Posted by Rebekah on .
Use the following reactions to arrange the elements A, B, C, and D in order of their redox activity, from greatest to least.
2 A + B2+ > 2 A+ + B
B + D2+> B2+ + D
A+ + C >No reaction
2 C + B2+ > 2 C+ + B
Can someone explain?
thanks

Chem 
DrBob222,
I think the easy way to look at this is to study the activity series. Here is a link.
http://www.files.chem.vt.edu/RVGS/ACT/notes/activity_series.html
An example or two. Any metal will oxidize any ION below it in the series; therefore, note that Li will oxidize H^+ or Cu^+2 or Ag^+. Therefore, Li goes ABOVE H and Cu. Cu will not react with HCl, for example, to give H2 gas; therefore, Cu is BELOW H so we would place the metals as Li, H, Cu.
It works the same way with A, B, C, and D EXCEPT that instead of giving the placement in the table the problem gives you the reactions.
2A + B^2+ ==> 2A^+ + B
Note that A is oxidized and B^2+ is reduced. Therefore the abbreviated table is
A
B
B + D^2+ ==> B^2+ + D
B is oxidized, D^2+ is reduced.
So
B
D is the placement.
A^+ + C ==> NR (you may see it better if we keep it consistent and write it as C + A^+ ==> NR) which means C is below A but last equation says it is above B. So I write it as
A
C
B
D
Check my thinking. 
Chem 
Rebekah,
That makes sense! Thanks:)