Standard reduction potential of `I_(3)^(-), I^(-) and Fe^(3+), Fe^(2+)` are `0.54 and 0.77 V`, respectively . Calculate the equilibrium constant for the reaction. `2Fe^(3+) + 3I^(-)

Answer

To calculate the equilibrium constant for the reaction:

\[ 2Fe^{3+} + 3I^{-} \rightleftharpoons 2Fe^{2+} + I_{3}^{-} \]

we will follow these steps:

### Step 1: Identify the Standard Reduction Potentials
We are given the standard reduction potentials:
- For \( Fe^{3+} + e^{-} \rightarrow Fe^{2+} \): \( E^\circ = 0.77 \, V \)
- For \( I_{3}^{-} + 2e^{-} \rightarrow 3I^{-} \): \( E^\circ = 0.54 \, V \)

### Step 2: Write the Half-Reactions
The half-reactions can be written as:
1. Reduction of \( Fe^{3+} \):
\[ Fe^{3+} + e^{-} \rightarrow Fe^{2+} \quad (E^\circ = 0.77 \, V) \]
2. Reduction of \( I_{3}^{-} \):
\[ I_{3}^{-} + 2e^{-} \rightarrow 3I^{-} \quad (E^\circ = 0.54 \, V) \]

### Step 3: Determine the Oxidation Potentials
For the reaction, we need the oxidation potentials:
1. Oxidation of \( Fe^{2+} \):
\[ Fe^{2+} \rightarrow Fe^{3+} + e^{-} \quad (E^\circ = -0.77 \, V) \]
2. Oxidation of \( I^{-} \):
\[ 3I^{-} \rightarrow I_{3}^{-} + 2e^{-} \quad (E^\circ = -0.54 \, V) \]

### Step 4: Combine the Half-Reactions
To find the overall cell potential, we need to combine the half-reactions. The balanced overall reaction is:
\[ 2Fe^{3+} + 3I^{-} \rightleftharpoons 2Fe^{2+} + I_{3}^{-} \]

### Step 5: Calculate the Cell Potential
The cell potential \( E_{cell} \) can be calculated using the formula:
\[ E_{cell} = E_{cathode} - E_{anode} \]

Where:
- \( E_{cathode} \) is the reduction potential of \( Fe^{3+} \) (0.77 V)
- \( E_{anode} \) is the reduction potential of \( I_{3}^{-} \) (0.54 V)

Thus:
\[ E_{cell} = 0.77 \, V - 0.54 \, V = 0.23 \, V \]

### Step 6: Relate Cell Potential to Equilibrium Constant
We use the Nernst equation to relate the cell potential to the equilibrium constant \( K \):
\[ E_{cell} = \frac{0.0592}{n} \log K \]

Where \( n \) is the number of moles of electrons transferred in the balanced equation. In our case, \( n = 6 \) (2 for \( Fe^{3+} \) and 4 for \( I^{-} \)).

### Step 7: Solve for the Equilibrium Constant \( K \)
Rearranging the equation gives:
\[ \log K = \frac{n \cdot E_{cell}}{0.0592} \]

Substituting the values:
\[ \log K = \frac{6 \cdot 0.23}{0.0592} \]

Calculating:
\[ \log K = \frac{1.38}{0.0592} \approx 23.29 \]

Thus:
\[ K = 10^{23.29} \approx 6.25 \times 10^{7} \]

### Final Answer
The equilibrium constant \( K \) for the reaction is:
\[ K \approx 6.25 \times 10^{7} \]

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