Question

Write balanced chemical equations for action of potassium permanganate on:
1. Hydrogen
2. Warm conc.sulphuric acid
Explain why $M{n^{2 + }}$ ion is more than $M{n^{3 + }}$? (Given:$Mn \to Z = 35$)

Answer 1

Hint: We know that a balanced chemical equation contains the same number of atoms for each element that takes place in the chemical reaction. We have to calculate the number of atoms for a given element by multiplying the coefficient of any formula having that element by the subscript of the element in the formula. If we notice an element is seen in more than one formula on a given side of the equation, we have to compute the number of atoms and then add them together.

Complete answer:
Action of potassium permanganate with hydrogen
When we react potassium permanganate with hydrogen the products formed are manganese dioxide, potassium hydroxide, and water. We can write the chemical equation as,
$KMn{O_4} + {H_2} \to Mn{O_2} + KOH + {H_2}O$
We can see that the above reaction is unreacted.
We need two moles of potassium permanganate and three moles of hydrogen are required in the reactant side. We need two moles of manganese dioxide, two moles of potassium hydroxide, and two moles of water in the product side to make this reaction a balanced one.
We can write the balanced equation as,
$2KMn{O_4} + 3{H_2} \to 2Mn{O_2} + 2KOH + 2{H_2}O$
Action of potassium permanganate with warm. Sulphuric acid
When we react potassium permanganate with warm sulfuric acid the products formed are potassium sulfate, manganese sulfate, water and hydrogen. The reaction of potassium permanganate with warm sulfuric acid takes place in the presence of air. We can write the chemical equation as,
$6KMn{O_4} + 9{H_2}S{O_4} \to 3{K_2}S{O_4} + 6MnS{O_4} + 9{H_2}O + 5{O_2}$
We know that electronic configuration of $Mn$ is $\left[ {Ar} \right]3{d^5}4{s^2}$.
We can write the electronic configuration of $M{n^{2 + }}$ as $\left[ {Ar} \right]3{d^5}4{s^0}$.
We can write the electronic configuration of $M{n^{3 + }}$ as $\left[ {Ar} \right]3{d^4}4{s^0}$.
From the electronic configurations of $M{n^{2 + }}$ and $M{n^{3 + }}$, we can observe that the number of unpaired electrons in $M{n^{2 + }}$ is five and the number of unpaired electrons in $M{n^{3 + }}$ is four. The maximum value for a transition metal ion in d-orbital is five orbitals and it has half-filled electronic configuration. This is the reason why $M{n^{2 + }}$ is more stable than $M{n^{3 + }}$.

Note: We have remembered that based on the law of conservation of mass, matter cannot be created or destroyed. Therefore, the mass of the products in a chemical reaction is equal to mass of the reactants. The purpose of balancing chemical equations is to have the same number of each type of element on each side of a chemical equation. Balancing chemical equations also helps to determine the stoichiometric relationship between the substances/elements/compounds.