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 use the standard electrode potentials provided for the half-reactions involved.

### Step 1: Identify the half-reactions and their standard electrode potentials

The standard electrode potentials given are:
- For the reduction of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \):
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad E^\circ = +0.77 \, \text{V} \]
- 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

1. **For \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \)** (reduction):
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (1 \text{ electron}) \]
We need 2 moles of \( \text{Fe}^{3+} \) to balance with 2 moles of \( \text{Fe}^{2+} \):
\[ 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \quad E^\circ = +0.77 \, \text{V} \]

2. **For \( \text{Sn}^{4+} \) to \( \text{Sn}^{2+} \)** (reduction):
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \quad E^\circ = +0.15 \, \text{V} \]

### Step 3: Combine the half-reactions

The overall reaction can be obtained by combining the two half-reactions:
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \quad (E^\circ = +0.15 \, \text{V}) \]
\[ 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \quad (E^\circ = +0.77 \, \text{V}) \]

### Step 4: Calculate the standard EMF of the overall reaction

The standard EMF of the overall reaction is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Here, the cathode is the reduction of \( \text{Sn}^{4+} \) (which has a lower potential) and the anode is the reduction of \( \text{Fe}^{3+} \):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Sn}^{4+/Sn^{2+}}} - E^\circ_{\text{Fe}^{3+/Fe^{2+}}} \]
\[ E^\circ_{\text{cell}} = 0.15 \, \text{V} - 0.77 \, \text{V} \]
\[ E^\circ_{\text{cell}} = -0.62 \, \text{V} \]

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

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