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In a redox
reaction, one species acts as a reducing agent, the other as an oxidizing
agent. The strength of these agents can be determined from their standard
potentials, the same table you look up the value of E0 in. If you
have a redox reaction, you can compare the oxidizing strength of the two agents;
whichever is stronger will be the oxidizing agent. (And will thus be reduced in
the reaction.)
The rules to determine strength is
- The more positive Ered0 is, the stronger the
oxidizing agent
- The more positive Eox0 is, the stronger the reducing
agent
Looking down the table in your book on page 504, the largest
value of Ered0 is for the reaction
- F2(g) + 2e- -> 2F-(aq)
Fluorine
is the most powerful oxidizing agent around. In the same vein,
- Li(s) -> Li+ + e-
has the largest
Eox, +3.040 V, and lithium is a very powerful reducing agent. Strong
oxidizing agents are poor reducing agents and vice-versa.
Example: Which of the following is the most powerful oxidizing agent?
Solution: Look at the values for Ered for the reduction of
each of the above species: the largest value is the most powerful oxidizing
agent
- Cr+3(aq) +3 e- -> Cr(s)
Ered = -0.744 V
- Br2(l) + 2e- -> 2Br-(aq)
Ered = +1.077 V
- Cu(s) -> Cu+2 + 2e- Ered
= -0.339 V
Since bromine has the largest Ered, it is the
most powerful oxidizing agent. Copper metal is next, followed by
Cr+2 ions, the weakest oxidizing agent. (And thus the strongest
reducing agent.) |