# Activity Series

1. Dec 16, 2004

### Raza

Can someone explain what is an Activity Series? I looked at sites that is explaining it but I do not get it. Please try to make it simple as possible because I am really dumb.
Thanks

2. Dec 16, 2004

### HallsofIvy

Staff Emeritus
I have absolutely no idea! I have worked in mathematics and dabbled in physics for many years and have never seen the phrase "activity series"! Could you give us the details of whatever problem you are working on that you saw this?

3. Dec 16, 2004

### dextercioby

I guess he might be talking about the 4 Radioactivity Series,or probably about Beketov-Volta Series???Or World Series???

Daniel.

4. Dec 16, 2004

### Raza

Last edited: Dec 16, 2004
5. Dec 16, 2004

### dextercioby

Man,i'm gooooooooood!!!!!!!! I knew it was about Beketov-Volta series.
Okay,here's the deal.It is basically made following two concepts:electronegativity/electropositivity and electrochemical potential;I honestly do not know what grade are u in,and how much inorganica chemistry u know.I'll then assume the worst:assimptotically going to zero. :tongue2:

U know that elements are being classified in metals nonmetals (there's another group,but let's not do into much detail,just the basics).Metals have the property of ineteracting with acids.Hope u know what acids are.These acids may have a rather complicated chemical/molecular structure,but they all share the same elementydrogen.Whethe it's only on atom (HCl,HF,HI,HBr) or two_{2}S,H_{2}Se,H_{2}SO_{4},...it doesn't matter.What it does is the fact that,when they interact with metals,that hydrogen atom (involved in a chemical bond (called covalent)) can be substituted by the metal's atom:E.g.
$$2HCl+Zn \rightarrow ZnCl_{2}+H_{2}$$.
Metals are classified in a ranking according to the criterior of hox easily they can substitute Hydrogen (if u know a bit about chemical bonds,then u might know that Hydrogen is univalent (in a chemical bond it contributes with only one electron,because,...that's what he hasne lousy electron),so some metal atoms take out one atom (Li,Na,K,Rb,Cs),some 2 (Zn,Cu,Mg,Ca,Fe,...),some three (Al,Fe,Sc,Cr,Au) and that's it.No metal atom can pull more than 3 atoms of Hydrogen from any acid.The more atoms (of H) a metal takes,the harder it is to do so.Because in the acid's molecule Hydrogen is tied up pretty well,ad the more atmos u wann take out,the more energy u need to have:chemical energy that is.Electropositivity is defined at the capacity to take out hydrogen atoms from acids.As i (hopefully made u realize) said,the easiest it is for the atoms which are univalent and those are alkali metals (Li,Na,K,Rb,Cs).Then it's a bit harder for the ones which can pull out 2 atoms:Ca,Mg,Sr,Fe,(II)u,Zn.And least for the ones with three atoms,like the case of Fe(III),Al (III).That series u have seen with metals only is not how i pictured so far,but a bit different:there are exception for this rules:there are metals which cannot remove hidrogen atoms from any acid,no matter how strong that acid may be:Those are Cu (copper),Hr(mercury),Ag (silver),Pt(Platinum),Au (gold).I'm afraid the reasons for this to happen do not seem dso obvious,not obvious at all.The key is,that those elements,unfortunately,must be memorized:there are not too many (i enumerated only 5,actuanlly there are VERY MANY),and u can do that.As for the fisrt part,there are some exceptions,for instance,Al can remove hydrogen (3 atom,remember ?) lot easier than Sn (tin) and Pb (lead),which have a smaller valence.This is not understandable theoretically,you need to convinceyourself that ot happens.Basically this scale is structred wrt to Hydrogen,but the order (e.g.Rubidium folllowed by Potassium (Kalium)) does not not only mean that one can remove H better than the other,but on e of them (the one with larger electropositivity) can remove the other from any combination (inorganic/organic it is irrelevant),as long it is in aquaous (with water) solution.Why it needs water??Well,because a certain phenomenon called "dissociation" occurs,but that's not on our agenda.
Let's return.We'veestablished the first Beketov-Volta series arranging elements on an axis in ascending/descending (it doesn't matter,really) order of how easily they can remove H atoms from acids (elsen how big the electropositivity is).But the order among them is made not arbitrariry,but by taking into account that metals can substitute one another from variious combinations.
So it should read (descending order of electropositivity/from right to left it gets harder to remove H atoms from acids) :Li-Cs-Rb-K-Sr-Ca-Na-Mg-Al-...-Pb-H-Cu-Hg-Ad-Pt-Au.Note that i specifically put H in the list (though obvious nonmetalic) as to depict the exact point in the series where left (in this case) is a metal that can remove H from acids,and to the right (in this case) is a metal which cannot do that.

For the nonmetals,the situation is reversed,the capability of accepting electrons is relevant and sets a criterion in this sense.The series respects the same procedure,according to valence:F,Cl,Br,I are the ones that accept the easiest the electron and therefore they should be atop of the others.Then nonmetals with 2 valence electrons:O,S,Se,Te,then 3:N,P,As,...Again 4 is not an acceptable number:neither Carbon,nor Silicium can accept 4 electons.But again,this rule is not rigid,ad that can be seen from the second element of the list,this time put again in descending order of electronegativity (a measure of easiness of accepting the electron/s);F-O-Cl-Br-I-S-Se-N-P.this time there are no elements to memorize,just that O and Cl have the same electronegativity (thesame ability of accepting electrons),and therefore,it's irrelevant whether they're ordered in one way or the other.

I know that those explanations were rather superficial and not thoroughly documented (i tried to come up with plausible explanations,not numbers or formulae,but hopefully it gives you a "basic",a "ground" to futher read about it and eventually understanding about these 2 very important series (for the analytical and inorganic chemistry mostly).

Daniel.

6. Dec 17, 2004

### Raza

Wow, Thank you very much, Daniel. I really understand it.