The following unbalanced equation describes the reaction that can occur when lead (II) sulfide reacts with oxygen gas to produce lead (II) oxide and sulfur dioxide gas: PbS + O2 PbO + SO2.
all i need help on is explaination of the electron transfers that would happen in this equation?
In order to explain the electron transfers that occur in the reaction, let's first assign oxidation numbers to the elements involved. Oxidation numbers are used to track the electron transfers in a chemical reaction.
In the compound lead (II) sulfide (PbS), lead (Pb) has an oxidation number of +2, while sulfur (S) has an oxidation number of -2.
In oxygen gas (O2), each oxygen atom has an oxidation number of 0.
In lead (II) oxide (PbO), lead (Pb) has an oxidation number of +2, and oxygen (O) has an oxidation number of -2.
In sulfur dioxide gas (SO2), sulfur (S) has an oxidation number of +4, and each oxygen (O) has an oxidation number of -2.
Let's analyze the electron transfers step by step:
1. Lead (II) sulfide (PbS) is made up of Pb2+ and S2-. In this reaction, the lead atom loses two electrons and is oxidized from +2 to 0, becoming lead (II) oxide (PbO). The sulfur atom gains two electrons and is reduced from -2 to 0, becoming sulfur dioxide (SO2).
Pb2+ --> Pb4+ + 2e- (oxidation)
S2- + 2e- --> S0 (reduction)
2. Oxygen gas (O2) does not undergo any changes in oxidation number since its oxidation number is already 0.
By balancing the equation, we can see that two electrons are transferred from the lead (II) sulfide to oxygen gas, leading to the formation of lead (II) oxide and sulfur dioxide:
PbS + O2 → PbO + SO2
Overall, lead (II) sulfide gets oxidized, losing electrons, while oxygen gas remains unchanged. This transfer of electrons is a crucial part of the reaction, as it allows the formation of the new compounds.