An aqueous solution containing `0.1M Fe^(3+)` and `0.01 M Fe^(2+)` was titrated with a concentrated solution of `NaOH` at `30^(@)C` , so that changes in volumes were negligible. Assuming that the new species formed during titration are `Fe(OH)_(3)` and `Fe(OH)_(2)` only.
Given `E^(c-)._(Fe^(3+)|Fe^(2+))=0.80V, `
`K_(spFe(OH)_(3))=10^(-37), ` and `K_(sp Fe(OH)_(2))=10^(-19)`
The redox potential of `Fe^(3+)|Fe^(2+)` electrode at `pH=6` is
(a)`0.8V`
(b)`0.5V`
(c)`0.2V`
(d)`0.1V`

Options

A. `0.8V`
B. `0.5V`
C. `0.2V`
D. `0.1V`

Explanation

To solve the problem, we need to calculate the redox potential of the Fe³⁺/Fe²⁺ electrode at pH = 6, considering the solubility product constants (Ksp) of the hydroxides formed during the titration. Here’s a step-by-step solution:

### Step 1: Identify the relevant equilibrium reactions
The relevant reactions for the formation of hydroxides are:
1. For Fe(OH)₃:
\[ \text{Fe}^{3+} + 3 \text{OH}^- \rightleftharpoons \text{Fe(OH)}_3 \quad (K_{sp} = 10^{-37}) \]
2. For Fe(OH)₂:
\[ \text{Fe}^{2+} + 2 \text{OH}^- \rightleftharpoons \text{Fe(OH)}_2 \quad (K_{sp} = 10^{-19}) \]

### Step 2: Calculate the concentration of OH⁻ ions at pH = 6
The concentration of hydroxide ions can be calculated from the pH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 6 = 8 \]
\[ [\text{OH}^-] = 10^{-8} \, \text{M} \]

### Step 3: Set up the expressions for Ksp
Using the Ksp expressions for both hydroxides:
1. For Fe(OH)₃:
\[ K_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3 = 10^{-37} \]
\[ [\text{Fe}^{3+}] \cdot (10^{-8})^3 = 10^{-37} \]
\[ [\text{Fe}^{3+}] \cdot 10^{-24} = 10^{-37} \implies [\text{Fe}^{3+}] = 10^{-13} \, \text{M} \]

2. For Fe(OH)₂:
\[ K_{sp} = [\text{Fe}^{2+}][\text{OH}^-]^2 = 10^{-19} \]
\[ [\text{Fe}^{2+}] \cdot (10^{-8})^2 = 10^{-19} \]
\[ [\text{Fe}^{2+}] \cdot 10^{-16} = 10^{-19} \implies [\text{Fe}^{2+}] = 10^{-3} \, \text{M} \]

### Step 4: Use the Nernst equation
The Nernst equation for the half-cell reaction is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} \right) \]
Where:
- \(E^\circ = 0.80 \, \text{V}\)
- \(n = 1\) (since one electron is transferred)

Substituting the concentrations:
\[ E = 0.80 - 0.0591 \log \left( \frac{10^{-3}}{10^{-13}} \right) \]
\[ = 0.80 - 0.0591 \log(10^{10}) \]
\[ = 0.80 - 0.0591 \times 10 \]
\[ = 0.80 - 0.591 \]
\[ = 0.209 \, \text{V} \]

### Step 5: Rounding off
Rounding off the value gives:
\[ E \approx 0.2 \, \text{V} \]

### Conclusion
The redox potential of the Fe³⁺/Fe²⁺ electrode at pH = 6 is approximately **0.2 V**. Therefore, the correct answer is (c) 0.2 V.