The standard reduction potential for `Fe^(2+) //Fe` and `Sn^(2+) //Sn` electrodes are `-0.44` and `-0.14 ` volt respectively. For the given cell reaction `Fe^(2+) + Sn rarr Fe + Sn^(2+) `, the standard ` EMF` is.

Options

A. `-0.30 V`
B. `-0.58 V`
C. `+0.58 V`
D. ` + 0. 3 0 V`

Explanation

To calculate the standard EMF (electromotive force) for the cell reaction \( \text{Fe}^{2+} + \text{Sn} \rightarrow \text{Fe} + \text{Sn}^{2+} \), we can follow these steps:

### Step 1: Identify the half-reactions
We have two half-reactions based on the given standard reduction potentials:
1. For iron:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (E^\circ = -0.44 \, \text{V}) \]
2. For tin:
\[ \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \quad (E^\circ = -0.14 \, \text{V}) \]

### Step 2: Determine the oxidation and reduction reactions
In the cell reaction, iron is being reduced (gaining electrons) and tin is being oxidized (losing electrons). Therefore, we can write:
- Reduction (gaining electrons):
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (E^\circ = -0.44 \, \text{V}) \]
- Oxidation (losing electrons):
\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \quad (E^\circ = +0.14 \, \text{V}) \quad \text{(reverse the sign)} \]

### Step 3: Calculate the standard EMF of the cell
The standard EMF of the cell can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Where:
- \( E^\circ_{\text{cathode}} \) is the reduction potential of the cathode (Fe).
- \( E^\circ_{\text{anode}} \) is the oxidation potential of the anode (Sn).

Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.44 \, \text{V}) - (-0.14 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = -0.44 \, \text{V} + 0.14 \, \text{V} \]
\[ E^\circ_{\text{cell}} = -0.30 \, \text{V} \]

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

---