The n-Factor and the Law of Equivalence in Redox

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle from the lab. In the flask, twenty five millilitres of ferrous sulphate, point one molar. In the burette, potassium permanganate, only point zero two molar. That is five times weaker. So you will need five times the volume, one hundred and twenty five millilitres, right? Hmm, let us not be so sure. The burette tells a very different story, and the reason is one small number called the n factor. Titration questions built on it come up often in JEE, so let's begin.

0:37What the n-factor counts

So, what is this n factor? The n factor, written with a small italic n, is the number of electrons gained or lost, per formula unit. That's it. Simple definition, okay? Let us start easy. Iron two plus becomes iron three plus. The two plus written above is the charge on the ion, and the arrow means goes to. How many electrons does each iron ion lose? Just one, yes! The charge goes from plus two to plus three, so one electron is lost. The n factor is one. Easy, right? Now oxalate, C two O four two minus, goes to two C O two. The small numbers written below are subscripts; they count atoms. Carbon is plus three in oxalate, and plus four in carbon dioxide. So the change is one, and the n factor is one, right? Oh no, careful! One oxalate ion has two carbon atoms, and each one changes by one. So per formula unit, the total is two. The n factor of oxalate is two. Always count every atom that changes. Got it? Same idea for potassium dichromate, K two Cr two O seven, in acid. Chromium comes down from plus six to plus three, a change of three, and there are two chromium atoms. Three times two: the n factor is six. Okay? And hydrogen peroxide, H two O two, has an n factor of two, whether it acts as an oxidant or as a reductant. With me?

2:08One compound, three n-factors: KMnO₄

Now, potassium permanganate, K M n O four. Manganese here is plus seven. Remember that number, okay? In acid, permanganate is reduced to manganese two plus. So what is its n factor? Five, exactly! Plus seven to plus two: five electrons gained per formula unit. Clear? But in a neutral or weakly alkaline medium, it stops at manganese dioxide, M n O two. Plus seven to plus four: n factor three. Right? And in a strongly alkaline medium, it only goes to manganate, M n O four two minus. Plus seven to plus six: n factor just one. Clear? So tell me, what is the n factor of potassium permanganate? Five? Three? One? See, there is no single answer! The n factor is not a fixed property of a compound. It depends on the reaction. So always write the product first, then find n. This is a very common trap. Got it?

3:13From n-factor to equivalents

Now, why do we need the n factor at all? For equivalents. The equivalent mass, capital E, equals capital M over n, where capital M is the molar mass. The horizontal bar means divided by, so we read it as M over n. Okay? Then, the number of equivalents is the mass, small m, over capital E, the equivalent mass. And that is the same as moles times n. Think about it: in a redox reaction, that is moles times electrons per formula unit. That is simply the moles of electrons transferred! And normality, capital N, equals molarity times n. Careful: here capital M means molarity, not molar mass. Same letter, different meaning. Clear?

4:00The law of equivalence

Now the big law. In any redox reaction, the reductant loses electrons and the oxidant gains them. Tell me, how do the electrons lost compare with the electrons gained? Equal, exactly! Every electron lost is gained by someone. And equivalents count moles of electrons. So, in any reaction, equivalents of oxidant equal equivalents of reductant. This is the law of equivalence. Okay? In a titration, that becomes N one V one equals N two V two, normality times volume on each side. The small one and two written below are subscripts for the two solutions. Right? Or, with molarity and the n factors, n one M one V one equals n two M two V two. Same law, okay? But many students drop the n factors and write M one V one equals M two V two. That works only when both n factors are equal. Otherwise, wrong answer! And that is exactly the mistake in our puzzle. Got it?

5:02Example 1: volume of 0.02 M KMnO4 for 25 mL of 0.1 M FeSO4

Now, remember our puzzle? Let's solve it. What volume of point zero two molar potassium permanganate, in acid, oxidises twenty five millilitres of point one molar ferrous sulphate? Capital M after a number means molar, and m L means millilitres. Pause the video, and try it first. Done? Okay, let's check. Some students write point zero two times V equals point one times twenty five, and get one hundred and twenty five millilitres. Correct? No, no, careful! That ignores the n factors. Permanganate in acid has n factor five. Iron two plus going to iron three plus has n factor one. Right? So use n one M one V one equals n two M two V two. With me? Put in the numbers. Five times point zero two times V, equals one times point one times twenty five. Clear? The left side is point one times V, and the right side is two point five. The double arrow means which gives. So V equals twenty five millilitres. See, puzzle solved! Permanganate is five times more dilute, but each formula unit takes five electrons, while each iron ion gives only one. So the same volume is enough. Beautiful, right?

6:20Example 2: n-factor of FeC2O4 vs Fe2(C2O4)3, equivalent mass

Now a harder one. Find the n factor and the equivalent mass of ferrous oxalate, F e C two O four, with molar mass one hundred and forty four, and of ferric oxalate, F e two, C two O four in brackets, whole three. The brackets mean the oxalate group is taken three times. Its molar mass is three hundred and seventy six. Both are oxidised by acidified potassium permanganate. Pause the video, and try it. Done? Okay, let's check. Ferrous oxalate first. Permanganate is a strong oxidant. So which parts of this compound get oxidised? Both parts, yes! Iron goes from plus two to plus three; that is one, shown in brackets. And the oxalate goes to two C O two; that is two, as we saw. Okay? So add them: n equals one plus two, three. Equivalent mass: one hundred and forty four over three, forty eight. Simple! Now ferric oxalate. Two irons and three oxalates. So two from the irons, plus six from the oxalates, n equals eight, right? Oh no, careful! What is the oxidation state of iron here? Plus three already. Ferric means iron three plus. So in this reaction iron does not change, and it adds nothing to n. Okay? Only the three oxalates change, and each one gives two. So n equals three times two, six. With me? Equivalent mass: three hundred and seventy six over six, sixty two point seven. So the answers are three and forty eight, then six and sixty two point seven. Always ask: which atoms actually change? Clear?

8:06Quick revision: n-factor, E = M/n, N = nM, n1M1V1 = n2M2V2

So, quick revision. The n factor is the number of electrons gained or lost per formula unit, and it depends on the reaction, so write the product first. Equivalent mass is M over n, equivalents are moles times n, and normality is molarity times n. And in any redox reaction, equivalents of oxidant equal equivalents of reductant: n one M one V one equals n two M two V two. Now go and try the practice problems on this page. You are ready for them. All the best!