The Nernst Equation

Transcript

0:00Introduction

Hello everyone! Quick puzzle. The table says the Daniell cell, zinc and copper, gives one point one zero volts. Now take zinc ions at zero point one molar, and copper ions at only zero point zero zero one molar. Will the voltmeter still read one point one zero volts? Hmm. That table value assumes one molar solutions. For other concentrations we need the Nernst equation, and it comes up often in JEE, so let's begin.

0:32The Nernst equation for one electrode

So, take any electrode, written as a reduction. Ox is the oxidised form, Red the reduced form, and the small letters in front are their coefficients. Small e with a minus is an electron, n counts them, and the arrow means gives. Okay? Now, the Nernst equation. E equals E naught, minus R T over n F, times the natural log of Red to the power b, over Ox to the power a. E naught, with the little circle, is the standard reduction potential. R is the gas constant, capital T the temperature in kelvin, capital F the Faraday constant, and square brackets mean molar concentration. Clear? At two hundred and ninety eight kelvin, the capital K, two point three zero three R T over F is zero point zero five nine one volts. The two point three zero three turns natural log into log to base ten. Got it? So, the working form. E equals E naught, minus zero point zero five nine one over n, times log of Red to the power b, over Ox to the power a. With me? Quick question. If every concentration is one molar, the capital M, what is E? E equals E naught, exactly! Log of one is zero. That is the standard value: two hundred and ninety eight kelvin, one molar, gases at one bar. Right?

1:56Two electrodes you will meet often

Now, two common electrodes. First, a metal in its own ions: M n plus, plus n electrons, gives the solid metal; the n plus up top is the ion's charge. And a pure solid has activity one. Okay? So the Red term is one, and E equals E naught, minus zero point zero five nine one over n, times log of one over the M n plus concentration. Clear? Look at the badge: log of one over x is minus log x. So what happens to the minus sign in front? It becomes a plus. Correct! E equals E naught, plus zero point zero five nine one over n, times log of the M n plus concentration. Careful: students often keep the minus sign. More metal ions, higher potential. Got it? Second, the hydrogen electrode: E naught is zero by definition, and n is two. E equals minus zero point zero five nine one over two, times log of p H two over H plus squared. The small p with H two below is the partial pressure of hydrogen, in bar. Okay? Keep the hydrogen at one bar. Log of one over H plus squared is two times p H, where p H is minus log of the H plus concentration. The twos cancel: E equals minus zero point zero five nine one times the p H value. With me?

3:20A whole cell: the reaction quotient Q

Now, the whole cell. E cell, with the small word cell below, equals E naught cell, minus zero point zero five nine one over n, times log Q; n is the number of electrons transferred in the cell reaction. We will read the graph on the right in a moment. Okay? And Q is the reaction quotient: products over reactants, each to the power of its coefficient. Clear? Pure solids and liquids have activity one, so they stay out of Q, okay? Gases go in as partial pressures, in bar. Got it? Now the graph: E cell in volts going up, and log of Q to base ten, the small ten below, going across. When Q equals one, what is E cell? E naught cell, yes! Log of one is zero, so both lines pass through the black dot on the dashed line, at E naught, one point one volts. Right? As Q grows, E cell falls in a straight line, slope minus zero point zero five nine one over n. For n equal to two, the orange line, that is minus zero point zero two nine five; the dark blue line, n equal to one, is twice as steep. With me?

4:31Example 1: Daniell cell EMF with 0.1 M Zn2+ and 0.001 M Cu2+

Now, our puzzle. Find the EMF, the cell voltage, of this cell: zinc, single bar, zinc two plus at zero point one molar, double bar for the salt bridge, copper two plus at zero point zero zero one molar, single bar, copper. E naught is one point one zero volts. Pause the video and try it. Done? Okay, let's check. The cell reaction: zinc plus copper two plus gives zinc two plus and copper. Two electrons move, so n is two. Okay? Now for Q: is it zinc over copper, or copper over zinc? Zinc over copper! Products over reactants, and zinc two plus is the product; the solid metals stay out. So Q is zero point one over zero point zero zero one: one hundred. Clear? Before calculating: Q is more than one. So will E be above or below one point one zero volts? Below, exactly, just like the graph! Zero point zero five nine one over two is zero point zero two nine five, and log of one hundred is two. So E equals one point one zero, minus zero point zero two nine five times two: one point zero four one volts. Puzzle solved! The table simply assumed one molar solutions.

5:53Example 2: MnO4- to Mn2+ at pH 2, oxidising power falls

Right? Now a harder one, JEE level. Permanganate, M n O four minus, is reduced to M n two plus; the slash separates the two forms. Both are at one molar, the p H is two, and E naught is one point five one volts. Find E, and say what it tells us about the oxidising power of permanganate. Pause the video and try it. Done? Okay, let's check. Permanganate, plus eight H plus, plus five electrons, gives M n two plus and four waters. So n is five. Okay? Now, where does the water go in the log term? Nowhere! Water is a pure liquid, activity one. So E equals one point five one, minus zero point zero five nine one over five, times log of M n two plus, over permanganate times H plus to the power eight. Watch that eight: forgetting it is the classic slip. Right? Both manganese species are at one molar. At p H two, H plus is ten to the power minus two molar, so log of one over H plus to the power eight is eight times two: sixteen. Clear? So E equals one point five one, minus zero point zero five nine one over five, times sixteen: one point three two volts. Got it? E fell from one point five one to one point three two volts. So is permanganate now a stronger or weaker oxidising agent? Weaker, correct! Lower reduction potential, less pull for electrons. So as p H rises, the oxidising power of permanganate falls. Quick check: eight H plus for five electrons, so E moves eight fifths of zero point zero five nine one volts per p H unit. Two units, same one point three two volts. That's it!

7:48Quick revision: Nernst at 298 K, log Q, what stays out of Q

So, quick revision. One: at two hundred and ninety eight kelvin, E equals E naught, minus zero point zero five nine one over n, times log of Red over Ox, with their powers. Two: for a cell, use log Q, and at Q equal to one, E cell equals E naught cell. Three: pure solids and liquids stay out; gases go in as pressures in bar. Now try the three practice problems on this page. You can do them!