To find the standard Gibbs free energy change (ΔG°) for the overall reaction given the half-reactions and their standard electrode potentials, we can follow these steps:
### Step 1: Identify the half-reactions and their potentials
The half-reactions provided are:
1. Oxidation of Zinc:
\[
\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-}, \quad E^{\circ} = 0.76 \, \text{V}
\]
2. Oxidation of Iron:
\[
\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^{-}, \quad E^{\circ} = 0.41 \, \text{V}
\]
### Step 2: Determine the overall cell reaction
The overall cell reaction can be written as:
\[
\text{Fe}^{2+} + \text{Zn} \rightarrow \text{Zn}^{2+} + \text{Fe}
\]
In this reaction, Zinc is oxidized and Iron(II) is reduced.
### Step 3: Calculate the standard cell potential (E°cell)
To find the cell potential, we need to use the reduction potential of Iron. The reduction potential for Iron is the negative of its oxidation potential:
\[
E^{\circ}_{\text{Fe}^{2+}/\text{Fe}} = -0.41 \, \text{V}
\]
Now, we can calculate the overall cell potential:
\[
E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = (0.41 \, \text{V}) - (0.76 \, \text{V}) = -0.35 \, \text{V}
\]
### Step 4: Calculate the number of moles of electrons (n)
From the half-reactions, we see that 2 moles of electrons are transferred in the overall reaction:
\[
n = 2
\]
### Step 5: Use the Gibbs free energy equation
The relationship between Gibbs free energy change and cell potential is given by:
\[
\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}
\]
Where:
- \( F \) (Faraday's constant) = 96500 C/mol
- \( E^{\circ}_{\text{cell}} = -0.35 \, \text{V} \)
### Step 6: Substitute the values into the equation
Substituting the values into the equation:
\[
\Delta G^{\circ} = -2 \times 96500 \, \text{C/mol} \times (-0.35 \, \text{V})
\]
Calculating this gives:
\[
\Delta G^{\circ} = 2 \times 96500 \times 0.35 = 67650 \, \text{J}
\]
### Step 7: Convert to kJ
To convert Joules to kilojoules:
\[
\Delta G^{\circ} = \frac{67650 \, \text{J}}{1000} = 67.65 \, \text{kJ}
\]
### Final Answer
Thus, the standard Gibbs free energy change (ΔG°) for the overall reaction is:
\[
\Delta G^{\circ} \approx 67.65 \, \text{kJ}
\]
---
