Question

Balance the following equation by ion electron method.
$FeC{l}_3+H_{2}S\to FeC{l}_2+HCl+S$.

Answer 1

$FeC{l}_3+H_{2}S\to FeC{l}_2+HCl+S$
Ionic equation, $F{e}^{3+}+H_{2}S\to F{e}^{2+}+H^{+}+S$
$1$st step: Splitting the redox reaction into two half reactions,
$\underset{\left(Oxidationhalfreaction\right)}{H_{2}S\to 2H^{+}+S}$ $\underset{\left(Reductionhalfreaction\right)}{F{e}^{3+}\to F{e}^{2+}}$
$2$nd step: Adding electrons to the side deficient in electrons,
$H_{2}S\to 2H^{+}+S+2e$
$F{e}^{3+}+e\to F{e}^{2+}$
$3$rd step: Balancing electrons in both the half reactions,
$H_{2}S\to 2H^{+}+S+2e$
$2F{e}^{3+}+2e\to 2F{e}^{2+}$
$4$th step: Adding both the half reactions,
$H_{2}S+2F{e}^{3+}\to 2H^{+}+S+2F{e}^{2+}$
Converting it into molecular form,
$H_{2}S+2F{e}^{3+}+6C{l}^{-}\to 2H^{+}+2C{l}^{-}+S+2F{e}^{2+}+4C{l}^{-}$
or $H_{2}S+2FeC{l}_3\to 2HCl+S+2FeC{l}_2$.