Why is reduction potential of water ( 2H₂O + 2e⁻ → H₂ + 2OH⁻ ) is -0.83 insted of 0, as in the case of (2H⁺ → H₂)?
Note that the actual potential for a particular redox reaction is not a fixed value, but depends on concentrations ( more exactly activities ) of reagents. The standard redox potentials are potentials with activities equal to 1,

If we consider reactions

$$\begin{align}\ce{
2 H+ + 2e-    &amp;&lt;=&gt; H2 \\
2 H2O + 2 e-     &amp;&lt;=&gt;  2 OH- + H2 \\
}\end{align}$$

their standard redox potentials imply 
$\rm{p}H=0$ for the former, $\rm{p}H=14$ for the latter.

For the former reaction and $\rm{p}H=14$, $E=-0.83\ \rm{V}$. As for $\ce{H+/H2}$ potential, at $\pu{p_{\ce{H_2}} = 1 atm}$

$$E_\ce{H2\ /H+}  = E^{\circ}_\ce{H2\ /H+} + 0.059 \log [\ce{H+}] = E^{\circ} - 0.059\  \rm{p}H$$

Thermodynamically, the equations they are equivalent, but kinetically, the latter reaction has in strongly alkaline environment the advantage of many concentration orders. And vice versa.

Elements can have multiple standard redox potentials, as such a potential can be expressed for any redox reaction (even if for some only formally from the thermodynamic data).

For example, some standard redox potentials for $\ce{Ag}$ are:

$$
\begin{array}{lcll}
\ce{Ag+ + e- &amp;&lt;=&gt;&amp; Ag(s)};\ &amp; E^\circ_{red} = \pu{0.80 V}\\
\ce{AgCl(s) + e- &amp;&lt;=&gt;&amp; Ag(s) + Cl-(aq)};\ &amp; E^\circ_{red} =\pu{ 0.22 V}
\end{array}
$$

(There are more; see the Standard electrode potential data table).

The latter can be calculated from the former using
$$E_{\ce{Ag/Ag+}} = E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+}$$

For the $\ce{Ag / AgCl / Cl-}$  system, we have to include into potential calculations the $\ce{AgCl}$ solubility constant.

$$\begin{align}
K_{\rm s, AgCl} &amp;= c_{\ce{Ag+}}\cdot c_{\ce{Cl-}} \\[1em]
E_{\ce{Ag/AgCl/Cl-}} &amp;= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+} =\\
  &amp;= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} - 0.059 \log {c_\ce{Cl-}}= \\
  &amp;= E^{\circ}_{\ce{Ag/AgCl/Cl-}} - 0.059 \log {c_\ce{Cl-}} \\
E^{\circ}_{\ce{Ag/AgCl/Cl-}} &amp;= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} =\\
 &amp;= 0.80 + 0.059\log(1.77\cdot10^{−10}) = \\
 &amp; = 0.22
\end{align}$$

Note that the $\ce{Ag/AgCl/Cl-}$ electrode is often used as the secondary standard of the electrode potential, aside of the calomel electrode $\ce{Hg/Hg2Cl2/Cl-}$, because maintaining of the primary potential standard electrode $\ce{Pt-H2/H+}$ is not convenient. The $\ce{Ag/AgCl/Cl-}$ electrode is used also as the electrode to measure $c_{\ce{Cl-}}$.