To solve the problem, we need to calculate the equilibrium constant (K) and the Gibbs free energy change (ΔG°) for the given reaction:
[tex]\[
Fe^{2+}(aq) + 2Cu^{+}(aq) \rightarrow Fe(s) + 2Cu^{2+}(aq)
\][/tex]
### Step-by-Step Solution:
1. Reduction Potentials:
The standard reduction potentials given are:
[tex]\[
E^\circ_{Fe^{2+}/Fe} = -0.440 \, V
\][/tex]
[tex]\[
E^\circ_{Cu^{2+}/Cu^+} = 0.153 \, V
\][/tex]
2. Determine the Cell Potential (E_cell):
- For this reaction, Fe^2+ is being reduced to Fe, and Cu^+ is being oxidized to Cu^2+.
- The standard reduction potential for Cu^+/Cu is 0.153 V and for Fe^2+/Fe is -0.440 V.
- The cell potential is calculated as:
[tex]\[
E_\text{cell} = E_\text{cathode} - E_\text{anode}
\][/tex]
- In this reaction, [tex]\( Cu \)[/tex] is acting as the anode (oxidation half-reaction), and [tex]\( Fe \)[/tex] is acting as the cathode (reduction half-reaction). So we use:
[tex]\[
E_\text{cathode} = E^\circ_{Fe^{2+}/Fe} = -0.440 \, V
\][/tex]
[tex]\[
E_\text{anode} = E^\circ_{Cu^{2+}/Cu^+} = 0.153 \, V
\][/tex]
- Therefore,
[tex]\[
E_\text{cell} = 0.153 \, V - (-0.440 \, V) = 0.593 \, V
\][/tex]
3. Equilibrium Constant (K):
- The number of electrons transferred (n) in the reaction is 2.
- Using the Nernst equation at equilibrium:
[tex]\[
E_\text{cell} = \frac{0.025693 \, V}{n} \ln K
\][/tex]
- Rearranging for [tex]\( K \)[/tex]:
[tex]\[
\ln K = \frac{n \cdot E_\text{cell}}{0.025693}
\][/tex]
- Substituting the known values:
[tex]\[
\ln K = \frac{2 \times 0.593}{0.025693} = 46.160432802708904
\][/tex]
- Taking the exponential (antilogarithm) to find [tex]\( K \)[/tex]:
[tex]\[
K = e^{46.160432802708904} \approx 1.1148623482737969 \times 10^{20}
\][/tex]
4. Gibbs Free Energy Change (ΔG°):
- The relationship between Gibbs free energy change and cell potential is given by:
[tex]\[
\Delta G^\circ = -nFE_\text{cell}
\][/tex]
- Where [tex]\( F \)[/tex] is the Faraday constant (96485 C/mol).
- Substituting the known values:
[tex]\[
\Delta G^\circ = - 2 \times 96485 \times 0.593 \approx -114431.20999999999 \, J/mol
\][/tex]
### Final Results:
1. Equilibrium Constant (K):
[tex]\[
K \approx 1.1148623482737969 \times 10^{20}
\][/tex]
2. Gibbs Free Energy Change (ΔG°):
[tex]\[
\Delta G^\circ \approx -114431.20999999999 \, J/mol
\][/tex]
Thus, ΔG° for this reaction would be less than zero, indicating that the reaction is spontaneous under standard conditions.
