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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 rarr; 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) rarr; PbSO4(s) + 4H2O(l)
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