[pmennen]

How does one pick the best reducer?

For example, reaction "e" of the 1989 AP exam is:
"A solution of tin(II) chloride is added to an acidified
solution of potassium permanganate".

So it's pretty clear that permanganate is going to be the
oxidizer (getting reduced to (Mn)2+). Since Halide ions
can be strong reducers, I assumed the chloride would get
oxidized yielding the net ionic equation:

(Cl)- + (MnO4)-  + H+  -->  (Mn)2+  + Cl2 + H2O

However (Sn)2+ can also be a reducer (getting oxidized
to (Sn)4+) yielding the net ionic equation:

(Sn)2+ + (MnO4)- + H+  -->  (Sn)4+ + (Mn)2+ + H2O

The answer sheet I have lists just the second one. Why
is that? Isn't my answer just as likely?

Similarly how does one pick the best oxidizer?
For example, reaction "c" of the same 1989 exam is:
"A stream of chlorine gas is passed through a solution
of cold, dilute sodium hydroxide".

So I figured chlorine can get oxidized by a reaction such as:

4(OH)-  + Cl2  -->  2(ClO)-  + 2H2O + 2e-

combined with the sodium ions being reduced giving the net
ionic equation:

OH-  + Cl2  +  Na+  -->  (ClO)-  +  Na  + H2O

However chlorine could be reduced as well as being oxidized
so the answer sheet gave the equation:

OH-  + Cl2  --> (ClO)-  + Cl-  + H2O

But how would one know that this is more likely than my answer?

Thanks
~Paul

[kyle1990]

Refer to a standard reduction potential table. The table is read as reduction potentials. Everything in the left column is arranged in order of decreasing oxidizing capability from top to bottom. Notice that fluorine is the best oxidizer on the table and thus at the top of the list. On the right hand column, the strongest reducers are on the bottom and generally include the alkali/alkaline metals. It can get a bit confusing if you don't have the jargon straight.
You can see from the table that Sn2+ is a weaker oxidizing agent than MnO4- so MnO4- will be reduced because it is the stronger oxidizing agent and Sn2+ will become oxidized.

hey