Gibbs Energy in Cells and Coupled Reactions

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. In your body's cells, glucose picks up a phosphate group: the phosphorylation of glucose. Its standard Gibbs energy change is plus thirteen point eight kilojoules per mole. Positive! So on its own, it is not spontaneous. Then how do your cells carry it out, all the time? Strange, isn't it? Today: Gibbs energy in cells, and coupled reactions that push uphill reactions forward. Both are useful for JEE, so let's begin.

0:35Galvanic cell: Gibbs energy and e.m.f.

So, first, a galvanic cell, a cell that gives us electricity. Its Gibbs energy change and its e m f are linked by one short formula: delta G equals minus n F E, the first line on screen. The little triangle is the capital Greek letter delta, meaning change in, and capital G is Gibbs energy. Capital E is the e m f of the cell, its electromotive force, measured in volts. Okay? Now the other two letters. Small n is the number of moles of electrons transferred in the reaction. And capital F is the Faraday constant: ninety six thousand five hundred coulombs per mole. On screen, capital C is for coulombs, and mol to the power minus one means per mole. Got it? Now think. Suppose the e m f, capital E, is positive. What is the sign of delta G then? Negative, yes, exactly! Because of that minus sign in front. The double arrow on screen means implies. So E greater than zero implies delta G less than zero: the cell reaction is spontaneous. Clear? Under standard conditions, we add the little circle on top, read as naught. Delta G naught equals minus n F E naught, where E naught is the standard e m f of the cell. With me? Now, minus delta G is the maximum electrical work the cell can give, at constant temperature and pressure, the capital T and P in brackets. But careful, only when the cell works reversibly, okay? Real cells work irreversibly. Current flows through the internal resistance of the cell, and part of the energy is lost as heat. So in practice, we get less work than minus delta G, right?

2:23Standard e.m.f. and the equilibrium constant

Now, one more link. From the last lesson, delta G naught also equals minus two point three zero three, times R T log K, okay? Capital R is the gas constant, capital T is the temperature in kelvin, and capital K is the equilibrium constant. Got it? So now we have two expressions for the same delta G naught. Set them equal: minus n F E naught equals minus two point three zero three, times R T log K, and the minus signs on both sides cancel. Right? Now divide both sides by n F, and put in the numbers. At two hundred and ninety eight kelvin, two point three zero three R T over F works out to zero point zero five nine volts, the capital V for volts. So, E naught equals zero point zero five nine over n, times log K, okay? But careful: this zero point zero five nine holds only at two hundred and ninety eight kelvin. Got it? Now tell me. If E naught is positive, is the equilibrium constant more than one, or less than one? More than one, correct! Positive E naught means log K is greater than zero, and so K is greater than one. Perfect!

3:40Coupled reactions: the total delta G decides

Now the second big idea: coupling. Suppose a reaction is non-spontaneous; on its own, it will not go. So we couple it to a strongly spontaneous reaction, okay? Then we add the two equations, and we add their delta G values too. If the total delta G is less than zero, the coupled reaction is spontaneous, uphill step and all. With me? Now, back to our puzzle. Phosphorylation of glucose: delta G naught is plus thirteen point eight, k J mol to the power minus one, that is, kilojoules per mole. Uphill, right? In cells, this step is coupled to the hydrolysis of ATP, its reaction with water. Delta G naught here is minus thirty point five kilojoules per mole. Strongly downhill, okay? So, think. What is delta G naught for the coupled reaction? And is it spontaneous? Yes, exactly, just add them! Plus thirteen point eight, plus minus thirty point five, gives minus sixteen point seven kilojoules per mole. Negative, so the coupled reaction is spontaneous. Puzzle solved: your cells pay for the uphill step with ATP hydrolysis. That's it! The same idea lets carbon reduce metal oxides at high temperature, as in our second example. Clear?

4:59Example 1: delta G naught of the Daniell cell, E = 1.10 V

Now, Example one, an easy one. Find delta G naught for the Daniell cell. Its standard e m f, E naught, is one point one zero volts. Pause the video, and try it. Done? Okay, let's check. First, the number of electrons. Many students simply put n equal to one. So what is n here? Two, correct! In the Daniell cell, zinc gives two electrons to the copper ions. Zinc, plus C u two plus, the copper ion with a two plus charge, gives Z n two plus, plus copper; the arrow means gives. Two electrons move, so n equals two. Okay? Now substitute. Delta G naught equals minus n F E naught: minus two, times ninety six thousand five hundred, times one point one zero. The cross means times. With me? Now, before you multiply, think about the unit. Does this product come out in joules, or in kilojoules? Joules, exactly! Coulomb times volt equals joule. So it is minus two hundred and twelve thousand three hundred joules per mole. Oh no, careful! If you write kilojoules here, your answer is a thousand times too big. Right? So divide by one thousand: delta G naught is minus two hundred and twelve point three kilojoules per mole. Negative, so the Daniell cell reaction is spontaneous, just as the positive E naught told us. Very good!

6:32Example 2: can carbon reduce ZnO at 1000 K?

Now Example two, a harder one. At one thousand kelvin, delta G naught is plus five hundred kilojoules for: two Z n O gives two Z n plus O two, zinc oxide breaking up. And it is minus four hundred kilojoules for: two C plus O two gives two C O, carbon forming carbon monoxide. Can carbon reduce zinc oxide at this temperature? Pause and try. Hint: couple the two reactions. Done? Okay, let's check. Write both steps with their delta G naught values. The zinc oxide step: plus five hundred kilojoules, uphill. Okay? The carbon step: minus four hundred kilojoules, downhill. With me? Now, minus four hundred is a big negative number, strongly spontaneous. So carbon reduces zinc oxide easily, right? No, no, careful! One strong step is not enough; we must couple them and look at the total. Add the two equations. The O two made in the first step is used up in the second, so it cancels. Overall: two Z n O plus two C gives two Z n plus two C O, got it? Now add the delta G naught values too. Plus five hundred, plus minus four hundred, gives plus one hundred kilojoules. Clear? Positive! So the coupled reaction is not spontaneous here. The answer is no, not at one thousand kelvin; a higher temperature is needed. Perfect!

8:05Quick revision: -nFE, 0.059/n log K and coupling

So, quick revision. One: for cells, delta G naught equals minus n F E naught, in joules first; divide by one thousand for kilojoules. Two: at two hundred and ninety eight kelvin, E naught equals zero point zero five nine over n, times log K, okay? Three: for coupling, add the delta G values; a negative total means the coupled reaction goes. Now go and try the practice problems on this page. You can do it. All the best!