Originally posted by CKTR:
Err, chemistry noob here, pls
help. For chemistry, how do you preduct the product of the
reactions if only given the reactants?
For example,
If Lead(II) sulfide is reacted
with hydrogen peroxide , why is the product lead(II) sulfate +
water but not lead (II) sulfate + hydrogen. (Adopted from CS TOH
Alevel practise questions)
Can anyone please tell me the
what are the steps of predicting the product of the
reaction?
Unlike O levels, now that you're entering A levels which is preparation for Uni level, there is no simple shortcut way of predicting the product of a chemical reaction. You must study all the different topics and properties of different species, and apply what you've learnt across all topics into answering a question.
H2O2 is a strong oxidizing agent, and hence is able to oxidize the sulfide ion to the sulfate(VI) ion, and H2O2 is reduced to H2O in the process. Due to the inert pair effect, it won't be easy to oxidize Pb2+ to Pb4+ or PbO2. And S2- can be oxidized to S, S2O3 2-, S4O6 2-, SO3 2-, S2O8 2-, or SO4 2- (there exist other sulfur containing ions with other OSes, but unless the question includes such data, stick to the common ions you're expected to be familiar with). And since the reduction potential of H2O2 to H2O is positive and of a sufficiently large magnitude, the cell potential for the combined redox reaction will be positive and hence thermodynamically favourable for the oxidation of S2- all the way to the maximum OS of +6 for sulfur (and SO4 2- is more stable and is hence the logical, most probable product, compared to S2O8 2- due to the peroxy / peroxo group with OS of -1 for oxygen, which you can deduce won't be present in the final product because the reduction of the peroxo group in the H2O2 reactant is the thermodynamic driving force behind the entire redox reaction in the first place).
As to your qn why you don't get H2(g), at A levels you have to be aware that to reduce H or H+ from H2O2 or H2O into H2 gas, you'll first need to either obtain a significantly large quantity of H+ (but neither H2O2 and H2O are sufficiently acidic to be readily deprotonated), or you'll need a stronger reducing agent than S2-, and no other reducible species that will take priority over such (in this case the peroxo group takes priority for reduction). Also, bear in mind you need a highly reactive (ie. electropositive) metal to reduce H+ (from H2O or H2O2) to H2(g), but in your CS Toh question, the reactant is Pb2+, not Pb(s). Pb2+ is already in a stable oxidized state, and hence is unable to reduce H2O or H2O2 into H2(g). And even if it was Pb(s), the oxidation potential of Pb to Pb2+ is not sufficiently positive to reduce H2O or H2O2 into H2(g), unless acidic H+(aq) is already present. Furthermore lead readily reacts with oxygen and/or water vapour in the atmosphere, to form a passivation layer of insoluble PbO(s), posing a further barrier to the reaction with H2O or H2O2. Strong acids are required to first protonate and dissolve away the PbO(s) passivation layer, to react with the Pb(s) underneath, and only then, would you be able to obtain some (not vigorous effervescence though) H2(g).