Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# Please explain how to determine what is oxidized and what is reduced in the following reaction: PbS + 4H2O2 --&gt; PbSO4 + 4H2O

Oxidation - involves the loss of electrons or hydrogen OR gain of oxygen OR increase in oxidation state.

Reduction - involves the gain of electrons or hydrogen OR loss of oxygen OR decrease in oxidation state.

We can look at the reactants and figure out which one of the reactant has oxidized and which one has got reduced.

PbS + 4H_2O_2 --> PbSO_4 + 4H_2O

our reactant is PbS, it takes up oxygen from H_2O_2 and changes to PbSO_4. It has undergone oxidation, or it has been oxidized. H_2O_2 has lost oxygen and has changed to water, H_2O. so it has been reduced.

In terms of electrons Pb goes from+2 oxidation state (OS) to +2 oxidation state. S goes from -2 OS to +6 (OS) so S has lost e-, S is oxidized O changes -2to form -1 , so O has gained e-,O is reduced

H2O2 loses one atom of oxygen to form H2O. Therefore, it is said that it is being reduced as it has lose an atom of oxygen. Therefore, PbS is being oxidized by receiving oxygen to form PbSo4.