PbO2 + 4H+ + SO42– + 2e– → PbSO4 +2H2O
How does the oxidation number for lead (Pb) change?
Pb goes from 4+ on the left to 2+ on the right.
PbO2. You know O is 2-, two of them makes 4- so Pb must be + in order for the comound to be zero. All compounds are zero.
On the right you know SO4^2- in PbSO4 is 2- so Pb must be 2+. You don't know how to determine oxidation states. Here is a very good site. Specifically you want to look at #2.
https://www.chemteam.info/Redox/Redox.html
In the given reaction, PbO2 is being reduced to PbSO4.
To determine the change in the oxidation number for lead (Pb), we need to compare the oxidation numbers of Pb in PbO2 and PbSO4.
In PbO2, the oxidation number of oxygen (O) is -2, and since there are two oxygen atoms in PbO2, the total oxidation number contribution from oxygen is -4.
The overall charge of the molecule is zero because there are no charges specified. Thus, the sum of the oxidation numbers of Pb and O should equal zero.
Let's assign the oxidation number of Pb in PbO2 as x.
The equation becomes:
x + (-4) = 0
Simplifying the equation:
x - 4 = 0
Adding 4 to both sides:
x = 4
Therefore, in PbO2, the oxidation number of lead (Pb) is +4.
Now let's determine the oxidation number of Pb in PbSO4.
In PbSO4, the oxidation number of oxygen (O) is -2, and since there are four oxygen atoms in PbSO4, the total oxidation number contribution from oxygen is -8.
The overall charge of the molecule is zero because there are no charges specified.
Let's assign the oxidation number of Pb in PbSO4 as y.
The equation becomes:
y + (-8) = 0
Simplifying the equation:
y - 8 = 0
Adding 8 to both sides:
y = 8
Therefore, in PbSO4, the oxidation number of lead (Pb) is +8.
To find the change in the oxidation number of Pb, we subtract the initial oxidation number from the final oxidation number:
Change in oxidation number of Pb = Final oxidation number - Initial oxidation number
Change in oxidation number of Pb = +8 - +4
Change in oxidation number of Pb = +4
So, the oxidation number for lead (Pb) increases by 4 in this reaction.
To determine how the oxidation number for lead (Pb) changes in the given equation, we can look at the oxidation states of Pb in both the reactants and the products.
In the reactant PbO2, the oxygen (O) has a -2 oxidation state. Since the overall charge of PbO2 is 0, the oxidation state of Pb must be +4 to balance out the -2 charge from each oxygen.
In the product PbSO4, the sulfur (S) has an oxidation state of +6, and the oxygen (O) has an oxidation state of -2. Again, the overall charge of PbSO4 is 0. So, to balance out the charges, the oxidation state of Pb in PbSO4 must be +2.
Therefore, in the given equation, the oxidation state of lead (Pb) changes from +4 to +2.