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 solve the problem, we need to determine the standard cell potential (E°_cell) for the given electrochemical reaction:

**Given Reaction:**
\[ \text{Fe}^{3+} + 2\text{I}^- \rightarrow \text{Fe}^{2+} + \text{I}_2 \]

**Given Electrode Potentials:**
1. For the reduction of iron:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}, \quad E^\circ = 0.771 \, \text{V} \]

2. For the reduction of iodine:
\[ \text{I}_2 + 2e^- \rightarrow 2\text{I}^-, \quad E^\circ = 0.536 \, \text{V} \]

### Step-by-Step Solution:

**Step 1: Identify the half-reactions and their potentials.**
- The reduction half-reaction for iron is:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}, \quad E^\circ = 0.771 \, \text{V} \]

- The oxidation half-reaction for iodine can be derived from the reduction half-reaction:
\[ 2\text{I}^- \rightarrow \text{I}_2 + 2e^-, \quad E^\circ = -0.536 \, \text{V} \]
(Note: The sign is flipped because we are considering the oxidation process.)

**Step 2: Write the overall cell reaction.**
- The overall cell reaction is:
\[ \text{Fe}^{3+} + 2\text{I}^- \rightarrow \text{Fe}^{2+} + \text{I}_2 \]

**Step 3: Calculate the standard cell potential (E°_cell).**
- The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]

- Plugging in the values:
\[ E^\circ_{\text{cell}} = 0.771 \, \text{V} - (-0.536 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 0.771 \, \text{V} + 0.536 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 0.771 + 0.536 = 1.307 \, \text{V} \]

**Step 4: Finalize the answer.**
- The standard cell potential for the reaction is:
\[ E^\circ_{\text{cell}} = 1.307 \, \text{V} \]