# How to calculate the equivalent weight of KMnO4?

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.)

• Hint - What are the two redox half-cell reactions? – MaxW May 8 at 17:29
• @MaxW Ummm...I think $$\ce{2 H+ + 2 Cl- -> Cl2 + 2e- + 2H+}$$ and $$\ce{K+ + MnO4- + 5e- -> Mn^2+ + K+ + 4O^2-}$$ – ami_ba May 8 at 17:31
• You can leave off the $\ce{2H+}$ and $\ce{K=}$ since they are on both sides of the equation. – MaxW May 8 at 17:33
• @MaxW Are not these two the Oxidation and Reduction halves? – ami_ba May 8 at 17:34
• Yes, that is right. So per $\ce{Cl-}$ atom how many electrons are exchanged? Per $\ce{Mn}$ atom how many electrons are exchanged? – MaxW May 8 at 17:36

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:

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