I want to figure out a way how to determine the products of a redox-reaction by looking at the documented Standard electrode potential (for example here). But with my "way" I have some trouble/contradictions with common redox reaction, in my case the reduction of $\ce{MnO4-}.
Here is my procedure for the reduction of $\ce{MnO4-}$:
Collect all available reactions with their Standard Potential:* \begin{align} \ce{MnO4- + 2H2O + 4e- &-> MnO2 + 4OH-}& E(\text{Red})&= \pu{0.83 V}\tag{a}\label{a}\\ \ce{MnO4- + H+ + e- &-> HMnO4^-} & E(\text{Red}) &= \pu{0.72 V}\tag{b}\label{b}\\ \ce{MnO4- + 8H+ + 5e- &-> Mn^2+}& E(\text{Red}) &= \pu{1.23 V}\tag{c}\label{c}\\ \ce{MnO4- + 4H+ + 3e- &-> MnO2 + 2H2O}& E(\text{Red}) &=\pu{1.6 V}\tag{d}\label{d} \end{align}
Then figure out the reaction with the highest reduction potential, because according to $E = E(\text{Red}) - E(\text{Ox})$, $E$ is maximal with the biggest $E(\text{Red})$ and the reaction is going to happen with the highest $E$.
In my example, $\eqref{d}$ has the highest potential, so according to my theory $\ce{MnO2}$ is formed in the reduction of $\ce{MnO4-}$ (in combination with for example the oxidation of $\ce{I-}$. But in all textbooks, internet sites, the reduction $\eqref{c}$ is named - not reaction $\eqref{d}$ which I had determined.
I am aware that these potentials are pH- and concentration dependent, I fit the $E(\text{Red})$-data with the Nernst-Equation to $\mathrm{pH} = 3$. Nevertheless, $E(\text{Red})$ of $\eqref{d}$ is always higher in $E(\text{Red})$ than $\eqref{c}$. According to my calculations (graph of $\eqref{d}$) which is dependent on the $\mathrm{pH}$ is always higher than $\eqref{c}$).
I have drawn a diagram which shows the $E(\text{Red})$ potential of the two reaction $\eqref{c}$ and $\eqref{d}$ (see legend). Reaction $\eqref{d}$ has always the higher standard potential - is the better oxidation reaction. Why does nevertheless reaction $\eqref{d}$ happen in acid solutions?
Functions calculated with Nernst-Equation: \begin{align} E(\text{pH}) &= 1.507 - \frac{0.059}{5} \cdot 8 \text{pH}\tag{c'}\\ E(\text{pH}) &= 1.7 - \frac{0.059}{3} \cdot 4 \text{pH}\tag{d'} \end{align} (other concentrations are assumed as equal)