Consider the following `E^o` values :
`E_(Fe^(3+)//Fe^(2+)) = + 0.77 V`
`E_(Sn^(2+)//Sn) =- 0.14 V`
Under standard conditions the potential for reaction
`Sn(s) +2Fe^(3+) (aq) rarr 2Fe^(2+) (sq) + Sn^(2+) (aq)` is.

Options

A. ` 0.63 V`
B. ` 1.40 V`
C. ` 1.68 V`
D. ` 0.91 V`

Explanation

To find the standard cell potential (E°cell) for the reaction:

\[ \text{Sn(s)} + 2\text{Fe}^{3+}(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Sn}^{2+}(aq) \]

we will follow these steps:

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

1. The reduction half-reaction for iron is:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (E^\circ = +0.77 \, \text{V}) \]

2. The oxidation half-reaction for tin is:
\[ \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \quad (E^\circ = -0.14 \, \text{V}) \]

### Step 2: Determine the oxidation and reduction processes.

- In the overall reaction, Sn is oxidized to Sn²⁺, and Fe³⁺ is reduced to Fe²⁺.
- Therefore, we need to reverse the standard reduction potential for the oxidation of Sn:
\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \quad (E^\circ = +0.14 \, \text{V}) \]

### Step 3: Write the overall cell reaction.

The overall cell reaction can be written as:
\[ \text{Sn}(s) + 2\text{Fe}^{3+}(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{Sn}^{2+}(aq) \]

### Step 4: Calculate E°cell.

The standard cell potential (E°cell) is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Where:
- \(E^\circ_{\text{cathode}} = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \, \text{V}\)
- \(E^\circ_{\text{anode}} = E^\circ_{\text{Sn}^{2+}/\text{Sn}} = +0.14 \, \text{V}\)

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.77 \, \text{V} - 0.14 \, \text{V} = 0.63 \, \text{V} \]

### Final Answer:
The standard cell potential for the reaction is:
\[ E^\circ_{\text{cell}} = 0.63 \, \text{V} \]