= Solution
Oxidize <zinc> at the <anode> and reduce nitrate to nitric oxide at the inert <platinum> <cathode>. Multiplying the oxidation <half-reaction> by three and the reduction <half-reaction> by two cancels six <electrons>:
$$
\boxed{\mathrm{3Zn(s)+2NO_3^-(aq)+8H^+(aq)\longrightarrow3Zn^{2+}(aq)+2NO(g)+4H_2O(l)}.}
$$
The <standard cell potential> is the cathode reduction potential minus the anode reduction potential:
$$
\boxed{E^\circ=0.957-(-0.762)=1.719\,\mathrm V.}
$$
\b[The written direction is spontaneous under standard conditions], since $\Delta_rG^\circ=-6FE^\circ<0$. Multiplying a <half-reaction> changes its extensive <Gibbs energy>, but does not multiply its intensive <electrode potential>.
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