What is the electrode potential `Fe^(3+)//Fe` electrode in which concentration of `Fe^(3+)` ions is 0.1M Given `E^(@)Fe^(3+)//Fe=+0.771V`

Options

A. `+0.79V`
B. `+0.75V`
C. `1.50V`
D. `+1.0V`

Explanation

To find the electrode potential of the `Fe^(3+)//Fe` electrode with a concentration of `Fe^(3+)` ions at 0.1 M, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the half-reaction
The half-reaction for the reduction of iron(III) ions to iron is:
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe (s)} \]

### Step 2: Identify the standard electrode potential
The standard electrode potential (\(E^\circ\)) for the reaction is given as:
\[ E^\circ = +0.771 \, \text{V} \]

### Step 3: Determine the reaction quotient (Q)
The reaction quotient \(Q\) for the reaction can be expressed as:
\[ Q = \frac{[\text{Fe}]}{[\text{Fe}^{3+}]} \]
Since the concentration of solid iron (\(\text{Fe}\)) is constant, it is considered to be 1. Therefore:
\[ Q = \frac{1}{[\text{Fe}^{3+}]} \]
Given that the concentration of \(\text{Fe}^{3+}\) is 0.1 M:
\[ Q = \frac{1}{0.1} = 10 \]

### Step 4: Use the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
where \(n\) is the number of moles of electrons transferred in the half-reaction. For this reaction, \(n = 3\).

### Step 5: Substitute values into the Nernst equation
Substituting the known values into the Nernst equation:
\[ E = 0.771 - \frac{0.0591}{3} \log(10) \]
Since \(\log(10) = 1\):
\[ E = 0.771 - \frac{0.0591}{3} \cdot 1 \]

### Step 6: Calculate the potential
Now calculate:
\[ E = 0.771 - \frac{0.0591}{3} \]
\[ E = 0.771 - 0.0197 \]
\[ E = 0.7513 \, \text{V} \]

### Step 7: Round the answer
Rounding to two decimal places, we get:
\[ E \approx 0.75 \, \text{V} \]

### Final Answer
The electrode potential for the `Fe^(3+)//Fe` electrode at a concentration of 0.1 M is approximately **0.75 V**.

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