If `E_(Fe^(2+))^(@)//Fe = -0.441 V`
and `E_(Fe^(3+))^(@)//Fe^(2+) = -0.771 V`
The standard `EMF` of the reaction
`Fe+2Fe^(3+) rarr 3Fe^(2+)`
will be:

Options

A. `0.111 V`
B. `0.330 V`
C. `1.653 V`
D. `1.212 V`

Explanation

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

### Step 1: Write the half-reactions
We have the following half-reactions based on the standard reduction potentials provided:

1. \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) with \( E^\circ = -0.441 \, \text{V} \)
2. \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) with \( E^\circ = +0.771 \, \text{V} \)

### Step 2: Reverse the first half-reaction
Since we need to convert \( \text{Fe} \) to \( \text{Fe}^{2+} \), we reverse the first half-reaction:

\[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \quad (E^\circ = +0.441 \, \text{V}) \]

### Step 3: Multiply the second half-reaction
We need 2 moles of \( \text{Fe}^{3+} \) to balance the reaction. Thus, we multiply the second half-reaction by 2:

\[ 2 \text{Fe}^{3+} + 2e^- \rightarrow 2 \text{Fe}^{2+} \quad (E^\circ = +0.771 \, \text{V}) \]

### Step 4: Combine the half-reactions
Now we can add the two half-reactions together:

\[ \text{Fe} + 2 \text{Fe}^{3+} \rightarrow 3 \text{Fe}^{2+} \]

### Step 5: Calculate the standard EMF
To find the standard EMF for the overall reaction, we use the formula:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]

Here, the reduction potential for \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \) is \( +0.771 \, \text{V} \), and the oxidation potential for \( \text{Fe} \) to \( \text{Fe}^{2+} \) is \( +0.441 \, \text{V} \).

Thus:

\[ E^\circ_{\text{cell}} = 2 \times 0.771 - 0.441 \]

Calculating this gives:

\[ E^\circ_{\text{cell}} = 1.542 - 0.441 = 1.101 \, \text{V} \]

### Final Answer
The standard EMF of the reaction \( \text{Fe} + 2 \text{Fe}^{3+} \rightarrow 3 \text{Fe}^{2+} \) is \( 1.101 \, \text{V} \).

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