Consider the following ` E^@` values . `E_(Fe^(3+)//Fe^(2+)^@ = + 0.77 V`,
`E_(Sn^(2+)//Sn)^@=- 0.14 V` The `E_(cell)^@` for the reaction ,
`Sn (s) + 2Fe^(3+)(aq) rarr 2 Fe^(2+)(aq)+ Sn^(2+)(aq)` is ?
(a)`-0.58 V`
(b)`-0.30 V`
(c)`+0.30V`
(d)`+0.58 V`

Options

A. `-0.58 V`
B. `-0.30 V`
C. `+0.30V`
D. `+0.58 V`

Explanation

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

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

we will follow these steps:

### Step 1: Identify the half-reactions
The overall reaction can be broken down into two half-reactions:

1. **Oxidation half-reaction (at the anode)**:
\[ \text{Sn (s)} \rightarrow \text{Sn}^{2+} (aq) + 2e^- \]

2. **Reduction half-reaction (at the cathode)**:
\[ \text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \]

### Step 2: Write down the standard reduction potentials
From the problem, we have the following standard reduction potentials:

- For the reduction of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \):
\[ E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \, \text{V} \]

- For the oxidation of \( \text{Sn} \) to \( \text{Sn}^{2+} \), we need to reverse the reduction potential:
\[ E^\circ(\text{Sn}^{2+}/\text{Sn}) = -0.14 \, \text{V} \quad \text{(given)} \]
Thus, the oxidation potential for Sn is:
\[ E^\circ(\text{Sn}/\text{Sn}^{2+}) = +0.14 \, \text{V} \]

### Step 3: Adjust the reduction half-reaction for stoichiometry
Since we have 2 moles of \( \text{Fe}^{3+} \) being reduced, we need to multiply the reduction half-reaction by 2:
\[ 2 \text{Fe}^{3+}(aq) + 2e^- \rightarrow 2 \text{Fe}^{2+}(aq) \]
The potential remains the same:
\[ E^\circ(\text{2Fe}^{3+}/\text{2Fe}^{2+}) = +0.77 \, \text{V} \]

### Step 4: Calculate the standard cell potential
Now we can calculate the standard cell potential using the formula:
\[ E_{cell}^\circ = E^\circ(\text{cathode}) - E^\circ(\text{anode}) \]
Substituting in the values:
\[ E_{cell}^\circ = E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\circ(\text{Sn}/\text{Sn}^{2+}) \]
\[ E_{cell}^\circ = 0.77 \, \text{V} - (-0.14 \, \text{V}) \]
\[ E_{cell}^\circ = 0.77 \, \text{V} + 0.14 \, \text{V} = 0.91 \, \text{V} \]

### Conclusion
The standard cell potential \( E_{cell}^\circ \) for the given reaction is \( +0.91 \, \text{V} \). However, since this value is not among the provided options, it indicates that there may have been a misunderstanding in the options given.