Why is the distance between plus or minus and the number different?

The reason is that \cee does a different scanning than usual and gets tricked by +, which does have a special meaning inside \ce or \cee, because it's used for reactions.

Indeed, if you try putting that part outside \cee (which it should be), the problem doesn't show.

Here's a better version.

\documentclass{article}

\usepackage[version=3]{mhchem}
\usepackage{siunitx}
\sisetup{retain-explicit-plus}

\begin{document}
\begin{align}
E&=-0.1\\
F&=+0.1
\end{align}

\begin{align}
  \cee{Fe^2+_{(aq)} + 2e- -> Fe_{(s)} &\ (reduction)}  & E^0 &= \SI{-0.41}{\V} \\
  \cee{Al_{(s)} -> Al^3+_{(aq)} + 3e- &\ (oxidation)} & E^0 &= \SI{+1.66}{\V} \\
  \cee{3Fe^2+_{(aq)} + 2Al_{(s)} -> 2Al^3+_{(aq)} + 3Fe_{(s)} & \ (overall)} &
    E_{\mathrm{cell}} &= \SI{+1.25}{\V}
\end{align}

\end{document}

+ and - have special meaning within \cee since it relates to charges of atoms/particles and is therefore treated differently. The - visually defaults to a unary minus while + defaults to a binary relation (for joining atoms in a reaction). In this instance one can force it to be unary using {+}, as in E^0 &= {+}1.66V.

Here is an alternative with some adjusted horizontal alignment:

\documentclass{article}

\usepackage[version=3]{mhchem}

\newcommand{\BRR}{\makebox{\phantom{\cee{2Al^3+_{(aq)} + 3Fe_{(s)}}}}}

\begin{document}

Adjusted alignments:
\begin{align}
  \cee{        Fe^2+_{(aq)} + 2e-} & \cee{ -> Fe_{(s)}                 } && \text{(reduction)} &             E^0 &= -0.41V \\
  \cee{                  Al_{(s)}} & \cee{ -> Al^3+_{(aq)} + 3e-       } && \text{(oxidation)} &             E^0 &= +1.66V \\
  \cee{ 3Fe^2+_{(aq)} + 2Al_{(s)}} & \cee{ -> 2Al^3+_{(aq)} + 3Fe_{(s)}} && \text{(overall)}   & E_{\text{cell}} &= +1.25V
\end{align}

Using \verb|\phantom|s:
\begin{align}
  \cee{ \llap{\cee{Fe^2+_{(aq)} + 2e-}} -> \rlap{\cee{Fe_{(s)}}} \BRR} & 
    \text{ (reduction)} &      E^0 &= -0.41V \\
  \cee{           \llap{\cee{Al_{(s)}}} -> \rlap{\cee{Al^3+_{(aq)} + 3e-}} \BRR} & 
    \text{ (oxidation)} &      E^0 &= +1.66V \\
  \cee{                     3Fe^2+_{(aq)} + 2Al_{(s)} -> 2Al^3+_{(aq)} + 3Fe_{(s)}} & 
    \text{ (overall)}   & E_{\text{cell}} &= +1.25V
\end{align}

\end{document}

Due to the right-left alignment of align, we push the reactions over by the width of the maximum of the right hand side (or 2Al^3+_{(aq)} + 3Fe_{(s)}).