The standard reduction potentials for this problem are:
Ag1+/Ag = 0.800 V vs NHE
MnO41-/Mn2+ = 1.49 V vs NHE
Consider the following electrochemical reaction at 25 oC:

8H1+ (1 M) + MnO41- (0.910 M) + 5Ag (s) Mn2+ (0.070 M) + 5Ag1+ (0.040 M) + 4H2O (l)

a. What is the Eo for this reaction?
o = correct check mark V (0.69)

b. What is the cell potential (Nernstian potential) under the conditions given?
cell = correct check mark V (0.7859)

c. What is the cell potential when 1/2 of the permanganate ion is consumed?
cell = wrong check mark V

Why the hell can't I get "c" right? Exactly as in the other problems, I halved the 0.910M of the permanganate ion and substituted the resulting new Q value into my equation, yet I keep getting some obviously wrong number.

I only need a hint at what, if anything, I'm doing wrong...not the answer.

[Edited on June 25, 2006 at 9:30 PM. Reason : I thought I had it at .7824, but no]

6/25/2006 9:17:48 PM

ok, past due, but i'd still like to know the answer to this one

6/26/2006 12:07:13 AM

that question is the only one I missed on that homework as well, couldn't figure it out.

I am so screwed for the final tomorrow

6/26/2006 8:17:35 PM

I'm just gonna make sure I can do that sample exam inside and out, and maybe pray to something

6/26/2006 8:49:23 PM

I'd recommend a reaction table. However, this problem seems to set up an impossible scenario. If 0.455 M permanganate is consumed, then that means 3.64 M H^1+ is consumed. The problem states that you only start out with 1M H^1+, so it's impossible to use more than you've got. Is it buffered at pH = 0 (H^1+ = 1M)? If so, then

.............8H^1+(aq) + MnO4^1-(aq) + 5Ag (s) --> Mn^2+(aq) + 5Ag^1+ (aq) + 4H2O (l)
initial.......1M.buffer.......0.910...............solid...........0.070.............0.040.............liquid
delta......-3.64.............-0.455.............-some........+0.455............+2.275............-some
_____________________________________________________________________________
final .......1M buffer........0.455.............less.............0.525.............2.315..............more

Then use the old Nerst equation:
Ecell = E^ocell - 0.0592/n*log Q

You've got E^0cell correct at 0.69 V
n = 5
Q is products/reactant for chemicals that are aqueous and gas
Q = [Mn^2+][Ag^1+]^5 /[MnO4^1-][H^1+]8 = (0.525)*(2.315)^5 / (0.455)*(1)^8

I don't have my scientific calculator on me right now, so I can't punch it in. I'm curious what the correct answer is, as the question doesn't make sense unless the solution is buffered at 1M H^1+. That or your MnO4^1- is 0.091, not 0.91.

Good luck tomorrow.

6/26/2006 9:44:19 PM