Question

Balance the following equations by ion electron method.
$H_{2}S+HN{O}_3\to H_{2}S{O}_4+N{O}_2+H_{2}O$.

Answer 1

$H_{2}S+HN{O}_3\to H_{2}S{O}_4+N{O}_2+H_{2}O$
Ionic equation,
$H_{2}S+N{O}_3^{-}\to N{O}_2+H_{2}O$
$1$st step: Splitting into two half reactions,
$\underset{\left(Oxidationhalfreaction\right)}{H_{2}S\to S{O}_4^{2-}}$; $\underset{\left(Reductionhalfreaction\right)}{N{O}_3^{-}\to N{O}_2+H_{2}O}$
$2$nd step: Add water to the side deficient in oxygen,
$H_{2}S+4H_{2}O\to S{O}_4^{2-}+10H^{+}$
$3$rd step: Add $H^{+}$ ions to the side deficient in hydrogen,
$N{O}_3^{-}+2H^{+}\to N{O}_2+H_{2}O$
$4$th step: Add electrons to the side deficient in electrons,
$H_{2}S+4H_{2}O\to S{O}_4^{2-}+10H^{+}+8e$
$N{O}_3^{-}+2H^{+}+e\to N{O}_2+H_{2}O$
$5$th step: Balancing electrons in both the half reactions,
$H_{2}S+4H_{2}O\to S{O}_4^{2-}+10H^{+}+8e$
$[N{O}_3^{-}+2H^{+}+e\to N{O}_2+H_{2}O]\times 8$
$6$th step: Adding both the half reactions,
$H_{2}S+4H_{2}O+8N{O}_3^{-}+6H^{+}\to S{O}_4^{2-}+8N{O}_2+8H_{2}O$
or $H_{2}S+8N{O}_3^{-}+6H^{+}\to S{O}_4^{2-}+8N{O}_2+4H_{2}O$
Converting it into molecular form,
$H_{2}S+8HN{O}_3\to H_{2}S{O}_4+8N{O}_2+4H_{2}O$.