Give one chemical reaction to show that :
(i) `Sn(II)` is a reducing agent whereas `Pb(II)` is not.
(ii) `Sn(II)` chloride is a reducing agent.

Explanation

(i) Both tin and lead show two oxidation state of `+2` and `+4` due to inrt pair effect . But the inert pair effect. But the inert pair effect is more prominaent in case of Pb than in Sn. In other words +2, oxidation state of Sn is less stable than its +4 oxidation state. Therefore, `Sn(II)` acts as a reducing agent and gets cor. erted into more stable `Sn (IV)` by losing two electrons. For example it reduces `Fe^(3+)` to `Fe^(2+)` ions.
`2Fe^(+) + Sn^(2+) rarr 2Fe^(2+) + Sn^(4+)`
In contrast`+2` oxidation state of Pb is more stable than its +4 oxidation state. In other words, `Pb(II)` does not lose electrons easily and hence does not acts as a redusing agent.
(ii) Being a reducing agent, tin (II) chloride reeduces mercuric chloride `(HgCl_(2))` to mercurous chloride `(Hg_(2)Cl_(2))` and ferric salt to ferrous salts.
`SnCl_(2) + 2HgCl_(2) rarr SnCl_(4) + Hg_(2)Cl_(2)`
`SnCl_(2) + 2FeCl_(3) rarr SnCl_(4) + 2 FeCl_(2)`