What is the stoichiometric coefficient of $\mathrm{SO}_2$ in the following balance reaction?…
What is the stoichiometric coefficient of $\mathrm{SO}_2$ in the following balance reaction?
$\mathrm{MnO}_4^{-}(a q)+\mathrm{SO}_2(g) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{HSO}_4^{-}(a q)$
(in acidic solution)
5
4
3
2
Solution
Oxidation half reaction:
$\mathrm{SO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{HSO}_4^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-}...(i)$
Reduction half reaction:
$\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_2 \mathrm{O}(l)...(ii)$
Multiply eq. (i) by 5 and (ii) by 2 and then add
$\begin{aligned}
2 \mathrm{MnO}_4^{-}(a q)+5 \mathrm{SO}_2(g) & +2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}^{+}(a q) \\
& \longrightarrow 2 \mathrm{Mn}^{+}(a q)+5 \mathrm{HSO}_4^{-}(a q)
\end{aligned}$
Thus, the stoichiometric coefficient of $\mathrm{SO}_2$ in the above balance equation is 5 .