Standard `E^0` of the half cell `Fe | Fe^(2+)` is `+0.44V` and standard `E^0` of half cell `Cu|Cu^(2+) is -0.32V`,then

Options

A. Cu oxidises `Fe^(2+) ion`
B. `Cu^(2+)` oxidises `Fe`
C. Cu reduces `Fe^(2+) ion`
D. `Cu^(2+)` reduces `Fe`

Answer

To solve the question regarding the standard electrode potentials of the half-cells \( \text{Fe} | \text{Fe}^{2+} \) and \( \text{Cu} | \text{Cu}^{2+} \), we will follow these steps:

### Step 1: Identify the Standard Electrode Potentials
We have the following standard electrode potentials:
- For the half-cell \( \text{Fe} | \text{Fe}^{2+} \): \( E^0 = +0.44 \, \text{V} \)
- For the half-cell \( \text{Cu} | \text{Cu}^{2+} \): \( E^0 = -0.32 \, \text{V} \)

### Step 2: Convert Oxidation Potentials to Reduction Potentials
The standard electrode potential given is for reduction. The reduction potential for each half-cell can be interpreted as follows:
- For \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \): \( E^0 = +0.44 \, \text{V} \) (this is already in reduction form)
- For \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \): \( E^0 = -0.32 \, \text{V} \) (this is also in reduction form)

### Step 3: Determine the Direction of Reactions
To understand which species can oxidize or reduce the other, we compare the reduction potentials:
- The higher the reduction potential, the more likely the species will be reduced.
- \( E^0 \) for \( \text{Fe}^{2+}/\text{Fe} \) is \( +0.44 \, \text{V} \) (more positive, more likely to be reduced).
- \( E^0 \) for \( \text{Cu}^{2+}/\text{Cu} \) is \( -0.32 \, \text{V} \) (less positive, less likely to be reduced).

### Step 4: Identify Oxidation and Reduction
Since \( \text{Fe} \) has a higher reduction potential, it will tend to be reduced, while \( \text{Cu}^{2+} \) has a lower reduction potential, meaning it can be oxidized:
- \( \text{Fe} \) can oxidize to \( \text{Fe}^{2+} \).
- \( \text{Cu}^{2+} \) can reduce to \( \text{Cu} \).

### Step 5: Conclusion
From the above analysis, we conclude that:
- \( \text{Cu}^{2+} \) will oxidize \( \text{Fe} \) to \( \text{Fe}^{2+} \).

Thus, the correct answer to the question is that **Copper (II) ion oxidizes Iron (Fe)**.