Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take some copper sulphate solution, dip two platinum electrodes into it, and pass a current for a few minutes. Copper metal comes out on one electrode. Fine. But now test the solution: it has become strongly acidic! We never added any acid. So where did it come from? Hold that thought. This is set three of mixed problems: the Nernst equation and Faraday's laws, the kind that come up often in JEE Main. So, let's begin.

0:34Idea 1: the Nernst equation for a cell

Idea number one: the Nernst equation for a whole cell. At two hundred and ninety eight kelvin, the capital K, E cell, with the small word cell below, equals E naught cell, minus zero point zero five nine one over n, times log Q, the log of the reaction quotient. E naught, the E with the little circle on top, is the standard value, and n is the number of electrons transferred. Okay? Now the Daniell cell. Zinc plus copper two plus gives zinc two plus, and copper; the two plus up top is the charge, and the arrow means gives. Two electrons move, so n is two. Right? E naught cell is cathode minus anode, both reduction potentials: copper's zero point three four, minus zinc's value, minus zero point seven six. That is one point one zero volts, at one molar, the capital M, and one bar. Clear? And Q is zinc two plus over copper two plus; the square brackets mean molar concentration. The solid metals stay out of it. Got it? Now the graph: E cell in volts going up, and log Q to base ten, the small ten below, going across. Suppose a Daniell cell reads more than one point one zero volts. Is Q more than one, or less? Less than one, exactly! E cell greater than one point one zero volts means Q less than one. At Q equal to one, log Q is zero, so the lines pass through the black dot on the dashed line, at E naught; to the left, log Q is negative and E cell rises. Our cell is the orange line, n equal to two, slope minus zero point zero two nine five; the dark blue line, n equal to one, is twice as steep. With me?

2:23Idea 2: from amperes to moles

Idea number two: counting electrons. Charge equals I times t: current in amperes times time in seconds, giving coulombs, the capital C on screen. One faraday, the capital F, is ninety six thousand five hundred coulombs: the charge on one mole of electrons. Okay? So moles of electrons equals I t over F, simple. And if each unit of product needs n electrons, moles of product equals I t over n F, so we just divide by n as well. Clear? Now, a lead storage battery delivers ten amperes, the capital A, for one hour, the small h, and we want the sulphuric acid it uses up. First the charge: ten times one, so ten coulombs, right? No, no, careful! Time must be in seconds. One hour is three thousand six hundred seconds, so the charge is ten times three thousand six hundred: thirty six thousand coulombs. Over ninety six thousand five hundred: zero point three seven three faraday. Got it? Now the discharge reaction. Lead, P b, plus lead dioxide, plus two sulphuric acid molecules, gives two lead sulphate and two waters; the small numbers below count atoms. Lead goes from zero to plus two, and the lead in lead dioxide from plus four to plus two. So two electrons, e minus, for every two acid molecules. With me? That is one mole of acid per faraday. So the acid used is zero point three seven three times ninety eight, its molar mass: thirty six point six grams. Right?

4:03Example 1: find Zn2+ concentration when E cell is 1.13 V

Now, example one. Zinc, a single bar for the phase boundary, zinc two plus at an unknown concentration x, then the double bar for the salt bridge, copper two plus at zero point one molar, a single bar, and copper. The cell reads one point one three volts. Find x. Pause the video and try it. Done? Okay, let's check. Before calculating, think about the graph. The cell reads more than one point one zero volts. So is x more than zero point one molar, or less? Less, exactly! Above E naught means Q below one, and here Q is x over zero point one, so x is less than zero point one molar. Keep that in mind as a check. Okay? Now Nernst. n is two, and zero point zero five nine one over two is zero point zero two nine five. So one point one three equals one point one zero, minus zero point zero two nine five times log of x over zero point one. With me? Move one point one zero across. Log of x over zero point one is minus zero point zero three over zero point zero two nine five: minus one point zero one seven, about minus one point zero two; the wavy equals sign means approximately. Negative, as predicted. Clear? Now take the antilog. x over zero point one is ten to the power minus one point zero one seven, about zero point zero nine six. Right? So x is about nine point six times ten to the power minus three molar. Careful: n is two, so always halve the zero point zero five nine one. Got it?

5:53Example 2: Cu deposited and pH after electrolysing CuSO4

Now example two: our puzzle from the start. Two hundred millilitres of zero point one molar copper sulphate, C u S O four, is electrolysed with platinum electrodes, P t, at one point nine three amperes for five hundred seconds. Find the mass of copper deposited, and the p H value afterwards, with the volume unchanged. Pause the video and try it. Done? Okay, let's check. Charge: one point nine three times five hundred, nine hundred and sixty five coulombs. Over ninety six thousand five hundred: zero point zero one faraday. Okay? So zero point zero one mole of copper is deposited, right? No, no, wait! At the cathode, each copper two plus ion takes two electrons. So moles of copper is zero point zero one over two: zero point zero zero five mole. Clear? With copper's molar mass, sixty three point five, the mass is about zero point three one eight grams. The solution held zero point zero two mole of copper ions, so plenty are left. With me? Now the anode, and our puzzle. What gets oxidised at the platinum anode? Water, correct! Sulphate ions are not oxidised at an inert anode, so water gives oxygen: two H two O gives O two, plus four H plus, plus four electrons. One H plus per electron, so zero point zero one faraday makes zero point zero one mole of H plus. That is where the acid came from. Puzzle solved! The volume stays zero point two zero zero litre. So the H plus concentration, in square brackets, is zero point zero one over zero point two zero zero: zero point zero five molar. Here we ignore the formation of HSO four minus, hydrogen sulphate. Okay? So p H is minus log of zero point zero five: about one point three. Copper out at one electrode, acid made at the other. That's it!

8:03Quick revision: Nernst, It/F (t in seconds), Pt anode acid

So, quick revision. One: at two hundred and ninety eight kelvin, E cell equals E naught cell, minus zero point zero five nine one over n, times log Q; above E naught means Q below one. Two: moles of electrons equal I t over F, with time in seconds; divide by n for the product. Three: at a platinum anode in sulphate solution, water gives oxygen, and one H plus per electron. Now go and try the practice problems on this page. All the best!