Posted by **Rebekah** on Sunday, May 8, 2011 at 6:08pm.

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**, Sunday, May 8, 2011 at 6:30pm
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**, Sunday, May 8, 2011 at 6:46pm
That makes sense! Thanks:)

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