The `E^@` values of the following reduction reactions are given `Fe_((aq))^(3+) +e rarr Fe_((aq))^(2+),E^@ = 0.771 V`
`Fe_((aq))^(2+) +2e rarr Fe_((s)) ,E^@ = -0.447 V` What will be the free energy change for the reaction `Fe_((aq))^(3+) +3e^(-) rarrFe_((s)) " " (1 F = 96485 "C mol"^(-1))`

Options

A. `+ 18.51 "kJ mol"^(-1)`
B. `+ 11.87 "kJ mol"^(-1)`
C. `-8.10 "kJ mol"^(-1)`
D. `-10.41 "kJ mol"^(-1)`

Explanation

To calculate the free energy change (ΔG) for the reaction:

\[ \text{Fe}^{3+}_{(aq)} + 3e^- \rightarrow \text{Fe}_{(s)} \]

we first need to determine the standard cell potential (E°) for the overall reaction. The standard cell potential can be calculated using the given half-reactions and their standard reduction potentials.

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

1. The first half-reaction is:
\[ \text{Fe}^{3+}_{(aq)} + e^- \rightarrow \text{Fe}^{2+}_{(aq)}, \quad E° = 0.771 \, \text{V} \]

2. The second half-reaction is:
\[ \text{Fe}^{2+}_{(aq)} + 2e^- \rightarrow \text{Fe}_{(s)}, \quad E° = -0.447 \, \text{V} \]

### Step 2: Combine the half-reactions

To find the overall reaction, we need to combine these half-reactions. We will multiply the first half-reaction by 2 to balance the number of electrons:

\[ 2 \left( \text{Fe}^{3+}_{(aq)} + e^- \rightarrow \text{Fe}^{2+}_{(aq)} \right) \quad \Rightarrow \quad 2\text{Fe}^{3+}_{(aq)} + 2e^- \rightarrow 2\text{Fe}^{2+}_{(aq)} \]

Now we can add this to the second half-reaction:

\[ 2\text{Fe}^{3+}_{(aq)} + 2e^- \rightarrow 2\text{Fe}^{2+}_{(aq)} \quad \text{and} \quad \text{Fe}^{2+}_{(aq)} + 2e^- \rightarrow \text{Fe}_{(s)} \]

This results in:

\[ 2\text{Fe}^{3+}_{(aq)} + 2e^- + \text{Fe}^{2+}_{(aq)} + 2e^- \rightarrow 2\text{Fe}^{2+}_{(aq)} + \text{Fe}_{(s)} \]

### Step 3: Calculate the overall standard cell potential (E°)

The overall cell potential (E°) is calculated by subtracting the reduction potential of the anode from the cathode.

1. The cathode (reduction) is:
\[ \text{Fe}^{2+}_{(aq)} + 2e^- \rightarrow \text{Fe}_{(s)} \quad E° = -0.447 \, \text{V} \]

2. The anode (oxidation) is:
\[ \text{Fe}^{3+}_{(aq)} + e^- \rightarrow \text{Fe}^{2+}_{(aq)} \quad E° = 0.771 \, \text{V} \]

To find the overall E° for the reaction:

\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = (-0.447) - (0.771) = -1.218 \, \text{V} \]

### Step 4: Calculate the free energy change (ΔG)

The relationship between the standard cell potential and the free energy change is given by the equation:

\[ \Delta G° = -nFE° \]

where:
- n = number of moles of electrons transferred (in this case, n = 3)
- F = Faraday's constant (96485 C/mol)
- E° = standard cell potential (-1.218 V)

Now substituting the values:

\[ \Delta G° = -3 \times 96485 \, \text{C/mol} \times (-1.218 \, \text{V}) \]

Calculating this gives:

\[ \Delta G° = 3 \times 96485 \times 1.218 \approx 352,000 \, \text{J/mol} \quad \text{or} \quad 352 \, \text{kJ/mol} \]

### Final Answer

The free energy change for the reaction is approximately:

\[ \Delta G° \approx 352 \, \text{kJ/mol} \]