contradiction between Gibbs energy variation and redox potential
Looking up the redox potentials in tables vs. Hydrogen as standard at zero volts the redox for Oxygen is 

$$\ce{O2 +  4H^+ + 4e &lt;=&gt; 2H2O} \qquad \qquad E^\circ = +1.229~\mathrm{V}$$

so the tables you use may be vs. calomel electrode as standard.

The redox potential for the nitrate reaction has to be constructed from others as 

$$\begin{align}
\ce{NO3^- + 3H^+ + 4e}\, &amp;\ce{&lt;=&gt; HNO2 + H2O} &amp; E_1^\circ &amp;= +0.934~\mathrm{V} \\
\ce{2HNO2 + 4H^+ + 4e}\, &amp;\ce{&lt;=&gt; N2O + 3H2O} &amp; E_2^\circ &amp;= +1.297~\mathrm{V} \\
\ce{N2O + 2H^+ + 2e}\, &amp;\ce{&lt;=&gt; N2 + H2O} &amp; E_3^\circ &amp;= +1.766~\mathrm{V}
\end{align}$$

The overall half-reaction, after multiplying the nitrate reaction by 2 (to balance) and adding is

$$\ce{2NO_3^- + 12H^+ + 10 e &lt;=&gt; N2 + 6H2O} \tag{1}$$

The redox potential for the reaction $(1)$ is calculated from $n_\mathrm{r}E_\mathrm{r}^\circ = n_1E_1^\circ + n_2E_2^\circ +....$ where the n's are the number of electrons in each half reaction and $n_\mathrm{r}$ for the new reaction. 

The $\ce{NO_3^- to N2}$ redox potential is therefore $1.246~\mathrm{V}$, calculated as

$$E_\mathrm{r}^\circ = \frac{(4 \times 0.934) + (4 \times 1.297) + (2 \times 1.766)}{10}$$ 

The two values, $1.229~\mathrm{V}$ and $1.246~\mathrm{V}$, would now seem to agree with your $\Delta G$ values. They may be the same within error as may your $\Delta G$ values.

(Redox Values are from the CRC Handbook of Chemistry and Physics, 75 ed.)