[antoinetta]

I've been trying to find the reduction potential OH- in a basic half-cell solution ... this may sound silly but is this the right equation I'm supposed to be using: O_{2 (g)} + 2 H₂O _(l) + 4 e^- => 4OH^- _(aq) EV= 0.401

[DevaDevil]

yes it is,

remember though that 0.40 V is the standard electrode potential, only valid when the concentration of OH^- is 1 M. and the Oxygen pressure 1 atm (activities of 1). If not, the Nernst equation will give the potential:

E = E^o + (RT/nF) ln (a_OH- / a_O2) = 0.401 + (RT/4F) ln ([OH^-]⁴ / [O₂])

F = Faraday constant
a = activity; concentration usually; =1 for a solvent (water in this case)

[antoinetta]

Ok, so by extension if I was to rank these according to their relative strength as reducing agents. For example, the most powerful reducing agent would be given rank ”1”, and the least ”6”.

H?   H₂ + 2e^- = 2H^-          EV= -2.25V
Br?   Br₂ + 2e^- = 2Br^-          EV= 1.07V
OH?    O₂ (g) + 2 H₂O (l) + 4 e- = 4OH- (aq)    EV= 0.401V
Ag+   Ag⁺ + e^- = Ag                                      EV = 0.80   (but this is switched to -0.80V)
Fe   Fe²⁺ + 2e^- = Fe                EV = -0.44V
Ca   Ca²⁺ + 2e^- = Ca              EV = -2.87V

 it will be:  Ca > H- > Ag+ > Fe > OH- > Br-


PS:  I asked a similar question called "Strength of Reduction Agent" under the General Chemistry Forum, if a moderator wouldn't mind deleting that topic, I would appreciate it

[DevaDevil]

Ag+   Ag⁺ + e^- = Ag        EV = 0.80   (but this is switched to -0.80V)

Ag⁺ is an "oxidising" agent in this reaction, not a reducing.

This is probably the reaction wanted:
Ag₂O₃(s) + 6H⁺ + 4e^? ? 2Ag⁺(aq) + 3H₂O       E = +1.67

here the silver ion is indeed the "reducing" agent

it will be:  Ca > H- > Ag+ > Fe > OH- > Br-

Well, seen as silver isn't at its place you get a different line of course, but your idea is correct

Ca > H- >  Fe > OH-  > Br- > Ag+