Can the thermodynamic predictions of redox reactions based on E and dG contradict each other? I am referring to the following three reactions. Two of these reactions are imaginary ($\ce{M}$ and $\ce{X}$ are imaginary).

$$\begin{alignat}{3}
\ce{X- + 2e- \;&amp;&lt;=&gt; X^3-} \qquad &amp;&amp;E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &amp;&amp;&amp;\text{(a)}\\
\ce{2H+ + MO^2+ + e- \;&amp;&lt;=&gt; M^3+ + H2O} \qquad &amp;&amp;E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &amp;&amp;&amp;\text{(b)}\\ 
\ce{8H+ + MnO4- + 5e- \;&amp;&lt;=&gt; Mn^2+ + 4H2O} \qquad &amp;&amp;E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &amp;&amp;&amp;\text{(c)}
\end{alignat}$$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations $\text{(a)}$ and $\text{(c)}$;

$$
\begin{align}
&amp;{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\
\ce{&amp;16H+ +5X^3- + 2MnO4- -&gt; 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\
&amp;E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\
&amp;\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F
\end{align}
$$

Equations $\text{(b)}$ and $\text{(c)}$;

$$
\begin{align}
&amp;{-}\text{(b)} \times 5 + \text{(c)}: \\
\ce{&amp;5M^3+ + H2O + MnO4- -&gt; Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\ 
&amp;E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\
&amp;\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F
\end{align}$$

My questions:


Are there any errors in above calculations? If so, please suggest corrections.
According to $E^\circ$ values, reaction $(2)$ is more feasible than reaction $(1)$. Am I correct?
According to $\Delta G^\circ$ values, reaction $(1)$ is more feasible than reaction $(2)$. Am I correct?
Is it useless to use $E$ values to predict the feasibilities of redox reactions? Should we always resort to $\Delta G$?
Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict?
If not how can these results be explained?
$\Delta G = -n {\mathcal F} \Delta {\mathcal E} $

$\mathcal F$ is Faraday's constant, which physicists currently believe has been constant throughout the universe for all time.  $n$ is the number of electrons transfered in an electrochemical reaction.  For a given reaction, $n$ is also constant in all places and times.

Thus, $\Delta G$ determines $\Delta {\mathcal E}$ and vice versa.  


  Are there any errors in above calculations? 


Yes, there must be.


  According to E∘ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  According to ΔG∘ values, reaction (1) is more feasible than reaction (2). Am I correct ?


Both of these cannot be correct.


  Is it useless to use E values to predict the feasibilities of redox reactions ? 


No, it is just as good and exactly as informative as $\Delta G$.


  Should we always resort to ΔG ?


No.  Feel free to resort to $\Delta G$ if you like units like kJ/mol instead of volts, but other than personal preference there is no reason to prefer one over another.


  Is it possible for predictions based on E values and ΔG values to contradict ?


No. 


  If not how can these results be explained ?


UPDATE: Arithmetic error.

The choice of $n$ is arbitrary.  Once you pick $n$ for a given half reaction, it determines the choice of $n$ for the other half reaction, but that's it.  If you then consider a separate cell with separate reactions, you could write that new cell's equation with an arbitrary $n$.  The $\textbf{mol}$ in $\Delta G$'s units of $\frac{\textrm{ kJ}}{\textbf {mol}}$ is $\textbf{mol}$ of reaction, and thus the value of $\Delta G$ depends on how you write the reaction.  For example you could write equation (b) + (c) as 

$$\ce{10M^3+ + 2 H2O + 2 MnO4- -&gt; 2 Mn^2+ + 10 MO^2+ +4 H+}$$

Now, $n$ is twice as large, so $\Delta G$ is twice as large.  But $\Delta \mathcal E$ is unchanged.  This is because $\Delta \mathcal E$ has a basis of "$\textbf{mol}$" of electrons, not of moles of a reaction of arbitrary scale.  In reality the "$\textbf{mol}$" of electrons is converted to a different units (Coulombs) to make things more confusing.  (Remember, a volt is a joule per coulomb.)

So contrary to what I said above, the sign of $\Delta G$ but not the magnitude of $\Delta G$ is fixed, meaning that whether a reaction is absolutely spontaneous can determined by either $\Delta G$ or $\Delta \mathcal E$, but to compare the relative spontaneity of two reactions, either the $n$ values of the two reactions must be the same and then $\Delta G$ values can be used, or else without worrying about $n$ you could just look at $\Delta \mathcal E$.