Name the oxidizing and reducing agents in the following
PbS(S) + 4H2O2(aq) = PbSO4(s) + 4H2O(l)

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Solution

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

Reactants:
Pb in PbS is +2
S in PbS is -2
H in H₂O₂ is +2
O in H₂O₂ is -1
Products:
Pb in PbSO₄ is +2
S in SO₄^2- is +6
H in H₂O is +1
O in H₂O is -2

So the oxidation state of Pb and H have remained constant.
The S went from -2 to +6, was oxidised and is the reducing agent
The O went from -1 to -2, was reduced and is the oxidising agent.

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