Passage - VI :
Industrially sulphuric acid is produced by the following steps:
Stage I : `S+O_(2) overset(Delta)to SO_(2)`
Stage II : `2SO_(2) +O_(2) overset(V_(2)O_(5))to 2SO_(3)`
Stage III : `SO_(3)+H_(2)O to H_(2)SO_(4)`
Since the reaction between `SO_(3) and H_(2)O` is violent,
`SO_(3)` is passed into `98% H_(2)SO_(4)` to produce oleum `(H_(2)S_(2)O_(7))`
Ionically the reaction between `H_(2)SO_(4)` & Mg may be presented as,
`Mg+2H_(3)O^(+) to Mg^(2+)+H_(2) uarr +2H_(2)O`
Therefore, in the given reaction -

Options

A. oxidising agent is `H^(+)(H_(2)O)`
B. reducing agent is `H^(+)(H_(2)O)`
C. oxidising agent is `SO_(4)^(2-)`
D. oxidising agent are `SO_(4)^(2-)` & `H_(3)O^(+)` both

Answer

A

Explanation

`H_(2)SO_(4)` act as oxidizing agent.
`H_(2)SO_(4)+2PCl_(5) to underset((x))(SO_(2)Cl_(2))+2POCl_(3)+2HCl`