The half cell reaction for rusting of iron are:
`2H^(+)+2e^(-)+(1)/(2)O_(2)rarrH_(2)O(l), E^(@)=+1.23V`
`Fe^(2+)+2e^(-)rarrFe(s), E^(@)=-0.44V`
`DeltaG^(@)` (in KJ) for the reaction is

Options

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

Answer

To calculate the change in standard Gibbs free energy (ΔG°) for the rusting of iron, we can use the following formula:

\[ \Delta G^{\circ} = -nF E^{\circ}_{\text{cell}} \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- \( E^{\circ}_{\text{cell}} \) = standard cell potential

### Step 1: Identify the half-reactions and their standard potentials
The half-cell reactions given are:
1. \( 2H^+ + 2e^- \rightarrow H_2O(l) \), \( E^{\circ} = +1.23 \, \text{V} \) (Cathode reaction)
2. \( Fe^{2+} + 2e^- \rightarrow Fe(s) \), \( E^{\circ} = -0.44 \, \text{V} \) (Anode reaction)

### Step 2: Determine the standard cell potential \( E^{\circ}_{\text{cell}} \)
The standard cell potential is calculated using the formula:

\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]

Substituting the values:

\[ E^{\circ}_{\text{cell}} = 1.23 \, \text{V} - (-0.44 \, \text{V}) = 1.23 \, \text{V} + 0.44 \, \text{V} = 1.67 \, \text{V} \]

### Step 3: Determine the number of electrons transferred \( n \)
From the half-reactions, we see that 2 electrons are involved in the overall reaction (as both half-reactions involve 2 electrons).

### Step 4: Calculate \( \Delta G^{\circ} \)
Now we can substitute the values into the ΔG° equation:

\[ \Delta G^{\circ} = -nF E^{\circ}_{\text{cell}} \]

Substituting the known values:

\[ \Delta G^{\circ} = -2 \times 96500 \, \text{C/mol} \times 1.67 \, \text{V} \]

Calculating this gives:

\[ \Delta G^{\circ} = -2 \times 96500 \times 1.67 = -322,180 \, \text{J} \]

### Step 5: Convert to kilojoules
To convert joules to kilojoules, we divide by 1000:

\[ \Delta G^{\circ} = -322.18 \, \text{kJ} \]

### Final Answer
Thus, the change in standard Gibbs free energy for the rusting of iron is approximately:

\[ \Delta G^{\circ} \approx -322 \, \text{kJ} \]

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