Concentration Cells and Applications

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take two nickel rods. Dip one in a dilute solution of nickel ions, and the other in a solution one hundred times more concentrated. Join them with a salt bridge and a voltmeter. Same metal on both sides, so the standard cell potential is zero. So the voltmeter reads zero, right? Not so fast! The needle moves. Let's find out why. Later, we even measure a solubility product with one! These cells do come up in JEE.

0:33Same electrode, two concentrations

Look at the figure: the Daniell cell from an earlier lesson. Zinc, Z n, is the negative anode; copper, C u, the positive cathode, each in its own sulphate solution. The salt bridge joins the beakers, V is the voltmeter, red arrows show the electrons, and below are the electrode reactions. Two different metals, right? Now, what if both sides have the same electrode, and only the concentrations differ: c one and c two? The subscripts label the sides; the slashed equals sign means not equal. That is a concentration cell. So, what is E naught cell here, the standard cell potential? The small circle, read naught, means standard. Zero, yes, exactly! E naught cell is E naught of the cathode minus E naught of the anode, both standard reduction potentials. Same electrode, so they cancel. Standard means one molar solutions and one bar gases; tables usually give values at two hundred and ninety eight kelvin. Clear? But in our puzzle, the needle moved. So the voltage comes from the concentration difference. The spontaneous change makes the two concentrations equal, and meanwhile electrons flow through the wire. Got it?

1:45The EMF of a concentration cell

Now let's find the voltage. M is the metal, and M n plus its ion, with charge n plus. Each single bar is a metal and solution boundary; the double bar is the salt bridge. Left, concentration c one, the anode; right, c two, the cathode. Okay? On the left, metal dissolves; on the right, ions deposit as metal. So overall, ions go from c two to c one; the arrow means goes to. So the reaction quotient, Q, is c one over c two. With me? Put this Q into the Nernst equation, with E naught zero. E equals zero, minus zero point zero five nine one over n, times log of c one over c two. Flip the fraction, and the minus goes: zero point zero five nine one over n, times log of c two over c one. This is at two hundred and ninety eight kelvin; log is to base ten. Okay? Now, a cell that works on its own has a positive E cell. So which side must be more concentrated, the anode or the cathode? The cathode, correct! If c two is bigger than c one, the log is positive, so E is positive; the double arrow means so. The concentrated side is the cathode. Careful, the trap: c one over c two gives a negative answer. Concentrated on top. Right? One condition: this is for a cation-reversible electrode, one that responds to a cation, like a metal in its own ions.

3:15Electrode concentration cell: two gas pressures

Clear? Now a second kind, the electrode concentration cell. P t is platinum. Hydrogen gas at pressure p one on the left, p two on the right, over one H plus solution. The two in H two is a subscript. Now careful. Which pressure goes on top inside the log, p one or p two? It is p one, the anode side! At the anode, hydrogen at p one is used up; at the cathode, hydrogen at p two is formed. So, once we flip the log, the reactant side goes on top, and here the reactant is hydrogen at p one, the anode side. Careful: in the metal cell it was the cathode side! E equals zero point zero five nine one over two, times log of p one over p two; n is two, two electrons per H two. Positive if p one is greater than p two. Got it?

4:07Applications: Ksp, Kw, pH and Ka from a measured EMF

Why are these cells useful? Measure E with a voltmeter, and the Nernst equation gives back a concentration, even a tiny one. So cells measure K s p, the solubility product; K w, the ionic product of water; pH; and K a, the acid dissociation constant. Okay? The key idea: dip an electrode in a saturated solution of a salt. Then K s p fixes the ion concentration around it. Clear? For silver chloride, K s p is silver ion concentration times chloride ion concentration; square brackets mean moles per litre. So silver ion concentration is K s p over chloride concentration. More chloride, fewer silver ions. Right?

4:51Example 1: EMF of Ni concentration cell, 0.001 M vs 0.1 M

Now, remember our puzzle? Here it is. Find the EMF, the cell voltage, of nickel, nickel two plus at zero point zero zero one molar, salt bridge, nickel two plus at zero point one molar, nickel. Capital M means molar. And which electrode is positive? Pause the video, and try it. Done? Okay, let's check. Concentrated over dilute: zero point one over zero point zero zero one is one hundred; log one hundred is two. So E is zero point zero five nine one times two, zero point one one eight volt, right? No, no, wait! You forgot n. Nickel two plus takes two electrons, so n is two. E equals zero point zero five nine one over two, times log of zero point one over zero point zero zero one. Okay? Zero point zero five nine one over two is zero point zero two nine five, times log one hundred, that is two. So E is zero point zero five nine volt, V for volt. Puzzle solved: same metal, E naught zero, and still a real voltage! Which electrode is positive? The nickel in zero point one molar, the concentrated side. That is the cathode. Right?

6:01Example 2: Ksp of AgCl from a cell EMF of 0.47 V

Now a harder one. Left: silver, A g, with solid silver chloride, the s in brackets means solid, in zero point one molar potassium chloride. Right: silver in zero point one molar silver nitrate. E is zero point four seven volt at two hundred and ninety eight kelvin. Find K s p of silver chloride. Hint: it is a silver concentration cell. Pause and try. Done? Okay, let's check. Silver on both sides, so a silver concentration cell; the silver ion is plus one, so n is one. On the right, silver nitrate gives zero point one molar silver ions. With me? And on the left? Solid silver chloride is present, so the solution is saturated. So what fixes the silver ion concentration there? K s p, exactly! So the left side is really dilute: it is the anode. Call its concentration silver ion L, subscript L for left. Concentrated over dilute: zero point four seven equals zero point zero five nine one, times log of zero point one over silver ion L, okay? Divide: zero point four seven over zero point zero five nine one is about seven point nine five; the wavy equals sign means approximately. So silver ion L is zero point one times ten to the power minus seven point nine five: about one point one times ten to the power minus nine molar. Tiny! Last step. Chloride is zero point one molar, from the potassium chloride. K s p is silver ion L times chloride: one point one times ten to the power minus nine, times zero point one, is one point one times ten to the power minus ten. We measured a solubility product with a voltmeter!

7:49Quick revision: E naught zero, EMF formula, gas cell, Ksp

So, quick revision. Same electrode on both sides: E naught cell is zero. E equals zero point zero five nine one over n, times log of c two over c one, at two hundred and ninety eight kelvin; the concentrated side is the cathode. Gas cell: zero point zero five nine one over two, times log of p one over p two. And a measured E gives K s p, pH and K a. Now try the practice problems on this page. All the best!