[Sheryl]

A cell is constructed at 25ºC.  One half-cell contains [Fe^3+]=1.00M and [Fe^2+]= 0.100M.  The other has [MnO4-]= 1.00x10^-2M and [Mn^2+] = 1.00x10^-4M in acidic solution where [H+] = 1.00x10^-3M.
Write a balanced redox equation for the reaction and calculate cell potential.

My answer for balance:
5[Fe^2+ --> Fe^3+(aq) + e-]    Eº= - 0.771V
MnO4-(aq) + 8H+(aq) +5e- --> Mn^2+(aq) + 4H2O(l)   Eº= 1.51V so balanced eq is:
5Fe^2+(aq) + MnO4-(aq) + 8H+(aq) --> 5Fe^3+(aq) + Mn^2+(aq) + 4H2O(l)

Cell potential:
Eºcell = 1.51 - (-0.771) = 2.281V
E = .0592/5 logQ    Q =  [1.00M]5 [1.00x10^-4M] / [0.100M]5 [1.00x10^-2] [1.00x10^-3]
E = .0592/ log 1x10^-6 =  -9.87 V

Am I on the right track with this?

Sheryl