Question

Give one chemical reaction to show that:
(i) Tin (II) is a reducing agent whereas Pb (II) is not.
(ii) Tin (II) reduces mercuric salt to mercurous salt.

Answer 1

(i) Both +2 and +4 oxidation states are adopted by the element tin. The same is true for Lead as well. But there is a difference between tin and lead regarding the context of oxidation states.

Tin (IV) oxidation state is more stable.  Hence, the conversion of tin (II) compounds to tin (IV) compounds is very easy as two electrons are lost easily by the tin(II). Hence, tin is a reducing agent.

(ii) This is best illustrated in the below chemical reaction where tin (II) ions reduces iron (III) to iron(II) ions. During this process, the tin (II) ions (less stable ions) are oxidized to the tin(IV) ions (more stable ions) .

$2{ Fe }^{ 3+ }+{ Sn }^{ 2+ }\rightarrow 2{ Fe }^{ 2+ }+{ Sn }^{ 4+ }$

(iii) But it is reverse in the case of Lead because the lead (II) is more stable than lead (IV). Thus, Lead will not be losing any electrons easily. Hence, lead cannot be a reducing agent.

(iv) As we have proved in (i) that tin is a reducing agent, it can reduce mercuric salts to mercurous salts. Thus is best illustrated in the below chemical reaction where tin(II) chloride reacts with mercuric chloride to form mercurous chloride

$Sn{ Cl }_{ 2 }+2{ Hg }Cl_{ 2 }\rightarrow { SnCl }_{ 4 }+{ Hg }_{ 2 }{ Cl }_{ 2 }$