Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You have one mole of iron two sulphate, and you oxidise it with potassium permanganate in acid. How many moles of permanganate do you need? One mole for one mole sounds fair, right? Hmm, not so fast, because the real answer is much smaller, and today we will see exactly why. This is Part A of the unit problem bank on the d and f block elements, at JEE Main level. We will build two tools, then solve two of its questions together, with a quick one on the way. So, let's begin.

0:37Tool 1: electrons taken by permanganate and dichromate

So, the first tool. Our two famous oxidising agents are K Mn O four, potassium permanganate, and K two Cr two O seven, potassium dichromate; the small numbers written low count the atoms. Manganese is in the plus seven state in permanganate, and chromium is plus six in dichromate, okay? In acid, M n O four minus, plus eight H plus, plus five e minus, gives M n two plus, plus four H two O; e minus is the electron, the arrow means gives, and the small raised signs are the charges. Manganese drops from plus seven to plus two, a gain of five electrons, so its n factor, the small n on screen, is five. Got it? Now tell me, is the n factor of permanganate always five, whatever the solution? No, it depends on the medium! In neutral solution, permanganate stops at M n O two, manganese dioxide: M n O four minus, plus two H two O, plus three e minus, gives M n O two, plus four O H minus. Manganese goes from plus seven to plus four, so n is three, and in strong alkali it is just one. So always check the medium. Clear? Now let us do dichromate: in acid, chromium goes from plus six to plus three, so it gains three electrons. So the n factor of dichromate is three, right? No, no, careful, count the chromium atoms! One dichromate ion has two chromium atoms, and each gains three electrons, so that makes six in total. C r two O seven two minus, plus fourteen H plus, plus six e minus, gives two C r three plus, plus seven H two O; so in acid, n is six for dichromate. With me? Here is a quick one on the way, from question four of Part A, okay? What is the equivalent mass of K two Cr two O seven in acid? Its molar mass, the capital M on screen, is two hundred and ninety four. Equivalent mass is the molar mass over the n factor. Two hundred and ninety four over six gives forty nine. And notice, we divided by six, not by three. Right?

2:57Tool 2: E naught and reducing power

Now the second tool. Capital E with a small raised circle is read E naught, the standard electrode potential. In the bracket, the slash separates M three plus from M two plus, where M stands for the metal. It tells us how easily M three plus takes one electron to become M two plus. Okay? If E naught is high, M three plus grabs that electron eagerly, so it is a stronger oxidising agent. Right? So let me ask you this: what does a low, more negative E naught tell you about M two plus? Yes, exactly, M two plus gives away its electron easily, so it is a stronger reducing agent! Low E naught, strong reducing agent. Got it? Here are the course values, in volts, written V at the end: titanium minus zero point three seven, vanadium minus zero point two six, chromium minus zero point four one, manganese plus one point five seven, iron plus zero point seven seven, and cobalt plus one point nine seven. Clear? Look at manganese: plus one point five seven is very high, and here is why. M n three plus is three d four; the raised four counts the d electrons. Taking one electron makes M n two plus, three d five, a half-filled set, which is very stable. So M n three plus is a strong oxidising agent. Keep this idea ready for example two. With me?

4:25Example 1: moles of KMnO4 to oxidise 1 mol FeSO4 in acid

Now, example one, which is question two of Part A, and it is our puzzle from the start. How many moles of K Mn O four oxidise one mole of F e S O four, iron two sulphate, in acid? Pause the video and try it first. Done? Okay, let's check. Iron two sulphate gives F e two plus, and permanganate turns it into F e three plus. So each F e two plus loses just one electron. Okay? Now, from tool one, permanganate in acid takes five electrons. So how many F e two plus ions can one permanganate ion oxidise? Yes, exactly, it is five! One M n O four minus takes five electrons, so it oxidises five F e two plus ions. Right? The course writes the full equation like this: M n O four minus, plus eight H plus, plus five F e two plus, gives M n two plus, plus five F e three plus, plus four H two O; one permanganate for five iron ions, just as we said. Clear? So the answer is five moles of permanganate, right? No, no, wait, that is upside down! Five F e two plus ions need one permanganate, so one F e two plus needs only one fifth of it. One over five gives zero point two. With me? So the answer is zero point two moles of potassium permanganate. Our puzzle is solved: one mole for one mole was wrong, because each permanganate takes five electrons while each iron gives only one. Very good!

6:08Example 2: why Cr2+ is a stronger reductant than Fe2+

Now example two, which is question five of Part A, a medium one. Why is C r two plus a stronger reductant, that means a stronger reducing agent, than F e two plus? Pause the video and try it first. Done? Okay, let's check. Use tool two and compare the two couples. For the chromium three plus, chromium two plus couple, E naught is minus zero point four one volt. Okay? For the iron three plus, iron two plus couple, E naught is plus zero point seven seven volt. Right? Iron's value is the bigger number. So F e two plus must be the stronger reducing agent, right? No, no, careful, that is the trap! For a reducing agent we want the lower value. Minus zero point four one is much lower, so C r two plus is oxidised more easily. Clear? But why is chromium's value so negative? Let me ask you this: when C r two plus loses an electron, which configuration does it reach? It becomes C r three plus, three d three, yes! C r two plus is three d four, and it loses one electron. Here t with a small two g below names a set of d orbitals, and the raised three means three electrons in it. This t two g three arrangement is stable, so C r two plus happily loses that electron. With me? So, oxidising C r two plus gives the stable d three ion, C r three plus. It is the same idea as manganese in tool two: the configuration explains the number. Very good!

7:49Quick revision: n-factors, electron balance, E naught rule

So, quick revision. Permanganate has an n factor of five in acid and three in neutral solution, while dichromate has six in acid. The electrons given by the reducing agent equal the electrons taken by the oxidising agent. And a lower E naught for the M three plus, M two plus couple means M two plus is a stronger reducing agent. Now go and try the practice problem on this page. All the best!