How to calculate the equivalent weight of KMnO4?

Question

Let us consider the following reaction

$$\ce{2 KMnO4 + 16 HCl → 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2}$$

Now, in order to calculate the equivalent mass of $\ce{KMnO4}$, first I need to calculate it's $n$-factor which turns out to be $5$ because the oxidation state of $\ce{Mn}$ in $\ce{KMnO4}$ is $+7$ whereas in $\ce{MnCl2}$ it is $+2$. And the $n$-factor of $\ce{HCl}$ is $1$. So, the equivalent weight of $\ce{KMnO4}$ is

$$\frac{\text{molecular weight}}{5}$$

So, ratio of moles of $\ce{KMnO4}$ reacting with $\ce{HCl}$ should be $1:5$.

But from the balanced reaction we can see that this ratio is $1:8$.

Where am I wrong? Is my understanding and hence calculation of $n$-factor and equivalent weight is wrong?

(Please pardon this childish fashion of writing the question. If there is a problem anywhere, please do comment before flagging or downvoting.)

Answer 1

Refer to the definition of equivalent (IUPAC Gold Book):

equivalent entity

Entity corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions.

In other words, $\pu{1 equiv}$ is the amount of substance reacting with $\pu{1 mol}$ of hydrogen atom. If hydrogen acts as a reducing or oxidizing agent, then either way $\pu{1 mol}$ of hydrogen atoms liberates or accepts $\pu{1 mol}$ of electrons:

$$ \begin{align} \ce{0.5 H2 &→ H+ + e-}\\ \ce{0.5 H2 + e- &→ H-} \end{align} $$

That's why an equivalent of a redox agent is its amount which liberates or accepts $\pu{1 mol}$ of electrons upon being oxidized or reduced, respectively, assuming electrons don't exist in solution on its own for a significant period of time.

Note that $\ce{KMnO4}$ participates in redox reaction

$$\ce{2K\overset{+7}{Mn}O4 + 16 H\overset{-1}{Cl} = 2 KCl + 2 \overset{+2}{Mn}Cl2 + 8 H2O + 5 \overset{0}{Cl}_2}$$

and the half-reaction for the reduction of manganese is

$$\ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- → \overset{+2}{Mn}^2+ + 4H2O}$$

Also note that $n$-factor of permangante here is not the number of protons $\ce{H+}$, which are also used up in water formation; rather it's the number of transferred electrons, e.g. $n = 5$ and

$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{5} = \pu{31.61 g mol-1}$$

Obviously, the equivalent mass of permanganate isn't a constant and depends on the $\mathrm{pH}$ of the reaction. For example, in neutral medium half-reaction appears as

$$\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- → \overset{+4}{Mn}O2 + 4 OH-}$$

and there is no explicitly shown protons to count at all! However, the $n$-factor is $3$, and the equivalent mass of permanganate would be a different value:

$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{3} = \pu{52.68 g mol-1}$$