Electrolysis in aqueous solution – which equations to use to predict product at each electrode?
The given potentials $E = 0.00{\text{ V}}$ and $E = -0.83{\text{ V}}$ for the reduction of water apply to different $\text{pH}$.

The thermodynamic relation of the potential $E$ to the composition of the solution is generally known as the Nernst equation:

$$E = E^\circ  + \frac{{0.059{\text{ V}}}}{z}\log \frac{{{{\prod\nolimits_i {\left[ {{\text{ox}}} \right]} }^{{n_i}}}}}{{{{\prod\nolimits_j {\left[ {{\text{red}}} \right]} }^{{n_j}}}}}$$

For the given reaction

$$\ce{2 H+ + 2e- &lt;=&gt; H2}$$

the Nernst equation reads

$$E = 0.00{\text{ V}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{{{{\left[ {{{\text{H}}^ + }} \right]}^2}}}{{\left[{{{\text{H}}_2}}\right]}}$$

Since

$${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$$

the potential $E$ depends on $\text{pH}$:

$$E = 0.00{\text{ V}} - 0.059{\text{ V}} \times {\text{pH}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{1}{{\left[ {{{\text{H}}_2}} \right]}}$$

Therefore, the potential is $E = 0.00{\text{ V}}$ at $\text{pH} = 0$ and $E = -0.83{\text{ V}}$ at $\text{pH} = 14$ respectively.