Why can we disregard stoichiometric coefficients when computing reduction potentials? A $\ce{Cr(s) | Cr^3+ || Cd^2+ | Cd(s)}$ galvanic cell has redox half reactions:
$$
\begin{align}
\ce{Cd^2+ + 2e- &amp;-&gt; Cd(s)} &amp;\quad E^\circ &amp;= \pu{-0.4 V}\tag{R1}\\
\ce{Cr^3+ + 3e- &amp;-&gt; Cr(s)} &amp;\quad E^\circ &amp;= \pu{-0.7 V}\tag{R2}
\end{align}
$$
Balanced reaction equation then is
$$\ce{3 Cd^2+ + 2 Cr(s) -&gt; 2 Cr^3+ + 3 Cd(s)}\tag{R3}$$
$$E_\mathrm{cell} = E_\mathrm{red} + E_\mathrm{ox} = \pu{-0.4 V} + \pu{0.7 V} = \pu{0.3 V}$$
Why is $E_\mathrm{cell}^\circ$ independent of molar coefficients $(3,2)?$
You have to be careful of the difference between $E^\circ$ and $E$. $E^\circ$ is the standard electrode potential at defined conditions, some of which are mentioned in GODPARTICLE's answer. Standard thermodynamic conditions are usually


A temperature of 298 K
A pressure of gaseous components of 1 atm (or 1 bar)
A concentration of 1 M


The reference reduction potentials you quote are all standard potentials. These are the contributions to the cell potential when the half-cells meet the criteria above. 

$E$ is the cell potential at all other conditions. The correction of $E^\circ$ to $E$ involves temperature, pressure, concentration, and stoichiometry. Behold the Nernst equation:

$$E=E^\circ - \dfrac{RT}{nF}\ln{Q}$$


$R$ is the ideal gas constant (this is thermo after all)
$n$ is the number of electrons transferred - here is the stoichiometry!
$F$ is the Faraday constant 
$Q$ is the reaction quotient - which accounts for changes in pressure and concentration