44 \, V\) (oxidation of Fe)

Thus,
\[ E^\circ_{cell} = 1.23 - (-0.44) = 1.23 + 0.44 = 1.67 \, V \]

### Step 4: Calculate ΔG using the formula.
The relationship between the Gibbs free energy change and the cell potential is given by:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
Where:
- \(n = 2\) (number of moles of electrons transferred)
- \(F = 96485 \, C/mol\) (Faraday's constant)
- \(E^\circ_{cell} = 1.67 \, V\)

Substituting the values:
\[ \Delta G^\circ = -2 \times 96485 \, C/mol \times 1.67 \, V \]
\[ \Delta G^\circ = -2 \times 96485 \times 1.67 = -322,310.1 \, J/mol \]

### Step 5: Convert ΔG to kJ/mol.
To convert Joules to kilojoules, divide by 1000:
\[ \Delta G^\circ = -322.31 \, kJ/mol \approx -322.2 \, kJ/mol \]

### Final Answer:
\[ \Delta G^\circ \approx -322.2 \, kJ/mol \]

---