`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 for the cell reaction \( \text{Fe}^{2+} + \text{Sn} \rightarrow \text{Sn}^{2+} + \text{Fe} \), we can follow these steps:

### Step 1: Identify the standard reduction potentials
We are given the standard reduction potentials for the two half-reactions:
- For iron: \( E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.75 \, \text{V} \)
- For tin: \( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.45 \, \text{V} \)

### Step 2: Write the half-reactions
The half-reactions can be written as follows:
1. Reduction of iron:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (E^\circ = -0.75 \, \text{V}) \]
2. Oxidation of tin:
\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \quad (E^\circ = +0.45 \, \text{V}) \text{ (change sign for oxidation)} \]

### Step 3: Calculate the standard EMF of the cell
The standard EMF of the cell (\( E^\circ_{\text{cell}} \)) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.75 \, \text{V}) + (+0.45 \, \text{V}) \]

### Step 4: Perform the calculation
Now, calculate the EMF:
\[ E^\circ_{\text{cell}} = -0.75 + 0.45 = -0.30 \, \text{V} \]

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