Question

for the following reaction, identify the element that was oxidized, the element that was reduced, and the reducing agent. Give an explanation for each answer
SnCl2+PbCl4-->SnCl4+PbCl2

Answer 1

oxidation states.
Sn= +2 , Pb= +4. after the reaction Sn= +4, Pb=+2

Then Sn lost 2e-. this is the oxidized element because lost e-.

That makes Pb the reduced element and Sn the reducing agent.

Answer 2

Breaking it down: We know that Chlorine has a constant oxidation state of -1. Now Sn is bonded with Cl in the reactants side as SnCl2, therefore it has an oxidation state of 2. Where as in the product side, it is bonded as SnCl4, therefore it's oxidation state is 4.

Since electrons carry a negative charge, the removal of 2 negatives makes 2 positives. Therefore 2+ - 2e- gives you 4+. OXIDATION is the loss of electrons. So, Sn having lost electrons is oxidized. A reducing agent reduces and gets oxidized itself, therefore Sn is also the reducing agent.

The element which has gained electrons is: As earlier, it has an oxidation state of 4 in the reactant stage and 2 in the product state. So, it has gained 2 electrons (negative) to reduce a positive 4 value. This is reduction (gaining of electrons), thus Pb has been reduced.

Answer 3

thank you guys so much :) @Guardian and @julian.pitalua