Calculate the equilibrium constant for the reaction` Fe + CuSO_4 ⇌ FeSO_4 + Cu` at 25°C. (Given E°(OP/Fe) = 0.5 V°, E°(OP/Cu) = -0.4 V)

Options

A. 3.46 × `10^30`
B. 3.46 × `10^26`
C. 3.22 × `10^30`
D. 3.22 × `10^26`

Answer

To calculate the equilibrium constant (Kc) for the reaction:

\[ \text{Fe} + \text{CuSO}_4 \rightleftharpoons \text{FeSO}_4 + \text{Cu} \]

at 25°C, we can follow these steps:

### Step 1: Identify the half-reactions
In this reaction, iron (Fe) is oxidized and copper (Cu) is reduced. The half-reactions can be written as:

- Oxidation half-reaction:
\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \]

- Reduction half-reaction:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

### Step 2: Determine the standard electrode potentials
From the problem, we have the following standard electrode potentials:

- \( E^\circ(\text{Fe}^{2+}/\text{Fe}) = 0.5 \, \text{V} \)
- \( E^\circ(\text{Cu}^{2+}/\text{Cu}) = -0.4 \, \text{V} \)

### Step 3: Calculate the standard cell potential (E°cell)
The standard cell potential can be calculated using the formula:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]

Here, the reduction potential for copper is -0.4 V, and the oxidation potential for iron is 0.5 V. Therefore:

\[ E^\circ_{\text{cell}} = (-0.4 \, \text{V}) - (0.5 \, \text{V}) = 0.5 + 0.4 = 0.9 \, \text{V} \]

### Step 4: Use the Nernst equation to relate E°cell to Kc
The Nernst equation at standard conditions is given by:

\[ E^\circ_{\text{cell}} = \frac{0.059}{n} \log K_c \]

where \( n \) is the number of moles of electrons transferred in the balanced equation. In this case, \( n = 2 \) (since 2 electrons are transferred).

Substituting the values we have:

\[ 0.9 = \frac{0.059}{2} \log K_c \]

### Step 5: Solve for Kc
Rearranging the equation to solve for \( K_c \):

\[ \log K_c = \frac{0.9 \times 2}{0.059} \]

Calculating the right-hand side:

\[ \log K_c = \frac{1.8}{0.059} \approx 30.51 \]

Now, converting from logarithmic form to exponential form:

\[ K_c = 10^{30.51} \approx 3.22 \times 10^{30} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction at 25°C is approximately:

\[ K_c \approx 3.22 \times 10^{30} \]

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