Question

Balance the following equations by ion electron method.
$Cu+HN{O}_3\to Cu(N{O}_3{)}_2+NO+H_{2}O$.

Answer 1

$Cu+HN{O}_3\to Cu(N{O}_3{)}_2+NO+H_{2}O$
Ionic equation,
$Cu+H^{+}+N{O}_3^{-}\to C{u}^{2+}+NO+H_{2}O$
$1$st step: Splitting into two half reactions,
$\underset{\left(Oxidationhalfreaction\right)}{Cu\to C{u}^{2+}};$ $\underset{\left(Reductionhalfreaction\right)}{N{O}_3^{-}+H^{+}\to NO+H_{2}O}$
$2$nd step: Adding $H^{+}$ ions to the side deficient in hydrogen,
$Cu\to C{u}^{2+};N{O}_3^{-}+4H^{+}\to NO+2H_O$
$3$rd step: Adding electrons to the side deficient in electrons,
$Cu\to C{u}^{2+}+2e$; $N{O}_3^{-}+4H^{+}+3e\to NO+2H_{2}O$
$4$th step: Balancing electrons in both half reactions,
$3Cu\to 3C{u}^{2+}+6e$ $2N{O}_3^{-}+8H^{+}+6e\to 2NO+4H_{2}O$
$5$th step: Adding both the half reactions,
$3Cu+2N{O}_3^{-}+8H^{+}\to 3C{u}^{2+}+2NO+4H_{2}O$
Converting it into molecular form,
$3Cu+2N{O}_3^{-}+8H^{+}+6N{O}_3^{-}\to 3C{u}^{2+}+6N{O}_3^{-}+2NO+4H_{2}O$
or
$3Cu+8HN{O}_3\to 3Cu(N{O}_3{)}_2+2NO+4H_{2}O$.