Answer to Question #31115 in Inorganic Chemistry for shreya

Question

pbs+ 4h2o2 = pbso4 + 4h2o2, which substance is reduced and which is oxidised?

Answer 1

Checking what the oxidation states of the elements are. H is +1 when not bound to a metal, O and S are -2 when bound to elements of lower electronegativity only.

With 2+1 ON in H₂O₂, the O₂ entity has to be -2, meaning O is -1. S is -2 in
PbS, so Pb is +2. Sulfate has a charge of -2, so Pb is still +2. O is -2 here,
so work out S. S = -[+2 - 4(-2)] = -(-6) = +6. H in H2O is still +1, so O is
-2.

H₂O₂ + PbS → PbSO₄ + H₂O
+1 -1 ...+2 -2 .. +2 +6 -2 .+1 -2

S goes from oxidation number -2 to +6 so it is oxidized and it is the reducing
agent

O goes from oxidation number -1 to - 2 so it is reduced and it is the
oxidizing agent

4H₂O₂(aq) + PbS(s) → PbSO₄(s) + 4H₂O(l)