The equilibrium constant of the following redox rection at 298 K is `1 xx 10^(8)`
`2Fe^(3+) (aq.) +2I^(-) (aq.) hArr 2Fe^(2+)(aq.) +I_(2) (s)`
If the standard reducing potential of iodine becoming iodide is +0.54 V. what is the standard reduction potential of `Fe^(3+)//Fe^(2+)` ?

Options

A. `+1.006V`
B. `-1.006V`
C. `+0.77V`
D. `-0.77V`

Explanation

To find the standard reduction potential of the Fe³⁺/Fe²⁺ couple, we can follow these steps:

### Step 1: Write the Nernst Equation
The Nernst equation relates the standard reduction potential (E°) to the equilibrium constant (K) of a redox reaction:
\[ E° = \frac{0.059}{n} \log_{10} K \]
where:
- \(E°\) is the standard cell potential,
- \(n\) is the number of moles of electrons transferred in the balanced equation,
- \(K\) is the equilibrium constant.

### Step 2: Identify the Values
From the problem:
- The equilibrium constant \(K = 1 \times 10^8\).
- The number of electrons transferred \(n = 2\) (since 2 moles of Fe³⁺ are reduced to 2 moles of Fe²⁺).

### Step 3: Substitute Values into the Nernst Equation
Substituting the values into the equation:
\[ E° = \frac{0.059}{2} \log_{10}(1 \times 10^8) \]

### Step 4: Calculate the Logarithm
Calculate the logarithm:
\[ \log_{10}(1 \times 10^8) = 8 \]

### Step 5: Calculate E°
Now substitute this value back into the equation:
\[ E° = \frac{0.059}{2} \times 8 \]
\[ E° = 0.0295 \times 8 = 0.236 \text{ V} \]

### Step 6: Relate the Standard Potentials
The standard potential of the cell can be expressed as:
\[ E°_{\text{cell}} = E°_{\text{reduced}} - E°_{\text{oxidized}} \]
In this case, the reduced species is \(Fe^{3+}/Fe^{2+}\) and the oxidized species is \(I^{-}/I_{2}\).

### Step 7: Substitute Known Values
We know:
- \(E°_{\text{oxidized}} = +0.54 \text{ V}\) (for iodine to iodide).
- \(E°_{\text{cell}} = 0.236 \text{ V}\) (calculated).

So we can write:
\[ 0.236 = E°_{Fe^{3+}/Fe^{2+}} - 0.54 \]

### Step 8: Solve for E°_{Fe^{3+}/Fe^{2+}}
Rearranging gives:
\[ E°_{Fe^{3+}/Fe^{2+}} = 0.236 + 0.54 \]
\[ E°_{Fe^{3+}/Fe^{2+}} = 0.776 \text{ V} \]

Thus, the standard reduction potential of the \(Fe^{3+}/Fe^{2+}\) couple is approximately **0.77 V**.

### Final Answer
The standard reduction potential of \(Fe^{3+}/Fe^{2+}\) is **0.77 V**.

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