Balance the following redox equation using the oxidation number change method. Describe each step that is involved Al(s) + PbSO4 - - > Al2(SO4)3 + Pb(s)

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To balance the redox equation using the oxidation number change method, follow these steps:

Step 1: Identify the oxidation states of the elements in the unbalanced equation.

The oxidation state of Al in Al(s) is 0 because it is an uncombined element.
The oxidation state of S in PbSO4 is +6 because it is in the sulfate ion (SO4)^2- where O has a -2 charge.
The oxidation state of Pb in PbSO4 is +2 because the overall charge of PbSO4 is 0.
The oxidation state of Al in Al2(SO4)3 is +3 because there are three sulfate ions with a -2 charge each.
The oxidation state of Pb in Pb(s) is 0 because it is an uncombined element.

Step 2: Determine the elements that are undergoing oxidation and reduction.

In this equation, Al is being oxidized from an oxidation state of 0 to +3, and Pb is being reduced from an oxidation state of +2 to 0.

Step 3: Write separate half-reactions for the oxidation and reduction processes.

Oxidation half-reaction:
Al(s) -> Al^3+ + 3e-

Reduction half-reaction:
Pb^2+ + 2e- -> Pb(s)

Step 4: Balance the atoms and charges in each half-reaction.

For the oxidation half-reaction, there are already equal numbers of atoms on both sides, but the charges are not balanced. To balance the charges, multiply the oxidation half-reaction by 2:

2Al(s) -> 2Al^3+ + 6e-

For the reduction half-reaction, the charges are not balanced. To balance the charges in the reduction half-reaction, multiply the reduction half-reaction by 3:

3Pb^2+ + 6e- -> 3Pb(s)

Step 5: Combine the half-reactions to cancel out the electrons.

To cancel out the electrons, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:

6Al(s) -> 6Al^3+ + 18e-
6Pb^2+ + 12e- -> 6Pb(s)

Step 6: Add the balanced half-reactions together, canceling out common species.

Combine the two half-reactions to obtain the balanced redox equation:

6Al(s) + 3PbSO4 -> 2Al2(SO4)3 + 6Pb(s)

Now the redox equation is balanced with respect to both atoms and charge.