The half-cell reactions of rusting of iron are
`2H^(+)+1/2O_(2)+2etoH_(2)O," "E^(0)=+1.23V`
`Fe^(2+)+2etoFe," "E^(0)=-0.44V`
`DeltaG^(0)` (in kJ) for the reaction is

Options

A. `-76`
B. `-322`
C. `-122`
D. `-176`

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the rusting of iron, we can use the relationship between Gibbs free energy and the standard cell potential (E°) given by the formula:

\[ \Delta G° = -nFE° \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately 96500 C/mol)
- \( E° \) = standard cell potential in volts

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

The half-cell reactions provided are:

1. Reduction half-reaction:
\[ 2H^+ + \frac{1}{2}O_2 + 2e^- \rightarrow H_2O, \quad E° = +1.23 \, V \]

2. Oxidation half-reaction:
\[ Fe^{2+} + 2e^- \rightarrow Fe, \quad E° = -0.44 \, V \]

### Step 2: Calculate the standard cell potential (E°)

To find the overall standard cell potential, we need to add the reduction potential of the reduction half-reaction and the oxidation potential of the oxidation half-reaction. However, since the oxidation potential is given as a negative value, we will take its absolute value:

\[ E° = E°_{\text{reduction}} + |E°_{\text{oxidation}}| \]
\[ E° = 1.23 \, V + 0.44 \, V = 1.67 \, V \]

### Step 3: Determine the number of moles of electrons (n)

From the half-cell reactions, we can see that 2 moles of electrons are transferred in the overall reaction (as indicated by the 2e^- in both half-reactions).

Thus, \( n = 2 \).

### Step 4: Use the formula to calculate ΔG°

Now we can substitute the values into the ΔG° formula:

\[ \Delta G° = -nFE° \]
\[ \Delta G° = -2 \times 96500 \, C/mol \times 1.67 \, V \]

Calculating this gives:

\[ \Delta G° = -2 \times 96500 \times 1.67 = -322,340 \, J/mol \]

To convert this to kilojoules:

\[ \Delta G° = -322.34 \, kJ/mol \]

### Final Answer

Thus, the value of \( \Delta G° \) for the rusting of iron is approximately:

\[ \Delta G° \approx -322 \, kJ \]