Standard oxidation potential of iron electrode is + 0.44 V. Calculate the potential of `Fe, FeSO_4(1 M)` at `25^@C`.

Answer

To calculate the potential of the Fe/FeSO4 electrode at 1 M concentration at 25°C, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write down the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
where:
- \( E \) is the cell potential,
- \( E^\circ \) is the standard electrode potential,
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T \) is the temperature in Kelvin,
- \( n \) is the number of moles of electrons transferred in the half-reaction,
- \( F \) is Faraday's constant (96500 C/mol),
- \( Q \) is the reaction quotient.

### Step 2: Identify the values
From the question:
- Standard oxidation potential \( E^\circ \) for the reaction \( \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \) is +0.44 V.
- Concentration of \( \text{Fe}^{2+} \) is 1 M.
- Temperature \( T = 25^\circ C = 298 \, K \).
- Number of electrons transferred \( n = 2 \) (since 2 electrons are involved in the oxidation of Fe to Fe²⁺).

### Step 3: Calculate the reaction quotient \( Q \)
For the reaction:
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \]
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Fe}^{2+}]}{[\text{Fe}]} \]
Assuming the concentration of solid iron \( \text{Fe} \) is constant and can be taken as 1, we have:
\[ Q = \frac{1}{1} = 1 \]

### Step 4: Substitute values into the Nernst equation
Now substituting the values into the Nernst equation:
\[ E = 0.44 \, V - \frac{(8.314 \, J/mol·K)(298 \, K)}{(2)(96500 \, C/mol)} \ln(1) \]

### Step 5: Simplify the equation
Since \( \ln(1) = 0 \):
\[ E = 0.44 \, V - 0 \]
Thus:
\[ E = 0.44 \, V \]

### Final Answer
The potential of the Fe/FeSO4 electrode at 1 M concentration at 25°C is **0.44 V**.

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