Making an electric cell

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Making an electric cell The zinc turns out to be the negative electrode (the black lead) as it is the more reactive of the two metals. You can check the order of reactivity in the reactivity series. Remember that metals react by losing electrons and turning into positive ions. It follows that a more reactive metal will lose electrons more readily than a less reactive one, and consequently be the negative electrode of the pair. BackgroundThe differing reactivities of metalsWhen metals react, they give away electrons and form positive ions. This particular topic sets about comparing the ease with which a metal does this to form hydrated ions in solution - for example, Mg2+(aq) or Cu2+(aq).We might want to compare the ease with which these two changes take place:Everybody who has done chemistry for more than a few months knows that magnesium is more reactive than copper. The first reaction happens much more readily than the second one. What this topic does is to try to express this with some numbers.Looking at this from an equilibrium point of viewSuppose you have a piece of magnesium in a beaker of water. There will be some tendency for the magnesium atoms to shed electrons and go into solution as magnesium ions. The electrons will be left behind on the magnesium.                                                                                      In a very short time, there will be a build-up of electrons on the magnesium, and it will be surrounded in the solution by a layer of positive ions. These will tend to stay close because they are attracted to the negative charge on the piece of metal.Some of them will be attracted enough that they will reclaim their electrons and stick back on to the piece of metal.A dynamic equilibrium will be established when the rate at which ions are leaving the surface is exactly equal to the rate at which they are joining it again. At that point there will be a constant negative charge on the magnesium, and a constant number of magnesium ions present in the solution around it.Simplifying the diagram to get rid of the "bites" out of the magnesium, you would be left with a situation like this:Don't forget that this is just a snapshot of a dynamic equilibrium. Ions are continually leaving and rejoining the surface.How would this be different if you used a piece of copper instead of a piece of magnesium?Copper is less reactive and so forms its ions less readily. Any ions which do break away are more likely to reclaim their electrons and stick back on to the metal again. You will still reach an equilibrium position, but there will be less charge on the metal, and fewer ions in solution.If we write the two reactions as equilibria, then what we are doing is comparing the two positions of equilibrium.The position of the magnesium equilibrium . . .. . . lies further to the left than that of the copper equilibrium.Notice the way that the two equilibria are written. By convention, all these equilibria are written with the electrons on the left-hand side of the equation. If you stick with this convention without fail, you will find that it makes the rest of this topic much easier to visualise.Everything else concerning
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electrode potentials is simply an attempt to attach some numbers to these differing positions of equilibrium.In principle, that is quite easy to do. In the magnesium case, there is a lot of difference between the negativeness of the metal and the positiveness of the solution around it. In the copper case, the difference is much less.This potential difference could be recorded as a voltage - the bigger the difference between the positiveness and the negativeness, the bigger the voltage. Unfortunately, that voltage is impossible to measure!It would be easy to connect a voltmeter to the piece of metal, but how ...

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