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.
3 answers
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.
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
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
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.
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.