Faraday's Laws of Electrolysis

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two beakers are joined in series, one after the other. One has silver nitrate solution, the other has copper sulphate solution. The same current flows through both, for the same time. So, the same mass of metal comes out in each beaker, right? Or at least the same number of atoms? Hmm. Hold that thought. Today, with Faraday's two laws of electrolysis, we will settle it properly. Questions on these laws come up often in JEE electrochemistry. So, let's begin.

0:38Charge, faradays and moles

So, everything in electrolysis starts with charge. Capital Q is the charge that flows, in coulombs, the capital C on screen. Capital I is the current, in amperes, written with a capital A, and small t is the time in seconds, the small s. Capital Q equals capital I times t. Okay? Now, the most important number of this chapter. One faraday, written capital F, is the charge on one mole of electrons. It is ninety six thousand four hundred and eighty five coulombs. The wavy equals sign means approximately: we usually round it to ninety six thousand five hundred. Right? So, tell me. If a charge of capital I times t flows, how many moles of electrons have passed? Yes, exactly! Divide by one faraday. Moles of electrons equals capital I times t, over capital F, really simple. Clear? Now, copper two plus ions are deposited as copper metal. How many electrons does each ion need? And so, how many moles of copper do we get? Two electrons for each ion, correct! So we get half as many moles of copper. In general, moles of product equals capital I times t, over n times capital F, where n is the number of electrons transferred for each ion: the n factor. For copper two plus, n is two. Got it?

2:04Faraday's first law

Now, Faraday's first law. The mass deposited, small m, equals capital Z times capital Q, that is, capital Z times capital I times t. So, double the charge, and you double the mass. With me? This capital Z has a name: the electrochemical equivalent. Capital Z equals capital E over capital F, which is also capital M, over n times capital F, okay? Here capital M is the molar mass, n is the same n factor as before, and capital E we will meet in a moment. Got it? Compare the two forms. Capital E equals capital M over n: the equivalent mass. Careful: in this chapter, capital E is a mass, not an electrode potential! Right? Now think. Put capital Q equal to one faraday in the first law. What mass comes out? Capital Z times capital F, and from the second line, that is just capital E, the equivalent mass! So, one faraday deposits one equivalent. Remember this line. Clear? Two values we need today. Copper two plus: sixty three point five over two, so capital E is thirty one point seven five. Silver plus takes just one electron, so its equivalent mass equals its molar mass, one hundred and eight. Okay?

3:23Faraday's second law

Now the second law. Pass the same charge through two different electrolytes. Then m one over m two equals capital E one over capital E two. The small one and two below the letters just label the first and the second substance. So the masses are in the ratio of their equivalent masses. Got it? To get the same charge, connect the cells in series, one after the other. The same current flows through each, for the same time, so every cell carries the same charge. Right? So, copper and silver cells in series. The masses are in the ratio sixty three point five to one hundred and eight, the molar masses, right? No, no, careful! That is the trap. Copper needs two electrons per atom, silver only one. So the ratio is thirty one point seven five to one hundred and eight, the equivalent masses, not sixty three point five to one hundred and eight. Clear?

4:16Gases at STP, and current efficiency

Now, gases. Hydrogen gas, H two, needs two electrons for every molecule. So, one faraday gives how many moles of hydrogen? Half a mole, exactly! And at STP, half a mole of hydrogen is eleven point two litres, the capital L for litres. Okay? Oxygen, O two, forms at the anode, and four electrons are given up for every molecule. So one faraday gives a quarter of a mole: at STP, five point six litres. Right? Last, current efficiency. The theoretical mass comes from the first law; the actual mass is what we really collect. Current efficiency equals the actual mass over the theoretical mass, times one hundred percent. With me?

5:03Example 1: mass of Cu deposited by 2 A in 1 hour

Okay, Example one, an easy one. Read it with me: a current of two point zero amperes passes through copper sulphate solution for one hour, the small h. Find the mass of copper deposited. Pause the video and try it. Done? Okay, let's check. First, the time. Can we put t equal to one straight into capital I times t? Oh no, careful! Time must be in seconds. One hour is three thousand six hundred seconds. Okay? Now, the number of faradays is capital I times t, over capital F: two times three thousand six hundred, over ninety six thousand five hundred. That gives zero point zero seven four six faraday. This is the moles of electrons. Right? Copper two plus takes two electrons, the two e minus on screen, and the arrow shows it becoming copper metal. So the moles of copper are one half times zero point zero seven four six. Clear? Multiply by the molar mass of copper, sixty three point five. Small m equals two point three seven grams. That's it! Two point three seven grams of copper from one hour of current. Got it?

6:17Example 2: Ag and Cu cells in series, mass of Cu

Now Example two, a little harder, and it is our opening puzzle! Two cells, one with silver nitrate and one with copper sulphate, are connected in series. One point zero eight grams of silver is deposited. Find the mass of copper. Pause the video and try it. Done? Okay, let's check. Cells in series. So, which law do we use, and which masses go in the ratio? Faraday's second law, yes! Series means the same charge, and the double arrow means therefore. So, m copper over m silver equals capital E copper over capital E silver. The small Cu and Ag below the letters name the metal. Okay? Now the numbers. m copper over one point zero eight equals thirty one point seven five over one hundred and eight. Equivalent masses, not molar masses: remember the trap! Right? Cross multiply. m copper equals zero point three one seven five grams. Perfect! Now, our puzzle. Silver: one point zero eight over one hundred and eight, zero point zero one mole. Copper: zero point three one seven five over sixty three point five, zero point zero zero five mole. So, half as many copper atoms, because each copper ion needs two electrons. Not the same mass, not the same atoms, but the same charge! Clear?

7:46Quick revision: m = ZIt, equivalent masses, one faraday

So, quick revision. First law: small m equals capital Z times capital I times t, with capital Z equal to capital M, over n times capital F, and time in seconds. Second law: for the same charge, the masses are in the ratio of the equivalent masses. And one faraday, about ninety six thousand five hundred coulombs, deposits one equivalent. Now go and try the practice problems on this page. You can do them! All the best!