Will nitrate be able to oxidise mercury? We have the following data:

\begin{align}
\ce{NO3- + 3 e- &amp;-&gt; NO} &amp;\quad E^\circ_1 &amp;= \pu{0.96 V}\\
\ce{Hg^2+ + 2 e- &amp;-&gt; Hg} &amp;\quad E^\circ_2 &amp;= \pu{0.86 V}
\end{align}

By the reasoning given in my book, nitrate oxidises mercury:


  ... of two substances, the one whose reduction $E^\circ$ value is greater, will be able to oxidise the other substance the other.


Then I tried calculating $E^\circ$ for overall reaction also taking into account number of electrons in each reaction. Or that we calculate $\Delta G$ for 

$$\ce{NO3^- + Hg -&gt; NO + Hg^2+}$$

$$\Delta G^\circ = -0.96 \cdot 2F + 0.86 \cdot 3  = 0.66F$$

So, $\Delta G^\circ$ is positive and so nitrate does not oxidise mercury. Which is correct?
It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

\begin{align}
\ce{NO3− (aq) + 2 H+ + e− &amp;&lt;=&gt;  NO2 (g) + H2O} &amp;\quad &amp;E_0 = \pu{+0.80 V} \\
\ce{Hg2^2+ + 2 e− &amp;&lt;=&gt; 2Hg (l)} &amp;\quad &amp;E_0 = \pu{+0.80 V} \\
\ce{Hg^2+ + 2 e− &amp;&lt;=&gt; Hg (l)} &amp;\quad &amp;E_0 = \pu{+0.85 V} \\
\ce{2 Hg^2+ + 2 e− &amp;&lt;=&gt; Hg2^2+} &amp;\quad &amp;E_0 = \pu{+0.91 V} \\
\ce{NO3−(aq) + 4 H+ + 3 e− &amp;&lt;=&gt; NO (g) + 2 H2O (l)} &amp;\quad &amp;E_0 = \pu{+0.958 V} \\
\end{align}

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.