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 calculate the EMF for the reaction \( \text{Sn}(s) + 2\text{Fe}^{3+}(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Sn}^{2+}(aq) \) under standard conditions, we will use the standard reduction potentials provided.

### Step 1: Identify the half-reactions
The overall reaction involves two half-reactions:
1. The oxidation of Sn:
\[ \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) + 2e^- \]
The standard reduction potential for this half-reaction can be derived from the given \( E^\circ \) value for \( \text{Sn}^{2+}/\text{Sn} \):
\[ E^\circ_{\text{oxidation}} = -E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \, \text{V} \]

2. The reduction of \( \text{Fe}^{3+} \):
\[ \text{Fe}^{3+}(aq) + 3e^- \rightarrow \text{Fe}^{2+}(aq) \]
The standard reduction potential for this half-reaction is given as:
\[ E^\circ_{\text{reduction}} = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \, \text{V} \]

### Step 2: Calculate the standard cell potential
The standard cell potential \( E^\circ_{\text{cell}} \) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.77 \, \text{V} + (-0.14 \, \text{V}) = 0.77 \, \text{V} - 0.14 \, \text{V} = 0.63 \, \text{V} \]

### Step 3: Final Result
Thus, the EMF for the reaction under standard conditions is:
\[ \text{EMF} = E^\circ_{\text{cell}} = 0.63 \, \text{V} \]