`E^(o) ""_(RP)" for " Fe^(+2)//Fe and Sn^(+2)//Sn` electrodes are `- 0.75 and - 0.45` volt respectively. The standard emf for cell `F^(+2)+Sn rarr Sn^(+2) + Fe` is :

Options

A. `+ 0 . 30` V
B. `-0.58` V
C. `+ 0.58` V
D. `-0.30` V

Answer

To find the standard EMF (E° cell) for the cell reaction \( \text{Fe}^{2+} + \text{Sn} \rightarrow \text{Sn}^{2+} + \text{Fe} \), we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials
We have the following standard reduction potentials:
- For the reduction of \( \text{Fe}^{2+} \) to \( \text{Fe} \):
\[ E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.75 \, \text{V} \]
- For the reduction of \( \text{Sn}^{2+} \) to \( \text{Sn} \):
\[ E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.45 \, \text{V} \]

### Step 2: Determine which half-reaction is reduction and which is oxidation
In the given cell reaction:
- \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) (reduction, occurs at the cathode)
- \( \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \) (oxidation, occurs at the anode)

### Step 3: Write the formula for the standard EMF of the cell
The standard EMF of the cell can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

### Step 4: Substitute the values into the formula
Here, the cathode is the \( \text{Fe}^{2+}/\text{Fe} \) half-reaction and the anode is the \( \text{Sn}/\text{Sn}^{2+} \) half-reaction. Thus:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Fe}^{2+}/\text{Fe}} - E^\circ_{\text{Sn}^{2+}/\text{Sn}} \]
\[ E^\circ_{\text{cell}} = (-0.75 \, \text{V}) - (-0.45 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = -0.75 \, \text{V} + 0.45 \, \text{V} \]
\[ E^\circ_{\text{cell}} = -0.30 \, \text{V} \]

### Conclusion
The standard EMF for the cell reaction \( \text{Fe}^{2+} + \text{Sn} \rightarrow \text{Sn}^{2+} + \text{Fe} \) is:
\[ E^\circ_{\text{cell}} = -0.30 \, \text{V} \]