Can you explain what is being oxidized and what is being reduced in this equation? H2O2 + PbS --> PbSO4 + H2O
Please explain how to determine this. Also, how do I determine what the oxidizing agent and the reducing agent is? I understand that oxidation is the loss of electrons and reducction is when electrons are gained. However, I do not understand where they are being lost and gained. This is pretty confusing.
Thank you for your help.
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H2O2 + PbS --> PbSO4 + H2O
You are correct that oxidation is the loss of electrons and reduction is the gaining of electrons. In an unofficial sense, the loss of oxygen is reduction and the gaining of oxygen is oxidation. So the PbS is getting oxidized to PbSO4 (adding oxygen) and the H2O2 is getting reduced to water (loss of oxygen). But let's look at the oxidation state of each atom. PbS has Pb in the +2 state and S in the -2 state. PbSO4 has O in the -2 state, Pb in the +2 state, and S in the +6 state. So S goes from -2 to +6. In other words it loses electrons and is oxidized. H2O2 has H in the +1 state and O in the -1 state. H2O has H in +1 state and O in the -2 state. So O goes from -1 to -2. In other words it gains electrons and is reduced. H2O2 is the oxidizing agent and PbS is the reducing agent.
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