Consider the following
`E^(@)` values `E^(o)""_(Fe^(3+)//Fe^(2+))= +0.77 V" "E^(o)""_(Sn^(2+)//Sn)= 0.14V`
Under standard conditions the EMF for the reaction
`Sn(s) + 2Fe^(3+) (aq) rarr 2Fe^(2+)(aq)+Sn^(2+)(aq)` is :

Options

A. `0.63` V
B. `1.40` V
C. `0.91` V
D. `1.68` V

Answer

To solve the problem, we need to calculate the standard EMF (Electromotive Force) for the given redox reaction:

**Reaction:**
\[ \text{Sn}(s) + 2\text{Fe}^{3+}(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Sn}^{2+}(aq) \]

**Given Standard Reduction Potentials:**
1. \( E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \, \text{V} \)
2. \( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = +0.14 \, \text{V} \)

### Step 1: Identify the half-reactions
- The oxidation half-reaction involves tin (Sn) being oxidized to tin ions (\( \text{Sn}^{2+} \)):
\[ \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) + 2e^- \]

- The reduction half-reaction involves iron ions (\( \text{Fe}^{3+} \)) being reduced to iron ions (\( \text{Fe}^{2+} \)):
\[ \text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \]

### Step 2: Write the standard reduction potentials
- For the oxidation of Sn:
\[ E^\circ_{\text{oxidation}} = -E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \, \text{V} \]

- For the reduction of Fe:
\[ E^\circ_{\text{reduction}} = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \, \text{V} \]

### Step 3: Calculate the standard EMF of the cell
The overall standard EMF of the cell can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.77 \, \text{V} + (-0.14 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 0.77 \, \text{V} - 0.14 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 0.63 \, \text{V} \]

### Final Answer:
The EMF for the reaction under standard conditions is:
\[ \boxed{0.63 \, \text{V}} \]