Posted by Anna on Tuesday, April 15, 2008 at 7:03pm.
Reaction 1.
Cu (strongest reducing agent)
I-
Ag
Br- (weakest reducing agent)
Reaction 2:
Zn(strongestreducing agent)
Pb
Cu
Ag(weakest reducing agent)
Reaction 3:
Cl2 (strongest oxidizing agent)
Br2
I2 (wekest oxidizing agent)
Using these results, how am I supposed to prepare a single table of relative strengths of oxidizing and reducing agents?
What happens if you put 1 and 2 together?
#1.
Cu (strongest reducing agent)
I-
Ag
Br- (weakest reducing agent)
#2.
Zn(strongestreducing agent)
Pb
Cu
Ag(weakest reducing agent)
So it seems to me, forgetting #3 for the moment, that it's almost done for you.#2 tells you Zn, and Pb are at the top (above Cu and Ag) and follow that from #2 with Cu, I^-, Ag, Br^-. Then a look at #3 tells you that Br2 and I2 are already in the series as I^- and Br^-. The only one left for you to do is Cl^-. Since Cl2 is stronger than Br2 and I2, wouldn't Cl^- be the weakest of the three. Check my thinking.
One other quick note. Make sure you want to list these with the strongest reducing agent at the top and the weakest at the bottom; otherwise, you need to reverse the order.
One other quick note. Make sure you want to list these with the strongest reducing agent at the top and the weakest at the bottom; otherwise, you need to reverse the order.
So Cl2
Br2
I2
becomes
I2
Br2
Cl2
?
With them listed as
Zn
Pb
Cu etc, we are looking at the half reactions as
Zn ==> Zn^+2 + 2e
Pb ==> Pb^+2 + 2e
Cu ==> Cu^+2 + 2e
These are written as oxidations and Zn is the strongest reducing agent with Pb and Cu following in order. Written as reductions (oxidizing agents) these would be Cu^+2 + 2e ==> Cu
Pb^+2 + 2e ==> Pb.
I don't remember the instructions from the original problem so I don't know how you want them arranged. The way the tables are arranged with the reactions, I THINK (but you must be the judge of that) that Zn, Pb, Cu, Ag, etc are the way it is to be done.
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