The standard electrode potentials (reduction) of `Pt//Fe^(3+),Fe^(+2)` and `Pt//Sn^(4+),Sn^(+2)` are `+0.77V` and `0.15V` respectively at `25^(@)C`. The standard `EMF` of the reaction `Sn^(4+) +2Fe^(2+) rarr Sn^(2+) +2Fe^(3+)` is

Options

A. `-0.62V`
B. `-0.92V`
C. `+0.31V`
D. `+0.85V`

Explanation

To find the standard EMF of the reaction \( \text{Sn}^{4+} + 2\text{Fe}^{2+} \rightarrow \text{Sn}^{2+} + 2\text{Fe}^{3+} \), we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
The given standard electrode potentials are:
1. For the reduction of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \):
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad E^\circ = +0.77 \, \text{V} \]
2. For the reduction of \( \text{Sn}^{4+} \) to \( \text{Sn}^{2+} \):
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \quad E^\circ = +0.15 \, \text{V} \]

### Step 2: Write the balanced half-reactions for the overall reaction.
To balance the overall reaction, we need to multiply the half-reaction of \( \text{Fe}^{3+} \) by 2:
\[ 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \quad E^\circ = +0.77 \, \text{V} \]

### Step 3: Combine the half-reactions.
Now, we can combine the half-reactions:
- The oxidation half-reaction (for \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \)) is reversed:
\[ 2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2e^- \quad E^\circ = -0.77 \, \text{V} \]
- The reduction half-reaction remains as is:
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \quad E^\circ = +0.15 \, \text{V} \]

### Step 4: Calculate the standard EMF of the overall reaction.
The standard EMF of the overall reaction is calculated by adding the standard potentials of the half-reactions:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.15 \, \text{V} - 0.77 \, \text{V} = -0.62 \, \text{V} \]

### Final Answer:
The standard EMF of the reaction \( \text{Sn}^{4+} + 2\text{Fe}^{2+} \rightarrow \text{Sn}^{2+} + 2\text{Fe}^{3+} \) is \( -0.62 \, \text{V} \).