The direct current of `1.25A` was passed through `200mL` of `0.25M Fe_(@)(SO_(4))_(3)` solution for a period of 1.1 hour. The resulting solution in cathode chamber was analyzed by titrating against `KMnO_(4)` solution 25Ml permaganate solution was required to reach the end point. Determine molarity of `KMnO_(4)` solution.

Explanation

To determine the molarity of the KMnO4 solution used to titrate the Fe²⁺ ions produced from the electrolysis of Fe₂(SO₄)₃, we can follow these steps:

### Step 1: Calculate the total charge passed through the solution.
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I = 1.25 \, \text{A} \) (current)
- \( t = 1.1 \, \text{hours} = 1.1 \times 3600 \, \text{seconds} = 3960 \, \text{seconds} \)

Calculating the total charge:
\[ Q = 1.25 \, \text{A} \times 3960 \, \text{s} = 4950 \, \text{C} \]

### Step 2: Calculate the number of moles of electrons transferred.
Using Faraday's law:
\[ n = \frac{Q}{F} \]
where:
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)

Calculating the number of moles of electrons:
\[ n = \frac{4950 \, \text{C}}{96500 \, \text{C/mol}} = 0.0513 \, \text{mol} \]

### Step 3: Convert moles of electrons to millimoles.
Since \( 1 \, \text{mol} = 1000 \, \text{mmol} \):
\[ n = 0.0513 \, \text{mol} \times 1000 = 51.3 \, \text{mmol} \]

### Step 4: Relate moles of Fe²⁺ produced to moles of electrons.
The reduction of Fe³⁺ to Fe²⁺ involves the transfer of 1 mole of electrons per mole of Fe³⁺ reduced. Therefore, the moles of Fe²⁺ produced is equal to the moles of electrons transferred:
\[ \text{Moles of Fe}^{3+} \text{ reduced} = 51.3 \, \text{mmol} \]

### Step 5: Set up the titration reaction with KMnO₄.
The balanced reaction for the titration is:
\[ \text{MnO}_4^- + 5 \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5 \text{Fe}^{3+} \]
From this reaction, we see that 1 mole of KMnO₄ reacts with 5 moles of Fe²⁺.

### Step 6: Calculate the milliequivalents of Fe²⁺.
The milliequivalents of Fe²⁺ produced:
\[ \text{Milliequivalents of Fe}^{2+} = 51.3 \, \text{mmol} \]

### Step 7: Calculate the milliequivalents of KMnO₄.
Using the stoichiometry from the titration:
\[ \text{Milliequivalents of KMnO}_4 = \frac{\text{Milliequivalents of Fe}^{2+}}{5} = \frac{51.3}{5} = 10.26 \, \text{meq} \]

### Step 8: Calculate the molarity of KMnO₄.
Using the formula:
\[ \text{Milliequivalents} = \text{Molarity} \times \text{Volume (mL)} \]
where the volume of KMnO₄ solution used is 25 mL:
\[ 10.26 = M \times 25 \]
Solving for M:
\[ M = \frac{10.26}{25} = 0.4104 \, \text{M} \]

### Final Answer:
The molarity of the KMnO₄ solution is approximately **0.41 M**.

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