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Step-by-Step Solution
Step 1: Identify the Known Standard Reduction Potentials
We are given two half-reactions:
1) $ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}; \quad E_{(\text{Fe}^{3+}/\text{Fe}^{2+})}^0 = +0.77 \text{ V} $
2) $ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}; \quad E_{(\text{Fe}^{2+}/\text{Fe})}^0 = -0.47 \text{ V} $
Step 2: Express Each Half-Reaction in Terms of Gibbs Free Energy
The change in standard Gibbs free energy $ \Delta G^0 $ for an electrochemical reaction is given by the relation:
$ \Delta G^0 = - n F E^0 $
where:
$ n $ is the number of moles of electrons transferred
$ F $ is the Faraday constant
$ E^0 $ is the standard reduction potential for the half-reaction
For the first half-reaction:
$ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+};\quad n = 1,\; E^0 = +0.77 \text{ V} $
Hence,
$ \Delta G_1^0 = - (1) F (0.77) = -0.77\,F. $
For the second half-reaction:
$ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe};\quad n = 2,\; E^0 = -0.47 \text{ V} $
Hence,
$ \Delta G_2^0 = - (2) F (-0.47) = +0.94\,F. $
Step 3: Combine the Two Half-Reactions to Get the Overall Reaction
When we combine the two half-reactions, we want the overall reaction:
$ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}. $
Let $ E_3^0 $ be the standard reduction potential for this overall reaction. Then its Gibbs free energy change is:
$ \Delta G_3^0 = - (3) F E_3^0. $
The total Gibbs free energy change for the overall reaction is the sum of $ \Delta G_1^0 $ and $ \Delta G_2^0 $:
$ \Delta G_3^0 = \Delta G_1^0 + \Delta G_2^0. $
Step 4: Calculate the Overall Standard Reduction Potential
Substituting the values:
$ \Delta G_1^0 + \Delta G_2^0 = -0.77\,F + 0.94\,F = 0.17\,F. $
Hence,
$ -3\,F\,E_3^0 = 0.17\,F. $
Solve for $ E_3^0 $:
$ E_3^0 = - \frac{0.17}{3} = -0.057\text{ V}. $
Therefore, the standard reduction potential for the reaction $ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} $ is $ -0.057 \text{ V}. $
