The following unbalanced equation describes the reaction that can occur when (II) sulfide reacts with oxygen gas to produce lead (Il) oxide and sulfur dioxide gas: PbS+O2---> PbO + SO2 Balance the equation and describe in words the electron transfers that takes place.
Pb goes from +2 on the left to +2 on the right; no change.
O2 on the left goes to O on PbO with -2 and to SO2 at -2 EACH.
To balance the given equation, we need to ensure that the same number of each type of atom appears on both sides of the equation.
Let's balance the equation step-by-step:
1. Write down the unbalanced equation:
PbS + O2 → PbO + SO2
2. Balance the lead (Pb) atoms by placing a coefficient of 2 in front of PbS:
2 PbS + O2 → PbO + SO2
3. Balance the sulfur (S) atoms by placing a coefficient of 2 in front of SO2:
2 PbS + O2 → PbO + 2 SO2
4. Balance the oxygen (O) atoms by placing a coefficient of 3 in front of O2:
2 PbS + 3 O2 → 2 PbO + 2 SO2
Now, the equation is balanced.
Regarding the electron transfers during the reaction:
- In the reactants, the sulfur atom in PbS is in the -2 oxidation state, while the oxygen atoms in O2 are in the 0 oxidation state.
- In the products, the lead atom in PbO is in the +2 oxidation state, and the sulfur atom in SO2 is in the +4 oxidation state.
During the reaction, there is a transfer of electrons from the sulfur atom in PbS to the oxygen atoms in O2. This transfer of electrons results in the oxidation of sulfur from -2 to +4 and the reduction of oxygen from 0 to -2.
To balance this chemical equation, we need to make sure that the same number of atoms of each element are present on both sides of the equation.
The unbalanced equation is: PbS + O2 → PbO + SO2
Let's start by looking at the elements one at a time:
1. Lead (Pb): There is one lead (Pb) atom on the left side, and one on the right side. It is already balanced.
2. Sulfur (S): There is one sulfur (S) atom on the left side, and one on the right side. It is balanced.
3. Oxygen (O): There are two oxygen (O) atoms in the reactant (O2), and two on the product side (PbO and SO2). It is balanced.
Now that all the atoms are balanced, let's write the balanced equation:
PbS + O2 → PbO + SO2
To describe the electron transfers that take place, we need to look at the oxidation states of the elements involved.
1. In PbS (lead sulfide), lead is in the +2 oxidation state (Pb^2+) and sulfur is in the -2 oxidation state (S^2-).
2. In O2 (oxygen gas), oxygen is in the 0 oxidation state (O^0).
3. In PbO (lead(II) oxide), lead is in the +2 oxidation state (Pb^2+) and oxygen is in the -2 oxidation state (O^2-).
4. In SO2 (sulfur dioxide), sulfur is in the +4 oxidation state (S^4+) and oxygen is in the -2 oxidation state (O^2-).
During the reaction:
- Oxygen gets reduced from an oxidation state of 0 in O2 to -2 in PbO and SO2.
- Sulfur gets oxidized from an oxidation state of -2 in PbS to +4 in SO2.
Therefore, the electron transfer in this reaction involves the reduction of oxygen and the oxidation of sulfur.