Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Two electrolytic cells are joined in series, one with silver nitrate and one with copper sulphate. The same current flows through both, for the same ten minutes. Same charge, so the same mass of metal is deposited in each cell, right? Hmm, hold on to your answer. You have finished six chapters on redox reactions and electrochemistry, and now it is time to test yourself. This chapter gives you three sets of mixed problems in the JEE Main style. Let us see how to use them, and then solve this exact puzzle together. So, let's begin.

0:40How to use this chapter

So, what is this chapter about? JEE questions on this unit move quickly between six ideas, all mixed together. Oxidation numbers, titrations, cell potentials, okay? Then the Nernst equation, conductance, and Faraday's laws. Six ideas in all. Right? Quick question. Our puzzle from the start, the two cells in series: which of these six ideas is at work there? Faraday's laws, yes, exactly! Charge, and the mass deposited. But see, the question never tells you that. Spotting the idea is your job. So this chapter gives you three sets. Set one is easy, Set two is medium, and Set three is hard, with twelve solved problems in each. Clear? Then each set ends with two practice problems for you. Got it? And the real past questions? They are at the end of Chapters one to six. The problems here are original, written in the same style. With me?

1:43Values to take, unless stated

Now, the values you need. Unless stated, take capital F, the Faraday constant, as ninety six thousand five hundred coulombs per mole. The C is coulombs, and mol to the power minus one means per mole. It is the charge on one mole of electrons. Okay? Next, two point three zero three R T over capital F equals zero point zero five nine one volt, at two hundred and ninety eight kelvin. R is the gas constant, and capital T is the temperature in kelvin. You will use this number in the Nernst equation problems. Right? And standard potentials: take them from Chapter three. E with the little circle on top, read E naught, means a standard potential, and these are all written as reduction potentials. Got it? Quick check. Standard means which conditions? Two hundred and ninety eight kelvin, one molar for solutions, and one bar for gases. Correct! Capital M means molar, moles per litre. Clear? And remember: unless stated. If a question gives its own value, use that one. Okay?

2:55Name the idea first: two from Set 1

Now, the real skill in a mixed set: name the idea before you calculate. Two quick ones from Set one. Find the oxidation number of manganese, M n, in K two M n O four, potassium manganate. The small numbers at the bottom count the atoms. So, what do you get? Plus six, yes! Two potassiums give plus two, four oxygens give minus eight, and x is the oxidation number of manganese. Two plus x minus eight equals zero; the double arrow means so, and x is plus six. Oxidation numbers, Chapter one. Okay? Next, find E naught for this cell: nickel, N i, on the left, and silver, A g, on the right. The single vertical line is a phase boundary, and the double line is the salt bridge. The small two plus and plus at the top are the charges, and the slash pairs each ion with its metal. Nickel is minus zero point two five, and silver is zero point eight zero volt. So, zero point eight zero minus zero point two five: zero point five five volt, right? No, no, careful! That is the trap. E naught cell equals E naught right minus E naught left, both as reduction potentials, with their signs. So zero point eight zero, minus minus zero point two five in brackets: one point zero five volt. Cell potentials, Chapter three. Name the idea, and each is one line. Got it?

4:32Example 1: volume of K2Cr2O7 for FeSO4, using n-factors

Now, Example one, from Set one. What volume of zero point one molar potassium dichromate, K two C r two O seven, oxidises thirty millilitres of zero point one molar ferrous sulphate, F e S O four, in acid? The m L is millilitres. Pause the video, and try it first. Done? Okay, let's check. Both solutions are zero point one molar. So we need the same thirty millilitres of dichromate, right? No, no, careful! Molarity alone is not enough; count the electrons. That is the n factor: the electrons gained or lost per formula unit. Dichromate, C r two O seven two minus, changes to C r three plus; the arrow means changes to. Two chromium atoms, each going from plus six to plus three. Two times three: n factor six. Okay? And the ferrous ion, F e two plus, goes to F e three plus. Just one electron, so n factor one. Right? Now the law of equivalence from Chapter two: the equivalents of oxidant equal the equivalents of reductant. n one M one V one equals n two M two V two. The small one and two below name the two solutions, and V is the volume. Clear? So, six times zero point one times V equals one times zero point one times thirty. V is five millilitres. Only five, not thirty! Titrations, Chapter two, and the key was the n factor. With me?

6:10Example 2: Ag and Cu deposited by cells in series

Remember our puzzle? Here it is, from Set two. Three point zero amperes flows for ten minutes through two cells in series: one with silver nitrate, A g N O three, and one with copper sulphate, C u S O four. Capital A is amperes. Find the masses of metal deposited. Pause the video, and try it first. Done? Okay, let's check. First, the charge. Capital Q equals I times t: the current in amperes, times the time in seconds. Ten minutes is six hundred seconds, so three point zero times six hundred: eighteen hundred coulombs. The cells are in series, so this same charge passes through both. Okay? Now divide by capital F, the Faraday constant. Eighteen hundred over ninety six thousand five hundred: zero point zero one eight six five faraday. One faraday is one mole of electrons, so these are moles of electrons. Right? Now the puzzle. Same electrons through both cells, so the same number of moles of silver and copper, right? Not quite! The silver ion, A g plus, takes just one electron, the e minus on screen, to become silver. So zero point zero one eight six five moles of silver, times one hundred and eight grams per mole: two point zero one grams. Got it? But the copper ion, C u two plus, needs two electrons for every atom. So only half as many moles of copper: zero point zero one eight six five over two, times sixty three point five. Zero point five nine two grams. Clear? So, same charge, but not the same mass! This is Faraday's second law: the mass of silver over the mass of copper is one hundred and eight over one, divided by sixty three point five over two. That is molar mass over electrons per ion, for each metal. The wavy equals sign means approximately: silver gets about three point four times the mass of copper. Puzzle solved! With me?

8:25Quick revision: name the idea, key values, count electrons

So, quick revision. Six ideas, three sets from easy to hard, and name the idea before you calculate. Take capital F as ninety six thousand five hundred coulombs per mole, and zero point zero five nine one volt at two hundred and ninety eight kelvin, unless stated. And count electrons: the n factor in titrations, and n in Faraday's laws. Now open Set one, and begin. All the best!