# Question fd517

Dec 6, 2015

$\text{1.8 N}$

#### Explanation:

In order to be able to determine the normality of the potassium permanganate, ${\text{KMnO}}_{4}$, solution, you first need to write a balanced equation for this reduction half-reaction.

Start by assigning oxidation numbers to the atoms that take part in the reaction

$\stackrel{\textcolor{b l u e}{+ 7}}{{\text{Mn") stackrel(color(blue)(-2))("O"_4^(-)) -> stackrel(color(blue)(+4))("Mn") stackrel(color(blue)(-2))("O}}_{2}}$

As you can see, manganese is being reduced from a $\textcolor{b l u e}{+ 7}$ oxidation state on the reactants' side to a $\textcolor{b l u e}{+ 3}$ oxidation state on the products' side.

This means the each manganese atom will gain $3$ electrons.

stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 3"e"^(-) -> stackrel(color(blue)(+4))("Mn") "O"_2

You don't need to worry about balancing the oxygens, that is not important for finding the normality of the potassium permanganate solution.

So, in the context of a redox reaction, normality is defined as the number of equivalents of an oxidizing (or reducing) agent per liters of solution.

Here an equivalent is simply the number of moles of electrons that are being transferred in the redox reaction. The relationship between normality and molarity can be written like this

$\textcolor{b l u e}{\text{normality" = "no. of Eq" xx "molarity}}$

In this case, one mole of permanganate ions will react with three moles of electrons. This means that the permanganate ion's normality is equal to

N = 3 * "0.60 M" = color(green)("1.8 N")#