Which element is oxidized and which is reduced in the following reactions?
(d) PbS (s) + 4 H2O2 (aq) → PbSO4 (s) + 4 H2O (l)
So here in the equation that we're dealing with is lead 2 sulphide solid plus 4 moles of hydrogen peroxide produces lead 2 sulphate plus 4 moles of water as a liquid. This is lead 2 so it's plus 2 for its oxidation number, this here therefore is -2, hydrogen here is plus 1 because it's connected to a non metal and this is a peroxide and remember when it's a peroxide oxygen's oxidation number will be -1.
Now here this is still lead 2 so it's plus 2, oxygen here is not a peroxide or a superoxide so it's -2, we don't know what sulphur is and we don't have a rule for it so we have to solve. So we'd say we have a lead which is +2 plus X for the sulphur plus 4 oxygen's, each one is -2, equals the charge in my molecule which is 0. So 2 plus X minus 8 equals 0, X minus 6 equals zero, X minus 6 equals zero. So sulphur's oxidation number here is +6, hydrogen here is +1 and oxygen here is not a peroxide or a superoxide so it's -2. So now we're going to talk about who's oxidation number have changed.
Well we're going to say that S2- within this compound goes from being -2 to +6. So what happened to its oxidation number? It increased, it went up therefore S2- has been oxidised. On the other hand, the oxygen in H2O2 goes from being -1 to being -2, so what happened to its oxidation number? Its oxidation number decreased, therefore the oxygen within H2O2 has been reduced. They're asking for the particular element so we're saying within H2O2, the oxygen is what's being reduced.