Reaction : `2Fe^(3+)+3I^(-) hArr 2Fe^(2+) + I_3^(-)` The standard reduction potentials in acidic conditions are 0.77 V and 0.54 V respectively for cathodic and anodic reactions. The equilibrium constant for the reaction is approximately . (Given `10^(7.79) = 6.26 xx10^7)`

Options

A. `6.26xx10^(-7)`
B. `5.33xx10^(-4)`
C. `6.26xx10^(7)`
D. `5.33xx10^(4)`

Answer

To find the equilibrium constant (Kc) for the given reaction, we can follow these steps:

### Step 1: Identify the Reaction and Standard Reduction Potentials
The reaction is:
\[ 2Fe^{3+} + 3I^{-} \rightleftharpoons 2Fe^{2+} + I_3^{-} \]

The standard reduction potentials are given as:
- For the cathodic reaction (reduction): \( E_{cathode} = 0.77 \, V \)
- For the anodic reaction (oxidation): \( E_{anode} = 0.54 \, V \)

### Step 2: Calculate the Standard Cell Potential (E°cell)
Using the formula:
\[ E°_{cell} = E_{cathode} - E_{anode} \]

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

### Step 3: Determine the Number of Electrons (n) Transferred
From the balanced equation, we can see that:
- 2 moles of \( Fe^{3+} \) are reduced to \( Fe^{2+} \), which involves 2 electrons.
- 3 moles of \( I^{-} \) are oxidized to \( I_3^{-} \), which involves the transfer of 2 electrons.

Thus, the number of electrons transferred \( n = 2 \).

### Step 4: Use the Nernst Equation to Relate E°cell and Kc
The Nernst equation in terms of the equilibrium constant is:
\[ E°_{cell} = \frac{0.059}{n} \log K_c \]

Substituting the known values:
\[ 0.23 = \frac{0.059}{2} \log K_c \]

### Step 5: Solve for log Kc
Rearranging the equation:
\[ \log K_c = \frac{0.23 \times 2}{0.059} \]

Calculating the right side:
\[ \log K_c = \frac{0.46}{0.059} \approx 7.79 \]

### Step 6: Calculate Kc
To find \( K_c \):
\[ K_c = 10^{7.79} \]

Using the provided information:
\[ K_c \approx 6.26 \times 10^{7} \]

### Conclusion
Thus, the equilibrium constant \( K_c \) for the reaction is approximately \( 6.26 \times 10^{7} \).