Gibbs Energy and Combining Electrode Potentials

Transcript

0:00Introduction

Hello everyone! Quick puzzle. Iron three plus takes one electron and becomes iron two plus, at plus zero point seven seven volts. Then iron two plus takes two more and becomes iron, at minus zero point four four volts. So, iron three plus all the way to iron? Just add them: plus zero point three three volts, right? The real answer is not even positive! Today: why potentials cannot simply be added, and what we add instead. This comes up often in JEE electrochemistry, so let's begin.

0:38Gibbs energy meets cell potential

So, energy meets voltage. Delta G equals minus n F E, okay? The triangle is capital delta, meaning change, so delta G is the change in Gibbs energy, and E is the cell potential. Next to it: delta G naught equals minus n F E naught. The little circle, read naught, means standard conditions: two hundred and ninety eight kelvin, one bar, and one molar. And every electrode potential is written as a reduction potential. With me? Capital F is the Faraday constant: ninety six thousand four hundred and eighty five coulombs per mole, the charge on one mole of electrons. Capital C is for coulombs, and mol to the power minus one means per mole. We round it to ninety six thousand five hundred; the wavy equals sign means approximately. Okay? And n is the number of electrons transferred, in the equation as written. Now tell me. With n and capital F both positive, if E is positive, what is the sign of delta G? Negative, exactly! Because of that minus sign. So a spontaneous reaction has E greater than zero, and delta G less than zero. Clear? Now the Daniell cell, zinc and copper. Its E naught cell, with the small word cell below, is one point one zero volts. So, what is n here? Two, correct! Zinc loses two electrons, and copper two plus gains two. So delta G naught equals minus two, times ninety six thousand five hundred, times one point one zero: about minus two hundred and twelve kilojoules per mole. Negative, so spontaneous. Careful: the product is in joules, so divide by one thousand for kilojoules. Got it?

2:30Double the reaction: what changes?

Now, a small experiment. Silver ion, written A g plus, takes one electron, e minus, and becomes silver. The plus above is the charge, and the arrow means gives. Its E naught is zero point eight zero volts, so delta G naught is minus F, times zero point eight zero. Now double the equation. What happens to delta G naught, and to E naught? Delta G naught doubles, yes! Now n is two: minus two F, times zero point eight zero. Twice the silver, twice the energy, right? But E naught equals minus delta G naught, over two F, okay? The two F cancels top and bottom, and we get zero point eight zero volts again. Unchanged! So, delta G naught is extensive: it grows with the amount. E naught is intensive, like temperature: mix two cups of hot tea, and the temperature does not double.

3:33Adding half-reactions: add Delta G0, not E0

Clear? Now, the main event. Half-reaction one plus half-reaction two gives a new half-reaction, like iron three plus to iron in our puzzle. Can we simply add E naught one and E naught two? The small one and two below are just labels. No, no, careful! This is the common trap. In a new half-reaction the electrons do not cancel, and E naught, being intensive, does not add. The equals sign with a slash means not equal to. Right? So what does add? Delta G naught, which is extensive! Delta G naught three equals delta G naught one, plus delta G naught two. Got it? Write minus n F E naught for each. Minus n three F E naught three equals minus n one F E naught one, minus n two F E naught two, okay? Cancel minus capital F, and divide by n three. E naught three equals n one E naught one, plus n two E naught two, all over n three. Topper's trick: each n times E naught stands for a packet of energy; add the packets, then share them over the new electrons. With me? But wait. For a full cell, we simply took cathode minus anode, with no n anywhere. Why is that allowed? Because in a full reaction, the electrons cancel! Electrons lost at the anode are gained at the cathode, so the same n sits in every term and divides out. So E naught cell equals E naught cathode minus E naught anode, both as reduction potentials. Electrons cancel: subtract potentials. Electrons left over: go through delta G, always. Clear?

5:20Example 1: E0 of Fe3+/Fe from Fe2+/Fe and Fe3+/Fe2+

Now, our puzzle. E naught of the iron two plus, iron couple is minus zero point four four volts; of the iron three plus, iron two plus couple, plus zero point seven seven volts. The slash separates the oxidised form from the reduced form. Find E naught for iron three plus to iron. Pause the video, and try it first. Done? Okay, let's check. Iron three plus, plus one electron, gives iron two plus, so n is one. Delta G naught one is minus F times zero point seven seven: minus zero point seven seven F, okay? Next, iron two plus, plus two electrons, gives iron, so n is two. Delta G naught two is minus two F, times minus zero point four four. Minus times minus is plus: plus zero point eight eight F, right? Now add the two equations. Iron two plus cancels from both sides. How many electrons are left? Three, yes! One plus two; the electrons add up. Iron three plus, plus three electrons, gives iron, so n three is three. Got it? Now add the delta G values. Minus three F E naught equals minus zero point seven seven F, plus zero point eight eight F, which is plus zero point one one F, with me? Cancel capital F, and divide by minus three. E naught is minus zero point one one over three, about minus zero point zero three seven volts. Not plus zero point three three volts. Not even positive! Puzzle solved. If you fell for plus zero point three three at the start, now you know why.

7:05Example 2: E0 of MnO4-/Mn2+ from MnO4-/MnO2 and MnO2/Mn2+

Now a harder one, JEE level. In acid, E naught for permanganate, M n O four minus, to manganese dioxide, M n O two, is one point seven zero volts; and for M n O two to M n two plus, one point two three volts. Find E naught for permanganate to M n two plus. Pause and try. Hint: find each n first. Done? Okay, let's check. What is the oxidation number of manganese in each of the three? Plus seven, plus four, and plus two. Very good! In permanganate, four oxygens give minus eight and the ion is minus one, so manganese is plus seven. In M n O two, it is plus four. Okay? So the first step gains three electrons: n one is three. The second gains two: n two is two. Overall, five electrons. Right? Three times one point seven zero, plus two times one point two three, all over five. That is five point one zero, plus two point four six, over five. Clear? Seven point five six over five: about one point five one volts. And look, that is exactly the value listed for permanganate to M n two plus in the electrochemical series. See how fast that was?

8:32Quick revision: Delta G = -nFE, add nE0, cathode minus anode

So, quick revision. One: delta G naught equals minus n F E naught; when E naught is positive, delta G naught is negative, and the reaction is spontaneous under standard conditions. Two: for a new half-reaction, add the n times E naught values and divide by the new n. Never add potentials directly. Three: for a full cell, E naught cell is E naught cathode minus E naught anode. Now go and try the practice problems on this page. All the best!