Standard Electrode Potentials and the Electrochemical Series

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. You have a blue solution of copper sulphate, and a zinc vessel. Can you store the solution in it? And silver nitrate solution, in a copper vessel? Both vessels are solid metal, so it looks safe, right? One table of numbers will answer both questions in seconds. That table is the electrochemical series, and it comes up often in JEE electrochemistry. So, let's begin.

0:29The reference: standard hydrogen electrode

So, first, a question. Look, a voltmeter always needs two electrodes. So how can one single half-cell get a potential of its own? We fix one partner, and measure every half-cell against it: the standard hydrogen electrode, SHE, in the figure. Hydrogen gas, H two, at one bar, bubbles over platinised platinum, P t, in a solution of H plus ions. Its potential is taken as exactly zero volts, capital V, by definition. The E with a small circle on top is read E naught; the circle means standard. And the slash separates the oxidised form from the reduced form. Okay? Standard means unit activity for every species, the small a equal to one in the figure. So, one molar, capital M, for solutions, one bar for gases, and two hundred and ninety eight kelvin, capital K, for the temperature. With me? And every E naught is written as a reduction potential: the oxidised form gains electrons. Like the reaction at the bottom of the figure: two H plus, plus two electrons, e minus, in equilibrium with hydrogen gas. The double half-arrows mean equilibrium; a q in brackets means aqueous, and g means gas. Clear? Now a quick question. If I multiply a half-reaction by two, does its E naught also become double? No, it does not! E naught is an intensive property. Times two, the small times sign on screen, changes nothing. Remember this for Example two. Got it?

2:06The electrochemical series (E^∘ in volts)

Here is part of the electrochemical series, from the lowest E naught to the highest, in volts. The lowest, lithium plus over lithium, minus three point zero four. Zinc two plus over zinc, minus point seven six; the two plus above is the charge on the ion. Our reference, H plus over H two, zero. And copper two plus over copper, plus point three four. Okay? Going higher: iodine over iodide, I two over I minus, plus point five four. Iron three plus over iron two plus, plus point seven seven. Silver plus over silver, plus point eight zero. Bromine over bromide, plus one point zero nine. And the highest, fluorine over fluoride, plus two point eight seven. Right? Now think. Fluorine, F two, has the highest E naught here. So is F two a strong oxidising agent, or a weak one? Strong, yes! The strongest in this table. The rule: the higher, or more positive, the E naught, the stronger the oxidised form as an oxidising agent. Clear? The other end works the opposite way. The lower, or more negative, the E naught, the stronger the reduced form as a reducing agent. So lithium, the lowest here, is the strongest reducing agent here. With me? But wait. Lithium has a high ionisation energy; it does not give up its electron easily. So how can it be the strongest reducing agent? The answer is water. In water, the lithium ion, L i plus, has a very large hydration energy, and that makes up for it. So in water, lithium is the strongest reducing agent. Students often slip here. Okay?

3:55Using the series

Now, how do we use the series? First, E naught cell equals E naught cathode, minus E naught anode, both as reduction potentials. The small words below each E are subscripts naming the electrode. And the reaction is spontaneous if E naught cell is greater than zero. Clear? Quick check with the Daniell cell. Copper is the cathode, and zinc is the anode. So E naught cell is point three four minus point seven six, that is, minus point four two volts. Right? No, no, careful! The zinc anode is minus point seven six, and minus of a minus is plus. So point three four, minus, minus point seven six in brackets, gives one point one zero volts. Positive, so the Daniell cell reaction is spontaneous. Okay? Second use: displacement. So, a metal displaces from solution any metal with a higher E naught. Zinc, minus point seven six, displaces copper, plus point three four, but copper does not displace zinc. With me? Third: metals with a negative E naught liberate hydrogen gas from dilute acids. Copper, silver and gold, written C u, A g and A u, do not. Got it? And fourth: a species can oxidise any species whose couple has a lower E naught. Higher oxidises lower. Now our puzzle is easy. Right?

5:25Example 1: CuSO4 in a zinc vessel, AgNO3 in a copper vessel

Now, back to our puzzle from the start. Can a copper sulphate solution be stored in a zinc vessel? Can silver nitrate be stored in a copper vessel? Use the table. Pause the video, and try it first. Done? Okay, let's check. Zinc, minus point seven six, lies below copper two plus over copper, plus point three four. So zinc reduces the copper two plus ions; zinc displaces copper. Okay? So the zinc of the vessel itself would react, and go into the solution. Copper sulphate in a zinc vessel: no! Clear? Now silver nitrate in copper. Copper does not even liberate hydrogen from dilute acids. So a copper vessel should be safe, right? No, no, careful! Compare copper with silver, not with hydrogen. Copper, plus point three four, lies below silver plus over silver, plus point eight zero. So copper reduces silver plus. With me? Check with the formula: the silver couple is the cathode, copper is the anode. Point eight zero minus point three four is point four six volts, greater than zero. So silver nitrate in a copper vessel: no, again. No and no! Got it?

6:43Example 2: will Fe3+ oxidise iodide and bromide

Now a harder one. Will iron three plus, F e three plus, oxidise iodide, I minus? And will it oxidise bromide, B r minus? Pause and try. Hint: compare the couples in the table. Done? Okay, let's check. Iron three plus over iron two plus, point seven seven, lies above iodine over iodide, point five four. Higher oxidises lower, so iron three plus can oxidise iodide. Okay? Here is the reaction. Two F e three plus, plus two I minus, gives two F e two plus, plus I two. The arrow means gives. Iron three plus is reduced at the cathode; iodide is oxidised at the anode. Right? Now look at the equation. It has two iron three plus ions. So should we take two times point seven seven? No! E naught is intensive, remember? So E naught cell is point seven seven minus point five four: point two three volts, greater than zero. Yes, iron three plus oxidises iodide. Clear? Now bromide. Iron three plus over iron two plus, point seven seven, lies below bromine over bromide, one point zero nine. This time the iron couple is the lower one. With me? So E naught cell is point seven seven minus one point zero nine: minus point three two volts, less than zero. Not spontaneous, so no. Same iron three plus: yes for iodide, no for bromide. Got it?

8:18Quick revision: SHE, oxidising power, E cell sign

So, quick revision. E naught is a reduction potential against the SHE, which is zero volts, at one molar, one bar and two hundred and ninety eight kelvin; and it is intensive. Higher E naught, stronger oxidising agent; lower E naught, stronger reducing agent. And E naught cell, cathode minus anode, positive means spontaneous. Go and try the practice problems on this page. All the best!