Posted by Anonymous on Wednesday, December 2, 2009 at 8:27am.
Using the standard reduction potentials, calculate the equilibrium constant for each of the following reactions at 298 K.
10Br + 2MnO4 + 16H+ > 2Mn2+ + 8H20 + 5Br2
Work:
2Br > Br2 + 2e (This would be multiplied by 5 to balance out with the upper equation.)
MnO4 + 8H+ + 5e > Mn2+ +4H20 (This would be multiplied by 2 to balance out with the upper equation.)
The standard reduction potential for Br is +1.065.
The standard reduction potential for MnO4 is +1.51.

Chemistry  DrBob222P, Wednesday, December 2, 2009 at 11:27am
Look up that equation, nEF = lnK (I'm not at home and can't confirm that equation so check it out) but that is the one you want to use.

Chemistry  Anonymous, Wednesday, December 2, 2009 at 10:14pm
I looked back through my notes and found the following equations:
K = e^(delta G/RT)
Delta G = nFE
I know that E = standard reduction of the reduced  standard reduction of the oxidized
I've tried plugging everything I have in, but I'm not getting the right answer. I can't figure out what it is that I am doing wrong.
I have E = 1.51  1.065 and n = 10 
Chemistry  Anon, Friday, March 29, 2013 at 8:13pm
So I have that
1.51 V  1.065 V = .445 V = E
and that
ΔG =  10 mol * 96485 J / mol*V * .445 V
ΔG = 429358 J
so then plugging in:
K = e^((429358)/(8.314 J/mol*K * 298 K))
K = 1.82 * 10^75
The 'correct' answer appears to be 1.5 * 10^75 , but that's not far off, and my homework program accepted my answer as correct.