Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. The oxidation number of iron in F e three O four comes out as plus eight over three. That is a fraction! But electrons never come in thirds, right? So how many faradays turn one mole of it into iron? Hold that thought. Today: Part A of our redox and electrochemistry problem bank, at JEE Main level. We solve seven of its eight problems. Most of them hide a small trap. So, let's begin.

0:34Tool 1: the n-factor counts every atom

First tool: the n-factor. It is the number of electrons lost or gained per formula unit. And the equivalent mass, capital E here, equals the molar mass, capital M, over n. Careful, this E is a mass, not a potential. Okay? Problem two. F e S two, molar mass one hundred and twenty, is oxidised to F e three plus and S O two; the three plus up top is the charge, and the small numbers below count atoms. Find its equivalent mass. With me? Start with iron: it is plus two here, so the two sulphurs share minus two, minus one each. Iron goes from plus two to plus three; the arrow means changes to. One electron. Clear? Now sulphur. In S O two it is plus four. From minus one to plus four, that is five electrons. So n is one plus five: six, right? No, no, careful! Every F e S two has two sulphur atoms, and each one loses five electrons. Two times five: ten. Got it? So n is one plus ten: eleven. And capital E is one hundred and twenty over eleven, about ten point nine; the wavy equals means approximately. Okay?

1:53Tool 2: E naught cell from reduction potentials

Second tool: the cell potential. Problem three. Zinc, Z n, in zinc ions; the double bar is the salt bridge, and a single bar is a boundary. Then F e three plus and F e two plus in one solution, hence the comma, on inert platinum, P t, which just carries electrons. Anode left, cathode right. Got it? The standard potentials, E naught, the E with the small circle: F e three plus over F e two plus, plus zero point seven seven volt; zinc ion over zinc, minus zero point seven six volt. Both are reduction potentials. With me? So, E naught cell is E naught of the cathode, minus E naught of the anode, at two hundred and ninety eight kelvin, one molar, and one bar for gases. So zero point seven seven minus zero point seven six: zero point zero one volt, right? Oh no, careful! Zinc's value is minus zero point seven six. Minus a minus is plus: zero point seven seven plus zero point seven six, one point five three volts. Keep the sign with the number. Clear?

3:00Tool 3: one faraday, one mole of electrons

Third tool: the faraday. One faraday, capital F, is the charge on one mole of electrons, about ninety six thousand five hundred coulombs; capital C stands for coulomb. So faradays simply count moles of electrons. With me? Problem seven: how many faradays form one mole of oxygen gas, O two, from water? On screen, m o l is short for mole. Oxygen goes from minus two in water to zero in O two: two electrons. So two faradays, right? No, no, wait! O two has two oxygen atoms. The half reaction: two H two O gives O two, plus four H plus, plus four e minus, the electrons. Four electrons. Okay? So one mole of O two needs four moles of electrons: four faradays. Got it?

3:51Tool 4: molar conductivity and degree of dissociation

Fourth tool: conductance. Kappa, the Greek letter like a small k, is the conductivity, in siemens per centimetre. Capital lambda m, the upside down V with a small m, is the molar conductivity: one thousand kappa over c, with c in moles per litre. The thousand turns litres into cubic centimetres. Okay? Problem six. Point zero five molar, with kappa point zero zero five five siemens per centimetre. Point zero zero five five over point zero five: so capital lambda m is point one one, right? Oh no, careful! You forgot the thousand. One thousand times point zero zero five five is five point five, and over point zero five, that is one hundred and ten siemens centimetre squared per mole. Clear? Now a weak electrolyte. Its degree of dissociation, alpha, is capital lambda m over lambda m naught; the small circle again means naught. Lambda m naught is the limiting molar conductivity, at infinite dilution, found from Kohlrausch's law, as the bracket says. This works for weak electrolytes only, not strong ones. Right? Problem eight. Capital lambda m is thirty nine, and lambda m naught is three hundred and ninety. Find alpha. Point one, yes! Both are in siemens centimetre squared per mole, and thirty nine over three hundred and ninety is one tenth.

5:18Example 1: Nernst equation, Cu2+ concentration at 0.281 V

Got it? Now, example one: problem four. At what concentration of copper ions, C u two plus, written in square brackets, is the copper ion over copper potential zero point two eight one volt? Pause the video, and try it first. Done? Okay, let's check. C u two plus takes two electrons, so n is two. For a metal in its own ions at two hundred and ninety eight kelvin, the Nernst equation gives E equals E naught, plus zero point zero five nine one over two, times log of the copper ion concentration. Okay? E naught for copper is plus zero point three four volt, a standard reduction potential. And zero point zero five nine one over two is about zero point zero two nine five. Call the concentration c. With me? So, zero point two eight one equals zero point three four, plus zero point zero two nine five times log c. The potential is below E naught. Is c more than one molar, or less? Less, exactly! Zero point two eight one minus zero point three four is minus zero point zero five nine. Over zero point zero two nine five, that is minus two. Right? So c is ten to the power minus two: zero point zero one molar. Fewer copper ions, lower potential. Got it?

6:35Example 2: faradays to reduce one mole of Fe3O4 to iron

Now the harder one, example two: problem five, and it is our puzzle. How many faradays convert one mole of F e three O four to iron? Pause and try. Hint: count every iron atom. Done? Okay, let's check. Let x be the oxidation number of iron. Three x, plus four times minus two, equals zero. The double arrow means so: x is plus eight over three. That is the average for iron. Clear? Iron metal has oxidation number zero. So iron goes from plus eight over three to zero: eight over three faradays, right? No, no, wait! There are three iron atoms in every F e three O four. Three times eight over three: eight electrons. Okay? And the fraction? F e three O four is really F e O and F e two O three together; the dot joins them. One iron is plus two, two are plus three: two plus three plus three, eight again. Puzzle solved! Eight thirds is only an average; the electrons come out as a whole eight. With me? So one mole of F e three O four needs eight moles of electrons: eight faradays. Got it?

7:50Quick revision: count atoms, cell potential, conductivity

So, quick revision. Count every atom: F e S two has an n-factor of eleven, O two needs four faradays, and F e three O four needs eight. E naught cell is cathode minus anode, signs kept; and for a metal M in its own ions, M n plus, E equals E naught plus zero point zero five nine one over n, times log of their concentration. Capital lambda m is one thousand kappa over c, never forget the thousand; and for a weak electrolyte, alpha is capital lambda m over lambda m naught. Now try the practice problems on this page. You can do it. All the best!