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Chemguide: Core Chemistry 14 - 16 The reactivity series and displacement reactions This page discusses reactions between a metal and a solution of a salt of another metal lower in the reactivity series. It assumes that you have read the previous page in this series about oxidation and reduction. What is a salt? Common salt, which you might use to put on your chips, for example, is sodium chloride, NaCl. But there are lots of other salts. These are all salts:
. . . and many, many more. With the exception of ammonium salts, they all contain a metal and the rest of the compound comes from an acid - in the examples above:
In each case, the hydrogen in the acid has been swapped for a metal (or the ammonium group). Salts like these (and others) will play a major part in the basic chemistry you will come across in the early part of a course. A displacement reaction between iron and copper(II) sulfate solution I am going to introduce this with a piece of video. The quantities chosen give about a two times excess of iron. That gives a clearer picture of what happens in the reaction. You will have seen that the solution loses its colour, and that a pinky-brown solid is produced - the solid is copper of course. The equation for the reaction is
The state symbols are important because we need to show what is in solution and what is solid. The copper is obvious as one of the products because of its colour. And it is obvious that the copper(II) sulfate has gone because of the lack of blue colour in the solution. The iron has replaced the copper and formed iron(II) sulfate solution instead. We call this a displacement reaction - you can think of the copper as being pushed out of the compound by the iron. Iron(II) sulfate solution is a very, very pale green colour, but its colour darkens slowly on exposure to air. This video accurately shows the almost colourless appearance of freshly created iron(II) sulfate solution. Displacement reactions and the reactivity series You will remember that metals higher in the reactivity series can remove oxygen from the oxides of other metals below them in the reactivity series. These are also displacement reactions. The same thing applies here. A metal higher in the reactivity series can displace one further down from one of its salts. A displacement reaction between zinc and lead(II) nitrate solution Zinc is above lead in the reactivity series and you would expect the zinc to displace the lead to form metallic lead and zinc nitrate solution. The next video shows this happening. It also shows that copper, which is below lead in the reactivity series, has no reaction with lead(II) nitrate solution. The reaction that happened was
Displacement reactions and oxidation and reduction You may have noticed that near the beginning of the last video, the presenter mentioned the words oxidation and reduction, but never came back to them later. Let's take the last equation again:
How can this be a redox reaction? There certainly isn't any simple oxygen transfer. We need to take this equation to pieces and see what is happening at the ion level.Zn(s) + Pb2+(aq) + 2NO3-(aq) The lead(II) nitrate and the zinc nitrate are both ionic solids and in water their ions are free to move around. If you look at this carefully, you will see that the nitrate ions are spectator ions - they are completely unchanged during the reaction, and so we leave them out of the ionic equation.
The zinc has lost electrons to form a zinc ion - it has been been oxidised. (Remember OIL RIG). The lead(II) ions have gained electrons to form neutrally charged lead atoms - this is reduction. So the underlying chemistry in these reactions is exactly the same as in the reactions between metals and the oxides of metals lower in the reactivity series. The more reactive metal gets oxidised; the ion of the less reactive metal gets reduced. All that is different is that this time the ions are in solution whereas previously they were a part of a solid. | |
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Note: You may possibly have wondered why we have referred to the lead compound as lead(II) nitrate, whereas we just talked about zinc nitrate without any Roman numeral in brackets. Lead forms two different ranges of compounds - lead(II) compounds and lead(IV) compounds, and we need to distinguish between them. At this level, you probably won't come across any lead(IV) compounds, but we still need to be precise. Zinc only forms one range of compounds, and so there is no need to make any distinction. You could call them zinc(II) compounds because that is the charge on the zinc ions, but mostly we don't bother! | |
© Jim Clark 2020 |
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Cu(s) + FeSO4(aq)