Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Explanation

1) In the reaction between `HgCl_(2)andSnCl_(2),HgCl_(2)` act as oxidising agent and `SnCl_(2)` act as reducing agent. If `SnCl_(2)` is excess the product Hg in its lower oxidation state. But if `HgCl_(2)` is excess the product is `Hg_(2)Cl_(2)`
`HgCl_(2)+underset("excess")(SnCl_(2))toHg+SnCl_(4)`
`underset("excess")(2HgCl_(2))+SnCl_(2)toHg_(2)Cl_(2)+SnCl_(4)`
2) In the reaction between phosphorous and chlorine phosphorous is reducing agent and chlorine is oxidising agent. If chlorine is in small amount the product is `PCl_(3)` but in the presence of excess chlorine `PCl_(5)` is the product.
`P_(4)+6Cl_(2)to4PCl_(3)`
`P_(4)+10Cl_(2)to4PCl_(5)`
3) When chlorine is passed into excess of liquid sulphate the product is sulphur monochloride `S_(2)Cl_(2)`. But if excess chlorine is passed until it is saturated, the product is `SCl_(2)`
`underset("excess")(2S)+Cl_(2)toS_(2)Cl_(2)`
`S+underset("excess")(Cl_(2))toSCl_(2)`