Question

A direct current of 1.25A was passed through 200ml of $F{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$solution for a period of 1.1 hour. The resulting solution in the cathode chamber was analyzed by titrating against the $KMn{{O}_{4}}$ solution. 25mL permanganate solution was required to reach the end point. Determine molarity of $KMn{{O}_{4}}$ solution.

Answer 1

In order to solve the question, you need the value of ${{E}^{0}}$. Use value of ${{E}^{0}}$ as 0.77V, for the conversion of $F{{e}^{3+}}$ to $F{{e}^{2+}}$.

Complete answer:
When electric current is passed through the $F{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ solution, a change in oxidation state occurs for Fe. It changes from the +3 oxidation state to the +2 oxidation state, by giving away an electron.
F{{e}^{3+}}$$$$+\text{ }{{e}^{-}}~->~$$$$F{{e}^{2+}}, E0= 0.77 V
Now as the current as Faradays is passed, it is to be noted that the charge is constant. Charge can be written in two ways: $Number\text{ }of\text{ }Electrons\times Faradays$ or $Current\times Time$
Equating both sides, we obtain that
$n\times F\text{ }=\text{ }I\times t$ , where n=number of electrons, F=Faraday, I=current and t=time
Taking F as 96500, we can find the number of electrons from the equation. So the number of electrons are:
$n\text{ }=\text{ }\left( 1.25\times 1.1\times 3600 \right)\div 96500$
So, $n\text{ }=\text{ }0.0513$……….(i)
So we have 0.0513 number of electrons. But you need to understand the fact that the number of electrons are actually the n mole of $F{{e}^{3+}}$ which got reduced in the reaction. So the molar equivalents of $F{{e}^{3+}}$and $KMn{{O}_{4}}$ will be the same. Now, the milli equivalent of $F{{e}^{3+}}$ will be 51.3. For finding out the milli equivalent for $KMn{{O}_{4}}$ ,
KMn{{O}_{4}}$$$$+\text{ }8{{H}^{+}}+\text{ }5{{e}^{-}}->\text{ }M{{n}^{2+}}+4{{H}_{2}}O
From the reaction , there is an exchange of 5 electrons, so the milli equivalent of $KMn{{O}_{4}}$ will be $molarity\text{ }KMn{{O}_{4}}\times 5\text{ }electron\times 25\text{ }volume$………….(ii)
Equating (i) and (ii), we have the resultant molarity of $KMn{{O}_{4}}$ as 0.41 molar.

Note:
Always check for the units in the question. The first priority should be to change the units to the SI system and then proceed. For convenience the value of F is taken to be 96500. For titration, you need to write the equations and balance them properly, or else there will be a problem in the calculation of the number of electrons exchanged.