Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Suppose a hydrogen electrode dips into an unknown solution, with hydrogen gas bubbling at one bar. Its electrode potential reads minus zero point two three six volt. From that one number, can you find the p H of that solution? Sounds strange, right? Hold that thought. This is set one of mixed problems on redox and electrochemistry, short numericals of the kind that come up often in JEE Main. So, let's begin.

0:32Idea 1: oxidation numbers and the n-factor

Idea number one: oxidation numbers. Find the oxidation number of manganese, M n, in K two M n O four; the small numbers below count the atoms. Call it x. Two potassiums give plus two; four oxygens give minus eight. The total for a neutral compound is zero, so two plus x minus eight equals zero. The double arrow means therefore: x is plus six. Okay? Now you do one. Chlorine, C l, in H C l O three. What is its oxidation number? Plus five, correct! Hydrogen is plus one, three oxygens give minus six, so one plus x minus six equals zero, and x is plus five. Clear? Now, dichromate: C r two O seven, with the two minus charge up top, changing into C r three plus; the arrow means changes into. Each chromium goes from plus six to plus three. So, how many electrons does one dichromate ion gain? Three, right? No, no, careful! Every dichromate ion has two chromium atoms. Two times three: six electrons. The electrons gained or lost per formula unit is the n-factor, so for dichromate going to C r three plus, it is six. Got it?

1:55Idea 2: the standard cell potential

Idea number two: the cell potential. Nickel, N i, dips in N i two plus ions; then the double bar, which is the salt bridge; then silver ions and silver metal, written A g. Each single bar is a boundary between phases. Anode on the left, cathode on the right. Okay? The standard potentials, E naught, the E with the little circle on top: nickel ion over nickel, minus zero point two five volt; silver ion over silver, zero point eight zero volt. The slash separates the oxidised form from the reduced form, and both are reduction potentials. Clear? So, E naught cell, with the word cell written small below, equals E naught of the cathode, minus E naught of the anode. Standard means two hundred and ninety eight kelvin, the capital K; one molar solutions, the capital M; and gases at one bar. With me? Now, the cathode is silver, zero point eight zero, and the anode is nickel, zero point two five. So zero point eight zero minus zero point two five: zero point five five volt, right? Oh no, careful! Nickel's value is minus zero point two five. Zero point eight zero minus minus zero point two five is zero point eight zero plus zero point two five: one point zero five volts. Keep the sign with the number. Got it?

3:23Idea 3: the Nernst equation for a metal electrode

Idea number three: what if the solution is not one molar? Take a metal in its own ions: M n plus, plus n electrons, gives the metal M; the n plus up top is the charge, and e minus is an electron. At two hundred and ninety eight kelvin, the Nernst equation gives E equals E naught, plus zero point zero five nine one over n, times log of the M n plus concentration, in square brackets. Okay? Now, silver dipping in silver ions of zero point zero zero one molar. One electron per silver ion, so n is one, and E naught is zero point eight zero volt. Right? So, E equals zero point eight zero, plus zero point zero five nine one, times log of ten to the power minus three. Clear? Before you calculate, tell me: will the answer be more than zero point eight zero volt, or less? Less, exactly! Log of ten to the power minus three is minus three, and three times zero point zero five nine one is zero point one seven seven. So, zero point eight zero minus zero point one seven seven: zero point six two three volt. Fewer silver ions, lower potential. With me?

4:42Example 1: volume of 0.1 M K2Cr2O7 for 30 mL of 0.1 M FeSO4

Now, example one. What volume of zero point one molar K two C r two O seven, potassium dichromate, oxidises thirty millilitres of zero point one molar F e S O four, ferrous sulphate, in acid? Pause the video and try it first. Done? Okay, let's check. Both solutions are zero point one molar. So we need the same volume, thirty millilitres, right? No, no, wait! In redox, equivalents match, not moles. Dichromate goes to C r three plus: n-factor six, from idea one. Ferrous ion, F e two plus, goes to F e three plus, losing one electron: n-factor one. Okay? Law of equivalence: equivalents of oxidant equal equivalents of reductant. So n one M one V one equals n two M two V two, n-factor times molarity times volume; the small one and two below label the two solutions. Clear? Dichromate side: six times zero point one times V, the volume we want. Iron side: one times zero point one times thirty. Right? So V is three over zero point six: five millilitres. Only five! Each dichromate ion takes six electrons, so it does six times the work. Got it?

6:08Example 2: pH from a hydrogen electrode reading of -0.236 V

Now the harder one, and it is our puzzle from the start. Here is a hydrogen electrode, with hydrogen at one bar; the small p with H two below is its partial pressure. It shows minus zero point two three six volt. What is the p H value? Pause the video and try it. Done? Okay, let's check. The electrode reaction: two H plus, plus two electrons, gives hydrogen gas. E naught is zero by definition, and n is two. Okay? Now Nernst. E equals minus zero point zero five nine one over two, times log of p H two, over the H plus concentration squared. Right? Hydrogen is at one bar. So, many students write E equals minus zero point zero five nine one over two, times the p H value. Is that correct? No, careful! Look at the square on H plus. At one bar, we need log of one over H plus squared, which is two times minus log of the H plus concentration: two p H, okay? That two cancels the two below, so E equals minus zero point zero five nine one times p H, right? Forget the square, and you get a p H of eight. Wrong! Clear? Now put in the reading: minus zero point two three six equals minus zero point zero five nine one times the p H value. Got it? The minus signs cancel, so p H is zero point two three six over zero point zero five nine one, about four point zero; the wavy equals sign means approximately. An acidic solution. One voltmeter reading gave us the p H value. Puzzle solved!

7:56Quick revision: n-factor, E0 cell, Nernst, E = -0.0591 pH

So, quick revision. The n-factor is electrons per formula unit, and in titrations, n one M one V one equals n two M two V two. E naught cell is E naught cathode minus E naught anode, signs kept. And at two hundred and ninety eight kelvin, a metal electrode follows E equals E naught, plus zero point zero five nine one over n, times log of the ion concentration; and a hydrogen electrode at one bar gives minus zero point zero five nine one times the p H value. Now go and try the practice problems on this page. You will find them easy. All the best!