Calculate the euilibrium constant for the reaction,
`2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_(3)^(-)`.
The standard reduction potential in acidic conditions are `0.77 V ` and `0.54 V` respectivelu for `Fe^(3+)//Fe^(2+)` and `I_(3)^(-)//I^(-)` couples.

Options

A. `4.25xx10^7`
B. `7.05xx10^5`
C. `6.25xx10^5`
D. `6.25xx10^7`

Explanation

To solve the problem of calculating the equilibrium constant for the reaction:

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

we will follow these steps:

### Step 1: Identify the half-reactions
The half-reactions for the given redox reaction are:
1. Reduction (cathode):
\[ Fe^{3+} + e^{-} \rightarrow Fe^{2+} \]
The standard reduction potential \( E^{\circ} \) for this reaction is \( 0.77 \, V \).

2. Oxidation (anode):
\[ I^{-} \rightarrow I_{3}^{-} + 2e^{-} \]
The standard reduction potential for the reverse reaction \( I_{3}^{-} + 2e^{-} \rightarrow 3I^{-} \) is \( 0.54 \, V \).

### Step 2: Write the balanced overall reaction
The balanced overall reaction is already given as:
\[ 2Fe^{3+} + 3I^{-} \rightleftharpoons 2Fe^{2+} + I_{3}^{-} \]

### Step 3: Determine the number of electrons transferred
In the oxidation half-reaction, 2 moles of electrons are transferred for every mole of \( I_{3}^{-} \) produced. Therefore, the total number of electrons transferred \( n \) is:
\[ n = 2 \]

### Step 4: Calculate the standard cell potential \( E^{\circ}_{cell} \)
Using the formula for the cell potential:
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]

Substituting the values:
\[ E^{\circ}_{cell} = 0.77 \, V - 0.54 \, V = 0.23 \, V \]

### Step 5: Use the Nernst equation to find the equilibrium constant
At equilibrium, the cell potential \( E_{cell} \) is 0. Therefore, we can use the Nernst equation:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log K_{c} \]

Setting \( E_{cell} = 0 \):
\[ 0 = 0.23 - \frac{0.0591}{2} \log K_{c} \]

Rearranging gives:
\[ \frac{0.0591}{2} \log K_{c} = 0.23 \]

### Step 6: Solve for \( \log K_{c} \)
Multiplying both sides by 2:
\[ 0.0591 \log K_{c} = 0.46 \]

Now, divide by \( 0.0591 \):
\[ \log K_{c} = \frac{0.46}{0.0591} \approx 7.796 \]

### Step 7: Calculate \( K_{c} \)
To find \( K_{c} \), we take the antilogarithm:
\[ K_{c} = 10^{7.796} \approx 6.25 \times 10^{7} \]

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

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