The reduction potential of a half-cell consisting of a Pt electrode immersed in `1.5 M Fe^(2+)` and ` 0.015 M Fe^(3+)` solutin at `25^@ C` is `(E_(Fe^(3+)//Fe^(2+))^@ = 0.770 V)` is .

Options

A. `0.652V`
B. `0.88V`
C. `0.710V`
D. `0.850V`

Explanation

To solve the problem, we need to calculate the reduction potential of the half-cell using the Nernst equation. Here's a step-by-step breakdown of the solution:

### Step 1: Identify the half-reaction
The half-reaction for the reduction of Fe³⁺ to Fe²⁺ is:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \]

### Step 2: Write down the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where:
- \( E \) = cell potential
- \( E^\circ \) = standard reduction potential
- \( n \) = number of electrons transferred
- \([\text{products}]\) = concentration of products
- \([\text{reactants}]\) = concentration of reactants

### Step 3: Identify the values
From the problem:
- \( E^\circ = 0.770 \, \text{V} \)
- Concentration of \(\text{Fe}^{2+} = 1.5 \, \text{M}\)
- Concentration of \(\text{Fe}^{3+} = 0.015 \, \text{M}\)
- \( n = 1 \) (since one electron is transferred in the reaction)

### Step 4: Substitute values into the Nernst equation
Now, substituting the values into the Nernst equation:
\[ E = 0.770 - \frac{0.0591}{1} \log \left( \frac{1.5}{0.015} \right) \]

### Step 5: Calculate the logarithm
Calculate the ratio:
\[ \frac{1.5}{0.015} = 100 \]
Now, calculate the logarithm:
\[ \log(100) = 2 \]

### Step 6: Substitute back into the equation
Now substitute back into the Nernst equation:
\[ E = 0.770 - 0.0591 \times 2 \]
\[ E = 0.770 - 0.1182 \]

### Step 7: Final calculation
Perform the final calculation:
\[ E = 0.770 - 0.1182 = 0.6518 \, \text{V} \]
Rounding to three significant figures, we get:
\[ E \approx 0.652 \, \text{V} \]

### Conclusion
The reduction potential of the half-cell is approximately \( 0.652 \, \text{V} \).

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