Electrode potential for the following half-cell reactions are
`Zn rarr Zn^(2+)+2e^(-), E^(@)=+0.76V,`
`Fe rarr Fe^(2+)+2e^(-), E^(@)=+0.44V`
The EMF for the cell reaction `Fe^(2+)+Zn rarr Zn^(2+)+Fe` will be

Options

A. `-0.32V`
B. `+1.20 V`
C. `-1.20 V`
D. `+0.32 V`

Answer

To calculate the EMF (Electromotive Force) for the cell reaction \( \text{Fe}^{2+} + \text{Zn} \rightarrow \text{Zn}^{2+} + \text{Fe} \), we will follow these steps:

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

The half-reactions and their standard electrode potentials are given as follows:
1. \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \), \( E^\circ = -0.76 \, \text{V} \) (oxidation)
2. \( \text{Fe}^{2+} + 2e^{-} \rightarrow \text{Fe} \), \( E^\circ = +0.44 \, \text{V} \) (reduction)

### Step 2: Convert the oxidation potential to reduction potential

For the zinc half-reaction, since it is being oxidized, we need to convert the oxidation potential to reduction potential:
- The reduction potential of zinc is \( E^\circ (\text{Zn}) = -E^\circ (\text{Zn} \rightarrow \text{Zn}^{2+}) = +0.76 \, \text{V} \).

### Step 3: Identify the cathode and anode

In the cell reaction:
- The cathode (where reduction occurs) is \( \text{Fe}^{2+} + 2e^{-} \rightarrow \text{Fe} \) with \( E^\circ = +0.44 \, \text{V} \).
- The anode (where oxidation occurs) is \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \) with \( E^\circ = +0.76 \, \text{V} \).

### Step 4: Calculate the EMF of the cell

The formula for calculating the EMF of the cell is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ (\text{Fe}) - E^\circ (\text{Zn}) = (+0.44 \, \text{V}) - (+0.76 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = +0.44 \, \text{V} - 0.76 \, \text{V} = -0.32 \, \text{V} \]

### Final Answer
The EMF for the cell reaction \( \text{Fe}^{2+} + \text{Zn} \rightarrow \text{Zn}^{2+} + \text{Fe} \) is \( -0.32 \, \text{V} \).

---