Given electrode potentials: `Fe^(3+) + e^(-) rarr Fe^(2+) : E^(o) =0.771 V and I_(2) 2e^(-) rarr 2I^(-), E^(o) = 0.556` V then `E^(o)` for the cell reaction `2 Fe^(3+) 2I ^(-) rarr 2 Fe^(2+) + l_(2)` will be:

Options

A. `(2xx 0.771 - 0. 536)= 1.006 ` Volt
B. `(0.771 - 0.5 xx 0.536) = 0. 503` Volt
C. `0.771 – 0.536 = 0.235` Volt
D. `0.536 – 0.771 = – 0.235` Volt

Answer

To find the standard cell potential \( E^\circ \) for the given cell reaction \( 2 \text{Fe}^{3+} + 2 \text{I}^- \rightarrow 2 \text{Fe}^{2+} + \text{I}_2 \), we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
The half-reactions given are:
1. \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) with \( E^\circ = 0.771 \, \text{V} \) (reduction)
2. \( \text{I}_2 + 2e^- \rightarrow 2 \text{I}^- \) with \( E^\circ = 0.556 \, \text{V} \) (reduction)

### Step 2: Write the oxidation half-reaction.
Since the iodine half-reaction involves reduction, we need to reverse it to represent oxidation:
- Oxidation: \( 2 \text{I}^- \rightarrow \text{I}_2 + 2e^- \)
- The standard potential for oxidation will be the negative of the reduction potential:
\[ E^\circ_{\text{oxidation}} = -0.556 \, \text{V} \]

### Step 3: Combine the half-reactions.
Now, we can combine the two half-reactions:
- Reduction: \( 2 \text{Fe}^{3+} + 2e^- \rightarrow 2 \text{Fe}^{2+} \) (multiplied by 2)
- Oxidation: \( 2 \text{I}^- \rightarrow \text{I}_2 + 2e^- \)

### Step 4: Write the overall cell reaction.
Adding the two half-reactions gives us:
\[ 2 \text{Fe}^{3+} + 2 \text{I}^- \rightarrow 2 \text{Fe}^{2+} + \text{I}_2 \]

### Step 5: Calculate the standard cell potential \( E^\circ_{\text{cell}} \).
The standard cell potential is calculated using:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.771 \, \text{V} + (-0.556 \, \text{V}) = 0.771 \, \text{V} - 0.556 \, \text{V} = 0.215 \, \text{V} \]

### Step 6: Final answer.
Thus, the standard cell potential \( E^\circ \) for the cell reaction is:
\[ E^\circ_{\text{cell}} = 0.215 \, \text{V} \]