Determine the equilibrium constant of the reaction at 298 K,
`2Fe^(3+) +Sn^(2+) hArr 2Fe^(2+) +Sn^(4+)`
From the obtained value of the equilibrium constant, predict whether `Sn^(2+)` ions can reduce `Fe^(3+)` to `Fe^(2+)` quantitatively or not.
Given: `E_(Fe^(3+),Fe^(2+)//Pt)=0.771V.E_(Sn^(4+)//Sn^(2+)//Pt)^(Theta)=0.150V`.
`[R=8.314JK^(-1)"mol"^(-1)]`

Explanation

To determine the equilibrium constant \( K \) for the reaction

\[ 2 \text{Fe}^{3+} + \text{Sn}^{2+} \rightleftharpoons 2 \text{Fe}^{2+} + \text{Sn}^{4+} \]

at 298 K, we can follow these steps:

### Step 1: Identify the half-reactions
The half-reactions involved in the overall reaction are:
1. Reduction half-reaction:
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (E^\circ = 0.771 \, \text{V}) \]
2. Oxidation half-reaction:
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \quad (E^\circ = 0.150 \, \text{V}) \]

### Step 2: Calculate the standard cell potential \( E^\circ_{\text{cell}} \)
The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is the reduction of \( \text{Fe}^{3+} \) and the anode is the oxidation of \( \text{Sn}^{2+} \).

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.771 \, \text{V} - 0.150 \, \text{V} = 0.621 \, \text{V} \]

### Step 3: Determine the number of electrons transferred \( n \)
From the balanced equation, we see that 2 electrons are transferred in the overall reaction:
\[ n = 2 \]

### Step 4: Calculate the Gibbs free energy change \( \Delta G^\circ \)
Using the relationship between cell potential and Gibbs free energy:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where \( F \) (Faraday's constant) is approximately \( 96500 \, \text{C/mol} \).

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 0.621 \, \text{V} = -120,000 \, \text{J/mol} \quad (\text{approximately}) \]

### Step 5: Relate \( \Delta G^\circ \) to the equilibrium constant \( K \)
Using the equation:
\[ \Delta G^\circ = -2.303RT \log K \]
We can rearrange this to find \( K \):
\[ \log K = -\frac{\Delta G^\circ}{2.303RT} \]

Substituting the values:
- \( R = 8.314 \, \text{J/(K mol)} \)
- \( T = 298 \, \text{K} \)

Calculating \( \log K \):
\[ \log K = -\frac{-120000}{2.303 \times 8.314 \times 298} \]
Calculating the denominator:
\[ 2.303 \times 8.314 \times 298 \approx 5700.4 \]
Thus,
\[ \log K \approx \frac{120000}{5700.4} \approx 21.0053 \]

### Step 6: Calculate \( K \)
Taking the antilog:
\[ K \approx 10^{21.0053} \approx 1.01 \times 10^{21} \]

### Conclusion
Since \( K \) is a very large number (\( K \approx 1.01 \times 10^{21} \)), this indicates that the reaction proceeds almost completely to the right. Therefore, \( \text{Sn}^{2+} \) ions can reduce \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \) quantitatively.

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