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

Question

In the align below the last element of the first row is slightly off compared to the next rows. It looks like the root cause is mostly the distance between minus and zero, but also between the equal sign and minus. Why are they so different?

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

Answer 1

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}

Answer 2

+ 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)}).