Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Here is a two electron cell whose standard potential is zero point five zero zero volt at two hundred and ninety eight kelvin. Warm it by just ten degrees, and it is zero point four nine zero volt. From these two readings, can we find the entropy change of the reaction? Measuring entropy with a voltmeter? Sounds strange, right? Hold on to that question. Today: the unit problem bank for redox reactions and electrochemistry, how to use it, and this exact problem, solved. Problems that join ideas like this come up often in JEE, so let's begin.

0:42How to use this chapter

So, let's open this chapter. Part A has eight solved problems, at JEE Main level. Okay? And the instruction: attempt them as a thirty minute test, after Chapters one to seven. Sit with a clock. Right? Now, a quick one for you. Eight problems, thirty minutes. How much time can you give each problem? Thirty over eight; that is the division sign on screen. Three point seven five, so a little under four minutes a problem. Clear? Then Part B: longer problems, at JEE Advanced level. Our second example today comes from there. Got it? And Part C checks the concepts that are often tested in statement form, like true or false. Let's try one right now. With me?

1:31Part C: one statement, one trap

Here is a statement in the Part C style. The anode is always the negative electrode. True or false? If you said true, careful! It is false. In a galvanic cell, the anode is negative. But in an electrolytic cell, the anode is positive, because it is joined to the positive terminal of the battery. Okay? So what stays the same? In both cells, oxidation takes place at the anode. Topper's trick: name an electrode by its reaction, never by its sign. Right? Now one more Part C statement. E cell is the cell potential: the E with the small word cell below it. If you double the whole cell reaction, does E cell double too? No, it stays the same! E cell is intensive: it does not depend on how much reaction you write. But delta G, with the little triangle delta meaning change, is the Gibbs energy change, and it equals minus n F E, okay? Here n is the number of electrons, and capital F is the Faraday constant. Double the reaction, and n doubles, so delta G doubles: it is extensive. Clear?

2:41Example 1: E0 of a Zn and Fe3+/Fe2+ cell, minus sign trap

Now, an easy problem from Part A, okay? Find E naught for this cell. The E with a small circle on top, E naught, is the standard value: one molar solutions and one bar for gases, usually quoted at two hundred and ninety eight kelvin. Read the cell with me: zinc, single bar, zinc two plus, double bar, iron three plus comma iron two plus, single bar, P t, platinum. Take the standard reduction potentials from the table in Chapter three. Pause the video, and try it first. Done? Okay, let's check. In cell notation, the anode is on the left. So zinc is the anode: the zinc two plus slash zinc couple, the slash separating the two forms, with E naught minus zero point seven six volt, the capital V for volt. Right? Now the right side, the cathode. The table has two iron values. Which one do we take: minus zero point four four, or plus zero point seven seven? Plus zero point seven seven, correct! There is no iron metal on the right. The comma means iron three plus and iron two plus sit in the same solution, and platinum is inert: it only carries the electrons. So the couple is iron three plus, iron two plus. Minus zero point four four belongs to iron two plus and iron metal. Got it? Now, E naught cell equals E naught cathode, minus E naught anode, both written as reduction potentials, the sign convention of our course. Okay? So, zero point seven seven minus zero point seven six. That is zero point zero one volt, right? No, no, wait! The anode value itself is minus zero point seven six. Subtracting a negative number means adding. So zero point seven seven, minus, minus zero point seven six, becomes zero point seven seven plus zero point seven six. With me? One point five three volts. And it is greater than zero, so the reaction runs as written: zinc is oxidised, and iron three plus is reduced to iron two plus. Perfect!

4:50Example 2: Delta G, Delta S, Delta H from two E0 readings

Now our puzzle from the start, as a hard problem from Part B, okay? Here is a two electron cell. Its E naught is zero point five zero zero volt at two hundred and ninety eight kelvin, the capital K for kelvin, and zero point four nine zero volt at three hundred and eight kelvin. Find delta G naught, delta S naught and delta H naught at two hundred and ninety eight kelvin. The small circle again means standard, S is entropy, and H is enthalpy. Pause the video and try it. Hint: watch the units. Done? Okay, let's check. Step one: how fast does E change with temperature? That is d E by d T, the temperature coefficient, at constant pressure, the small p in brackets. Zero point four nine zero minus zero point five zero zero, over three hundred and eight minus two hundred and ninety eight. Minus zero point zero one over ten: minus ten to the power minus three volt per kelvin. The wavy equals sign means approximately: from two readings, we take the slope as steady over these ten degrees. Okay? Step two: delta G naught equals minus n F E naught. Here n is two, and capital F is ninety six thousand five hundred coulombs per mole. The cross means times. Minus two, times ninety six thousand five hundred, times zero point five zero zero: minus ninety six thousand five hundred joules, that is minus ninety six point five kilojoules. Right? Now entropy. The cell potential falls as the temperature rises. So, what is the sign of delta S naught? Negative, exactly! Delta S naught equals n F times d E by d T, and n and F are positive. Two, times ninety six thousand five hundred, times minus ten to the power minus three: minus one hundred and ninety three joules per kelvin. Clear? Now enthalpy: delta H naught equals delta G naught, plus capital T, the temperature, times delta S naught. So, minus ninety six point five, plus two hundred and ninety eight times minus one hundred and ninety three? Oh no, careful! Look at the units. Delta G naught is in kilojoules, but delta S naught is in joules per kelvin. So in this formula, use delta S naught in kilojoules: minus zero point one nine three kilojoules per kelvin. Got it? Now, minus ninety six point five, plus two hundred and ninety eight times minus zero point one nine three. That is minus ninety six point five, minus fifty seven point five: minus one hundred and fifty four kilojoules. And our puzzle? Solved! Two readings, and n equals two, gave us all three quantities. That's it!

7:41Quick revision: E0 cell, Delta G = -nFE, dE/dT, match units

So, quick revision. Part A is a thirty minute test, Part B is JEE Advanced level, and in Part C, read every word. E naught cell equals E naught cathode minus E naught anode, both as reduction potentials. Delta G naught equals minus n F E naught, delta S naught equals n F times d E by d T, and delta H naught equals delta G naught plus T delta S naught, with the units matched. Now open Part A, start your clock, and begin. All the best!