The standard electrode potential of the half cells are given below :
`Zn^(2+)+ 2e^(-) rarr Zn(s), E^(o) = - 7.62 V `
`Fe^(2+) + 2e^(-) rarr Fe(s) , E^(o) = -7 .81 V`
The emf of the cell `Fe^(2+) + Zn rarr Zn^(2+) + Fe` will be :

Options

A. `1.54` V
B. `-1.54` V
C. `-0. 19` V
D. `+ 0 . 19 ` V

Answer

To find the emf of the cell reaction given by:

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

we will follow these steps:

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

The half-reactions are:
1. Reduction half-reaction:
\[ \text{Fe}^{2+} + 2e^{-} \rightarrow \text{Fe}(s), \quad E^{\circ} = -0.78 \, \text{V} \]
2. Oxidation half-reaction:
\[ \text{Zn}(s) \rightarrow \text{Zn}^{2+} + 2e^{-}, \quad E^{\circ} = -0.76 \, \text{V} \]

### Step 2: Reverse the oxidation half-reaction

Since oxidation occurs at the anode, we reverse the zinc half-reaction:
\[ \text{Zn}^{2+} + 2e^{-} \rightarrow \text{Zn}(s) \]
This means we take the standard electrode potential for this half-reaction as positive:
\[ E^{\circ}_{\text{oxidation}} = +0.76 \, \text{V} \]

### Step 3: Calculate the standard cell potential (emf)

The standard cell potential \( E^{\circ}_{\text{cell}} \) can be calculated using the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]

Here, the cathode is where reduction occurs (Fe) and the anode is where oxidation occurs (Zn):
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{Fe}^{2+}/\text{Fe}} - E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} \]
\[ E^{\circ}_{\text{cell}} = (-0.78 \, \text{V}) - (-0.76 \, \text{V}) \]
\[ E^{\circ}_{\text{cell}} = -0.78 + 0.76 \]
\[ E^{\circ}_{\text{cell}} = -0.02 \, \text{V} \]

### Step 4: Final answer

The emf of the cell reaction is:
\[ E^{\circ}_{\text{cell}} = -0.02 \, \text{V} \]