Products of Electrolysis

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Melt some common salt, sodium chloride, and pass a current through it. Shiny sodium metal forms at one electrode. Now dissolve the same salt in water, and pass the same current. At that electrode, no sodium at all! Just bubbles of hydrogen gas. Same salt, same current, and yet a different product. Strange, right? Today we learn to predict exactly what forms at each electrode. This comes up often in JEE electrochemistry. So, let's begin.

0:36Electrolytic cell: the current forces the reaction

So, first, what is electrolysis? Look at the figure. The battery symbol at the top is the external source, and the two grey rods are the electrodes. In an electrolytic cell, electrical energy from this external source drives a non-spontaneous redox reaction, one that would not go on its own. Okay? Now think. Here, the cathode is connected to the negative terminal of the source. Does that change which electrode does the reduction? No, it does not! Reduction still occurs at the cathode. In the figure, the cathode is on the left, with a minus sign in brackets: it is connected to the negative terminal. And see the arrow: the sodium ions, N a plus, move towards it. Right? And oxidation still occurs at the anode, on the right, with a plus sign in brackets: the positive terminal. The chloride ions, C l minus, move towards it. Got it? Now the simplest case: a molten salt. Only the salt's own ions are present. So the metal forms at the cathode, and the non-metal at the anode. Molten N a C l gives sodium, N a, at the cathode, and chlorine, C l two, at the anode. The small two below means two atoms in each molecule.

1:49In water: who is reduced at the cathode?

Clear? Alright, now the salt is dissolved in water, like in the figure: N a C l, with a q in brackets, which means aqueous. The sodium ions reach the cathode. But they are not alone there. Who else is sitting at the cathode, ready to take electrons? Water, yes, exactly! Water molecules are everywhere. So there is a competition, and the rule is simple: the species with the higher reduction potential is reduced. With me? Now, the ions of very active metals: N a plus, K plus, M g two plus, and A l three plus. The small plus signs up top are their charges. These ions are not discharged in water. Their reduction potentials are too low, so water wins. Okay? So water is reduced instead: two H two O, plus two e minus, gives H two, plus two O H minus. The e minus is an electron, and the arrow means gives. Those little circles at the cathode in the figure? Hydrogen bubbles! That is why our salt water gave hydrogen, not sodium. Right? But not every metal ion loses. Copper two plus, C u two plus, and silver plus, A g plus, are discharged, and the metal deposits on the cathode. Got it?

3:04In water: who is oxidised at the anode?

Now the anode. Take an inert anode of platinum, P t, which does not react itself. Here three halide ions, chloride C l minus, bromide B r minus, and iodide I minus, are oxidised in preference to water. Okay? For chloride, the reason is the overpotential of oxygen. Overpotential is an extra voltage, beyond the expected value, that a gas needs to form at an electrode. Oxygen needs a lot of it, so chlorine, C l two, forms instead. Clear? But sulphate, S O four two minus, and nitrate, N O three minus, are not oxidised. They just stay in the solution. So water is oxidised instead, and oxygen gas, O two, comes off. Right? Now a quick question. Make the sodium chloride solution very, very dilute. Do we still get chlorine at the anode? No, careful! In very dilute N a C l, oxygen forms. With so few chloride ions around, water is oxidised instead. Remember this trap, okay? And now our puzzle is completely solved. Concentrated brine, that is strong salt water, with platinum electrodes: hydrogen at the cathode, chlorine at the anode, and sodium hydroxide, N a O H, stays in the solution. That's it!

4:24When the electrode itself joins in

So far our electrodes were inert. Now take an active anode, made of copper, C u, or silver, A g. Then the anode metal itself dissolves: it is oxidised to its ions. Okay? This is the basis of electrorefining, which purifies a metal, and electroplating, which coats an object with a metal. With me? One more special case: a mercury cathode, H g. With brine, it gives sodium! The sodium dissolves in the mercury, as sodium amalgam. So on mercury, sodium beats water. Surprising, right? The reason: hydrogen, H two, has a high overpotential on mercury. Hydrogen finds it very hard to form there, so the sodium ions are discharged instead. Got it?

5:08Example 1: aqueous CuSO4 with Pt and with Cu electrodes

Now, Example one. Give the products of electrolysis of aqueous copper sulphate, C u S O four. Part a: with platinum electrodes. Part b: with copper electrodes. Pause the video, and try it first. Done? Okay, let's check. Part a, the cathode. Copper two plus is one of the ions that is discharged, remember? So copper metal deposits on the cathode. Right? Now the anode. Sulphate is not oxidised, so water is, and oxygen gas comes off. Now think: copper ions keep leaving the solution as copper metal. So what is left behind in the beaker? Yes, exactly! The sulphate stays behind, and the water oxidised at the anode gives hydrogen ions. So the solution becomes acidic: it slowly turns into sulphuric acid, H two S O four. Clear? Now part b, copper electrodes. The cathode still gets copper. But the anode is now active, so the copper anode itself dissolves: C u gives C u two plus, plus two e minus. No oxygen at all! So copper leaves the anode as fast as it deposits on the cathode. The square brackets mean concentration: the concentration of copper two plus stays constant. This is the idea behind electrorefining. With me?

6:32Example 2: order of deposition of Ag, Cu and Mg

Now a slightly harder one. Silver nitrate, A g N O three, copper nitrate, and magnesium nitrate are mixed in water and electrolysed. In the copper and magnesium formulas, the brackets with a two below mean two nitrate groups. In what order do the metals deposit? Pause and try. Hint: think about who wins at the cathode. Done? Okay, let's check. We need the standard reduction potentials, E naught: the E with the little circle on top, measured at two hundred and ninety eight kelvin and one molar. Silver, zero point eight zero volts. Copper, zero point three four volts. Magnesium, minus two point three seven volts. Okay? The highest E naught is reduced first. So silver deposits first, then copper. Right? And once the copper is used up, magnesium deposits next. That completes the order, doesn't it? No, no, careful! Magnesium two plus was in our list of very active metal ions. At minus two point three seven volts, it is not discharged from water. Water is reduced instead. Clear? So the order is: silver, then copper, and no magnesium at all. See how the one rule did the whole question?

7:45Quick revision: molten salt, then cathode and anode in water

So, quick revision. Molten salt: metal at the cathode, non-metal at the anode. In water, at the cathode, the higher E naught wins, and active metal ions leave it to water, which gives hydrogen. At the anode, chloride, bromide and iodide beat water, while sulphate and nitrate leave it to water, which gives oxygen. Now go and try the practice problems on this page. You will find them easy. All the best!