The following reaction occurs during rusting of iron
`2H^(+)+2e+(1)/(2)O_(2)rarrH_(2)O,E^(@)=+1.23V`
`Fe^(2+)+2erarr Fe(s), E^(@)=0.44V`
Calculate magnitude of `DeltaG^(@)(kJ)` for the net process
`Fe(s)+2H^(+)+(1)/(2)O_(2)rarrFe^(2+)+H_(2)O`

Answer

To calculate the magnitude of \(\Delta G^\circ\) for the net process of rusting of iron, we will follow these steps:

### Step 1: Identify the Half-Reactions
We have two half-reactions given:
1. Reduction of oxygen:
\[ 2H^+ + 2e^- + \frac{1}{2}O_2 \rightarrow H_2O, \quad E^\circ = +1.23 \, V \]
2. Reduction of iron:
\[ Fe^{2+} + 2e^- \rightarrow Fe(s), \quad E^\circ = +0.44 \, V \]

### Step 2: Write the Overall Reaction
The overall reaction for rusting of iron can be written as:
\[ Fe(s) + 2H^+ + \frac{1}{2}O_2 \rightarrow Fe^{2+} + H_2O \]

### Step 3: Determine the Oxidation and Reduction Potentials
- The oxidation potential for iron (since it is being oxidized from \(Fe\) to \(Fe^{2+}\)) is the negative of the reduction potential:
\[ E^\circ_{\text{oxidation}} = -E^\circ_{\text{reduction}} = -0.44 \, V \]

- The reduction potential for the oxygen reaction remains:
\[ E^\circ_{\text{reduction}} = +1.23 \, V \]

### Step 4: Calculate the Standard Cell Potential \(E^\circ_{cell}\)
The standard cell potential is calculated as follows:
\[ E^\circ_{cell} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} = 1.23 \, V - 0.44 \, V = 0.79 \, V \]

### Step 5: Use the Gibbs Free Energy Equation
The relationship between Gibbs free energy and cell potential is given by:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
Where:
- \(n\) = number of moles of electrons transferred (which is 2 for this reaction)
- \(F\) = Faraday's constant = 96500 C/mol
- \(E^\circ_{cell}\) = standard cell potential calculated above

### Step 6: Substitute Values into the Equation
Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, C/mol \times 0.79 \, V \]
Calculating this gives:
\[ \Delta G^\circ = -2 \times 96500 \times 0.79 = -152,870 \, J/mol \]

### Step 7: Convert to Kilojoules
To convert from joules to kilojoules:
\[ \Delta G^\circ = -152.87 \, kJ/mol \]

### Final Answer
The magnitude of \(\Delta G^\circ\) for the net process is:
\[ \Delta G^\circ = 152.87 \, kJ/mol \]

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