Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Dip one silver rod in a dilute solution of silver ions, and another in a more concentrated one, and join them with a salt bridge. The same electrode on both sides. So the voltmeter must read zero, right? Hmm, hold on to your answer. This is Set two of the mixed problems on electrochemistry, the medium set. Questions like these come up often in JEE Main. Let's begin.

0:30The Nernst equation and a concentration cell

First, our tool: the Nernst equation for a 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, okay? The little circle, read naught, means standard: one molar and one bar. n counts the electrons transferred, and Q is products over reactants. And this zero point zero five nine one holds only at two hundred and ninety eight kelvin. Clear? Now our puzzle cell: silver, A g, on both sides. The single bar is a phase boundary, and the double bar is the salt bridge. The silver ion, A g plus, is zero point zero one molar on the left, and zero point five molar on the right; capital M means molar. Same electrode, so E naught cell is zero. Right? But E naught assumes one molar on both sides, and here the concentrations differ. So, is E also zero? No, it is not! The difference in concentration alone drives it: a concentration cell. E equals zero point zero five nine one over n, times log of c two over c one. Got it? Here, c two, with the small two below, is the cathode side, on the right: the more concentrated one; c one is the left side. The silver ion takes one electron, so n is one. Clear? So, E equals zero point zero five nine one, times log of zero point five over zero point zero one: log fifty. That gives zero point one zero zero volt. Not zero at all! Puzzle solved. Same electrodes, different concentrations, and still a voltage. With me?

2:18From E0 to log K and delta G0

Next, two quick ones from Set two. At equilibrium, E cell is zero, and Q becomes K, the equilibrium constant. So the Nernst equation gives log K equals n E naught over zero point zero five nine one, at two hundred and ninety eight kelvin. Okay? Now take nickel, N i, plus two silver ions. What is n for this reaction? Two, correct! Nickel gives two electrons, and each silver ion takes one. The two half arrows mean equilibrium, and N i two plus is the nickel ion. E naught is one point zero five volt. Right? So log K equals two times one point zero five, over zero point zero five nine one: thirty five point five. Careful: we want log K itself, so stop right here; no antilog needed. Clear? Now the Gibbs energy. Delta G naught equals minus n capital F E naught; the triangle, delta, means change. Capital F is the Faraday constant, the charge on one mole of electrons. Got it? Next, a reaction with three electrons, and E naught of zero point five zero volt. Is delta G naught positive or negative? Negative, yes! Positive E naught, negative delta G naught: spontaneous under standard conditions. So, minus three, times capital F, ninety six thousand five hundred, times zero point five zero: minus one hundred and forty four thousand seven hundred and fifty joules, or minus one hundred and forty four point eight kilojoules. And careful: do not drop the n, the three here; students often do. With me?

4:09Example 1: EMF of Zn-Cu cell with 0.01 M Cu2+ ions

Example one, the easier one. Find the EMF, the cell voltage: zinc in one molar zinc ions, Z n two plus, and copper in zero point zero one molar copper ions, C u two plus. The picture shows this cell: zinc anode, copper cathode. Use the standard potentials from Chapter three. Pause the video, and try it first. Done? Okay, let's check. E naught cell is E naught cathode minus E naught anode, both as reduction potentials. Copper is plus zero point three four, zinc minus zero point seven six. So, zero point three four minus minus zero point seven six: one point one zero volt. Okay? The cell reaction: zinc plus copper two plus gives zinc two plus and copper; the arrow means gives. Two electrons flow from zinc to copper, the e minus arrows in the picture, so n is two. Right? Now, what about Q? Is it zero point zero one over one, or one over zero point zero one? One over zero point zero one! Q is products over reactants, and zinc two plus is the product. The square brackets mean concentration, and the solid metals stay out. So Q is one hundred. Got it? So E equals one point one zero, minus zero point zero two nine five, times log one hundred. That zero point zero two nine five is just zero point zero five nine one over two. Clear? Log one hundred is two. So, one point one zero, minus zero point zero two nine five times two: one point zero four one volt. Fewer copper ions at the cathode, so the voltage drops a little. That's it!

6:00Example 2: BaSO4 Ksp from conductivity, water subtracted

Example two, the harder one. Saturated barium sulphate, B a S O four, has conductivity kappa, the Greek letter like a small k, of three point zero six times ten to the power minus six. The water used has one point five zero times ten to the power minus six siemens per centimetre. Capital lambda m naught, the upside down V, is two hundred and eighty six siemens centimetre squared per mole. Find K with a small s p below, K s p, the solubility product. Pause the video, and try it. Done? Okay, let's check. So we put kappa equal to three point zero six times ten to the power minus six straight into the formula, right? No, no, wait! Water itself conducts a little. So kappa salt, with the small word salt below, is the measured kappa minus the water's kappa: one point five six times ten to the power minus six siemens per centimetre. That is the trap here. Okay? The solution is saturated, but so little dissolves that it is very dilute. So capital lambda m is approximately capital lambda m naught, the limiting value: two hundred and eighty six. The wavy equals sign means approximately. Right? So the solubility, s, equals kappa times one thousand, over capital lambda m naught; the one thousand turns cubic centimetres into litres. One point five six times ten to the power minus six, times one thousand, over two hundred and eighty six: five point four five times ten to the power minus six molar. Clear? Now, K s p. Barium sulphate gives one barium ion and one sulphate ion. So, in terms of s, what is K s p? s squared, yes! B a two plus and S O four two minus are each at concentration s, so K s p is s times s. Got it? Five point four five times ten to the power minus six, squared: two point nine seven times ten to the power minus eleven, about three times ten to the power minus eleven. Always subtract the water first. That's it!

8:22Quick revision: concentration cell, log K, delta G, Ksp

So, quick revision. One: a concentration cell has E naught cell zero, but E is not zero. Two: from E naught, log K is n E naught over zero point zero five nine one, and delta G naught is minus n capital F E naught. Three: subtract the water's kappa first, then s is kappa times one thousand, over capital lambda m naught. Now try the two practice problems on this page. All the best!