Question

Balance the following reaction
$Cu+HN{O}_3\to Cu(N{O}_3{)}_2+NO+H_{2}O$

Answer 1

Now,

$Cu\to C{u}^{2+}+2e^{-}⟶\left(1\right)$

${N{O}_3}^{-}+4H^{+}+3e^{-}⟶NO+2H_{2}O⟶\left(2\right)$

Equation (1) $\times 3+$ Equation (2) $\times 2$ (for equalising electrons)

$3Cu⟶3C{u}^{2+}+6e^{-}$

${2N{O}_3}^{-}+8H^{+}+6e^{-}⟶2NO+4H_{2}O$

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$3Cu+{2N{O}_3}^{-}+8H^{+}⟶{3Cu}^{2+}+2NO+4H_{2}O$

But $8H^{+}+{8N{O}_3}^{-}⟶8HN{O}_3$

The $6$ ${N{O}_3}^{2-}$ combine with $C{u}^{2+}$

$\therefore$ Balanced equation is,

$3Cu+8HN{O}_3⟶3Cu(N{O}_3{)}_2+2NO+4H_{2}O$