Given electrode potentials asre
`Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.771V`
` I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.536V`
`E^(c-)._(cell)` for the cell reaction,
`Fe^(3+)+2I^(c-) rarr Fe^(2+)+I_(2)` is

Options

A. `(2xx0.771-0.536)=1.006V`
B. `(0.771-0.5xx0.536)=0.503V`
C. `(0.771-0.536)=0.235V`
D. `(0.536-0.771)=0.236V`

Explanation

To find the cell potential \( E_{cell} \) for the reaction:

\[ \text{Fe}^{3+} + 2 \text{I}^- \rightarrow \text{Fe}^{2+} + \text{I}_2 \]

we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
- The reduction half-reaction for iron is:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \]
with a standard electrode potential \( E^\circ = 0.771 \, \text{V} \).

- The reduction half-reaction for iodine is:
\[ \text{I}_2 + 2e^- \rightarrow 2 \text{I}^- \]
with a standard electrode potential \( E^\circ = 0.536 \, \text{V} \).

### Step 2: Write the oxidation half-reaction for iodine.
Since iodine is being oxidized in the overall reaction, we need to reverse the reduction half-reaction:
\[ 2 \text{I}^- \rightarrow \text{I}_2 + 2e^- \]
The standard electrode potential for this oxidation reaction will be the negative of the reduction potential:
\[ E^\circ_{\text{oxidation}} = -0.536 \, \text{V} \]

### Step 3: Determine the cathode and anode reactions.
- **Cathode (reduction)**: The reduction of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \) occurs at the cathode.
- **Anode (oxidation)**: The oxidation of \( \text{I}^- \) to \( \text{I}_2 \) occurs at the anode.

### Step 4: Calculate the cell potential using the formula.
The cell potential \( E_{cell} \) can be calculated using the formula:
\[ E_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E_{cell} = 0.771 \, \text{V} - 0.536 \, \text{V} \]
\[ E_{cell} = 0.235 \, \text{V} \]

### Conclusion
The standard cell potential \( E_{cell} \) for the reaction is:
\[ \boxed{0.235 \, \text{V}} \]