In the following reaction, identify the speech oxidised, the species reduced, the oxidising agent and reducing agent.
(a). `4HCl+MnO_(2)=MnCl_(2)+2H_(2)O+Cl_(2)`
(b). `SnCl_(2)+2FeCl_(3)=SnCl_(4)+2FeCl_(2)`
(c ).`2H^(+)+Mg=Mg^(2+)+H_(2)`
(d).`H_(2)SO_(3)+2H_(2)S=3S+2H_(2)O`

Explanation

`(a).MnO_(2)+4HCl=MnCl_(2)+Cl_(2)+2H_(2)O`
Writing the oxidation number on various atoms,
`overset(+4 -2)(MnO_(2))+overset(+1 -1)(4HCl)=overset(+2 -1)(MnCl_(2))+overset(+0)(Cl_(2))+overset(+1 -2)(2H_(2)O)`
Ox. No of Mn changes from +4 to +2, i.e., decreases oxidation number. `MnO_(2)` is thus reduced. It acts as oxidising agent. Ox. no of Cl changes from -1 to 0, i.e. increases oxidation number. HCl is thus oxidised. It acts as a reduced agent.
(b). `SnCl_(2)+2FeCl_(3)=SnCl_(4)+2FeCl_(2)`
Writing the oxidation number on various atoms,
`overset(+2 -1)(SnCl_(2))+overset(+3 -1)(2FeCl_(3))=overset(+4 -1)(SnCl_(4))+overset(+2 -1)(2FeCl_(2))`
Ox, no of Sn changes from +2 to +4 i.e. increases in oxidation number of `SnCl_(2)` is thus oxidised or it acts as a reducing agent. Ox. no. of Fe changes from +3 to +2 i.e., decrease in oxidation number. `FeCl_(3)` os thus reduced or it acts as an oxidising agent.
(c ).`Mg+2H^(+) =Mg^(2+)+H_(2)`
Writing oxidation number of various atoms,

Mg is oxidised i.e. it acts as a reducing agent. `H^(+)` is reduced. i.e. it acts as an oxidising agent.
(d). `H_(2)SO_(3)+2H_(2)S=3S+3H_(2)O`
Writing oxidatoin number on various atoms,

`H_(2)SO_(3)` is reduced, i.e. it acts as oxidising agent. `H_(2)S` is oxidised i.e. it acts as reducing agent.