Mixed Problems: Set 2 (Dichromate, Permanganate and Other Compounds)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a green solution of potassium manganate, and simply bubble carbon dioxide gas through it. We add nothing but carbon dioxide, okay? And yet the green turns purple, the colour of permanganate! So where does the purple come from? This is Set two of the mixed problems on the d block: eight quick questions on dichromate, permanganate and other compounds, in the JEE Main style. Questions like these come up often. First we sharpen three tools with quick questions from the set, and then we solve two more problems, our puzzle included. So, let's begin.

0:42Manganese: oxidation states and n-factors

First tool: the oxidation states of manganese. Here is K Mn O four, potassium permanganate, and its ion, M n O four minus; the small four written low counts the oxygen atoms, and the raised minus is the charge. Let x be the oxidation state of manganese. Four oxygens at minus two each, and the total charge is minus one, so x plus four times minus two equals minus one. That gives x equal to plus seven; the double arrow means so. And permanganate is purple. Okay? Now manganate, M n O four two minus, with the raised two minus as its charge. The same working gives manganese plus six, and manganate is green. And in M n O two, manganese dioxide, manganese is plus four. So three states to remember: plus seven, plus six and plus four. Right? Now the n factor: the number of electrons gained or lost by one formula unit. In acid, permanganate goes to M n two plus; the arrow means goes to. Plus seven to plus two is five electrons, so the n factor is five. Clear? Now a quick question from Set two. What is the n factor of K Mn O four in neutral solution? Five again, right? No, no, careful, five is only for acid! In neutral or faintly alkaline solution, permanganate stops at M n O two. Plus seven to plus four is three electrons, so the n factor is three, and that is the answer in the set. With me? And in strongly alkaline solution, it only goes as far as manganate: plus seven to plus six, so the n factor is just one. So always read the medium first. Got it?

2:25Chromium at +6: chromyl chloride and dichromate

Second tool: chromium. Here is another quick question from the set. Find the oxidation state of chromium in C r O two C l two, chromyl chloride. Let us check it together, with x as the oxidation state of chromium. Two oxygens at minus two give minus four, and two chlorines at minus one give minus two. The molecule is neutral, so x minus four minus two equals zero, and x is plus six. Clear? Now K two Cr two O seven, potassium dichromate, where chromium is also plus six. In acid, one C r two O seven two minus ion takes fourteen H plus and six electrons, and gives two C r three plus ions and seven water molecules. Here H plus is the hydrogen ion, and e minus is the electron. Okay? Now let me ask you this. Each chromium goes from plus six to plus three, a change of three. So is the n factor of dichromate three? No, it is six, and that is a common trap! One dichromate ion has two chromium atoms, and each one takes three electrons. Two times three is six; the cross means times. And as plus six becomes plus three, orange turns green. Right?

3:43Counting electrons: H₂O₂ with permanganate

Third tool: counting electrons, again with questions from the set. Hydrogen peroxide, H two O two, is added to acidified K Mn O four. Which gas is evolved? Yes, it is oxygen, O two! Permanganate is the oxidising agent here, so the peroxide acts as the reductant, the reducing agent, and gives off oxygen gas. With me? Now the next question: how many moles of K Mn O four, in acid, oxidise one mole of H two O two? We count the electrons. Each H two O two loses two electrons, and each permanganate in acid takes five. Okay? So one mole of peroxide needs two point five moles of permanganate, right? Oh no, careful, that ratio is upside down! Let x be the moles of permanganate; the m o l on screen is short for mole. Electrons gained must equal electrons lost, so five times x equals two. Two over five gives zero point four mole. Clear? And the balanced equation agrees. Two M n O four minus, plus six H plus, plus five H two O two, give two M n two plus, plus eight water, plus five O two. Two permanganates for five peroxides, and two over five is zero point four again. Got it?

5:05Example 1: why green K2MnO4 turns purple with CO2

Now, Example one from Set two, and it is our puzzle! Why does a green solution of K two Mn O four, potassium manganate, turn purple when C O two, carbon dioxide, is bubbled through it? Pause the video, and try it first. Done? Okay, let's check. Carbon dioxide is an acidic oxide, so it uses up the alkali that keeps manganate stable. And in manganate, manganese is plus six, right between plus four and plus seven. So once the alkali is gone, what can it do? Yes, exactly, it goes both up and down! Manganate is stable only in strongly alkaline solution; in neutral or acid solution it does this. Three M n O four two minus, plus four H plus, give two M n O four minus, plus M n O two, plus two water. Right? Two manganese atoms go up from plus six to plus seven, as purple permanganate. One comes down from plus six to plus four, as M n O two. So no outside oxidising agent is needed, and that solves our puzzle! With me? So the answer is disproportionation, where one element, here manganese at plus six, is both oxidised and reduced in the same reaction.

6:17Example 2: moles of I2 from CuSO4 with excess KI

Okay? Now, Example two from Set two, a medium one. Copper sulphate, C u S O four, is treated with an excess of K I, potassium iodide. How many moles of iodine are liberated per mole of copper sulphate? Pause the video, and try it first. Done? Okay, let's check. Copper two plus oxidises the iodide. Many students say the copper goes down to copper metal, taking two electrons, so one mole of iodine per mole. Is that right? No, no, wait, the copper stops at plus one! Two C u two plus, plus four I minus, give two C u I, plus I two. C u I is copper one iodide, a white solid, so only two of the four iodide ions become iodine. Clear? Now count the electrons. Each copper goes from plus two to plus one, taking one electron. And two iodide ions give two electrons to make one I two. So it takes two coppers to make one iodine molecule. Right? So two moles of copper two plus give one mole of iodine. Per mole of copper sulphate, that is zero point five mole of iodine, not one. Always check the product before you count. Got it?

7:35Quick revision: n-factors, electron balance, CuI, manganate

So, quick revision. First, the n factor of permanganate is five in acid, three in neutral solution and one in strongly alkaline solution, and the n factor of dichromate is six. Second, electrons lost equal electrons gained, and check the product: copper stops at C u I, not copper metal. And third, manganate, at plus six, disproportionates once carbon dioxide removes the alkali, so green turns purple. Now go and try the practice problem on this page. You can do it. All the best!