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April 18, 2015

April 18, 2015

Posted by **Anonymous** on Wednesday, December 2, 2009 at 8:27am.

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 11:27amLook 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 10:14pmI 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 8:13pmSo 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.

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