Given this table:
Half - reactions Eo, V
Ag+ + e- -> Ag +0.80
Cd2+ + 2e- -> Cd -0.40
2F- -> F2 + 2e- -2.87
2I- -> I2 + 2e- -0.54
Na+ + e -> Na -2.71
Se2- -> 2Se + 2e- +0.67
I was wondering why Cd2+ is the strongest oxidizing agent, and that I- is the strongest reducing agent?
One thing I forgot to post was that for picking the strongest oxidizing agent and strongest reducing agent I was given 4 choices:
Cd2+
F-
I-
Na+
first remember what happens to an oxidizing agent: it gets reduced.
so the strongest oxidizing agent will be the thing most likely to get reduced.
looking at that list, you can forget about I- and F-...they are already in their reduced form.
good candidates for reduction would be Cd2+ and Na+...whichever one has the highest (ie. most positive) reduction potential is the one that is more driven to be reduced (which means it will act as the stronger oxiding agent)
the reduction potential value for Cd2+ = -0.40
the reduction potential value for Na+ = -2.71
obviously -0.40 is more positive, so Cd2+ is more driven to be reduced which would mean it would be the stronger oxidizing agent.