This oxidation-reduction reaction in acidic solution is spontaneous: 5 F e 2 + (

Question

This oxidation-reduction reaction in acidic solution is spontaneous:
5Fe2+(aq)+MnO?4(aq)+8H+(aq)?5Fe3+(aq)+Mn2+(aq)+4H2O(l).
A solution containing KMnO4 and H2SO4 is poured into one beaker, and a solution of FeSO4 is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and the two solutions are connected by a wire that passes through a voltmeter.

Part A - Calculate the emf of the cell under standard conditions.

-The answer for this is .74V

Part B - Calculate the emf of the cell at 298

K when the concentrations are the following: pH=0.80, [Fe2+]=0.10M , [MnO?4]=1.40M , [Fe3+]=3.0

Explanation / Answer

The complete reaction is: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

I looked up the Eo values for the two half-reactions:

Fe2+ ==> Fe3+ + e- . . . . . . . . . . . .Eo = -0.77 V
MnO4- + 8H+ ==> Mn2+ + 4H2O . . .Eo = +1.51 V
======================================
E cell = -0.77 V + 1.51 V = +0.74 V

Now use the Nernst equation to calculate the Eo under nonstandard conditions.
If pH = 0.9, [H+] = 10^-0.9 = 0.13 M

Q cell = [Fe3+]^5 [Mn2+] / [Fe2+]^5 [MnO4-][H+]^8 = (2.4 x 10^-4)^5 (2 x 10^-3) / (0.12)^5 (1.50)(0.13)^8
= 5.2 x 10^-10

E cell = Eo - 0.059/n log Q = 0.74 - 0.059/5 log 5.2 x 10^-10 = 0.74 - (0.0295)(-9.3) = 1.01 V

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