Nernst equation and equilibrium I am currently stuyding electrochemistry, and recently I've stumbled upon a problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate, but I obtained an unrealistic result and I became confused.

What I first did was write the half-reactions involved within the voltaic cell:
Anode (-): 5Fe2+(aq) ⇌ 5Fe3+(aq) + 5e- ε01= -0.77 V
Cathode (+): MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l) ε02= 1.5 V
Notice the fact that I obtained the first reaction, the oxidation, by reversing the equation for the reduction of Fe3+.
By combining both half-cell reactions, one can obtain the equation for the whole process:
5Fe2+(aq) + MnO4-(aq) +  8H+(aq) ⇌ 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) E0= 0.73 V

A positive standard potential will yield a negative value for the Gibbs free energy, so the reaction is product-favoured. The problem is concerned with calculating the concentration at equillibrium, corresponding to the equivalence point of our titration, so the overall Gibbs free energy will be 0, just as the non-standard potential, which we can calculate using the Nernst equation (considering 298.15 K of temperature and 1 atm of pressure). I'm also assuming that the reaction quotient is equivalent to the equilibrium constant, since we are at the equivalence point.

\begin{align*}
    E=E^{0}-\frac{0.0592}{5}\log{K}
\end{align*}
\begin{align*}
    E=0, E^{0}=\frac{0.0592}{5}\log{K} 
\end{align*}
I then obtained the mathematical expression for the equilibrium constant:
\begin{align*}
    K=10^{\frac{5E^{0}}{0.0592}}
\end{align*}
I then entered the values, and calculated K (substantially larger than 1):
\begin{align*}
    K=4.523*10^{61}
\end{align*}
However, this is the hurdle I have yet to overcome. I didn't quite understand what activities are in the Nernst equation (afaik they are not equilibrium concentrations), so I assigned a formula for K based on the equilibrium reaction:

\begin{align*}
    K=\frac{[Fe^{3+}]^{5}[Mn^{2+}]}{[Fe^{2+}]^{5}[MnO4^{-}][H^{+}]^{8}}
\end{align*}

The problem mentions the following concentrations for the titration:
Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
H2SO4: 1N (probably 0.1M for H+)
KMnO4: 1N (0.2M MnO4- following the same steps)

In my conception, the activities of the reagents in the equilibrium constant were their initial concentrations, and the activities of the products were their equillibrium concentrations. I then made an ice table, considering a concentration of x. By replacing those values into the equilibrium constant, I obtained:
\begin{align*}
    K=\frac{5x^{5}*x}{0.02^{5}*0.2*0.1^{8}}
\end{align*}
X was supposed to be the concentration of Fe^{2+}, however the answer I got is definitely not the right one.
\begin{align*}
    x=6726501.235
\end{align*}

Where did I go wrong?
problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate, 


There are several things which may be problematic in your approach.


For the stated problem you do not need to invoke Nernst equation. Simply use the balanced equation. 
When potentiometric titration curve are to be estimated, it is far more convenient would use the half-cell instead of a full reaction. I am assuming the redox electrode is dipping in iron solution.


\begin{align*}
    E=E^{0} -\frac{0.0592}{5}\log{[Fe^{2+}]/[Fe^{3+}]}
\end{align*}

where $E^{0}$ is the iron half cell value. You should not calculate equilibrium constants from potentiometric titrations. 

See page 17 for a full example Potentometric Titration clculation


I do not know when people will stop teaching normality. It is obsolete (18th century concept). Your normality conversions are incorrect.



  The problem mentions the following concentrations for the titration:
  Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
  H2SO4: 1N (probably 0.1M for H+)
  KMnO4: 1N (0.2M MnO4- following the same steps)


1 N sulfuric acid will be half the molarity =0.5 M

1 N permanganate will be 1/5th the molarity = 1/5 M

Check the concept of normality in your textbook.