Question

(i) Calculate oxidation number of $Cl$ in $C{l}_2{O}_7$.
(ii) $2FeC{l}_3+H_{2}S\to 2FeC{l}_2+2HCl+S$
Write the name of oxidising agent and reducing agent in above reaction:

Answer 1

(i) Let X be the oxidation number of $Cl$ in $C{l}_2{O}_7$.
The oxidation number of O is $-2$
$2X+7(-2)=0$
$2X-14=0$
$X=+7$
The oxidation number of $Cl$ in $C{l}_2{O}_7$ is $+7$.

(ii) $2FeC{l}_3+H_{2}S\to 2FeC{l}_2+2HCl+S$
The oxidation number of iron decreases from $+3$ (in $FeC{l}_3$) to $+2$ (in $FeC{l}_2$).
Hence, iron is reduced and $FeC{l}_3$ is the oxidising agent.
The oxidation number of sulphur increases from $-2$ (in $H_{2}S$) to 0 (in S).
Hence, sulphur is oxidised and $H_{2}S$ is the reducing agent.