Balancing redox equations using 1/2 reaction method

[veganazi]

Homework Statement

Balance PbO2(s) + Cl(-) → PbCl2(s) + O2(g)

Homework Equations

n/a

The Attempt at a Solution

Pb(4+) + 2e(-) → Pb(2+) and I'm stuck because there's no 2nd reaction.

 

[Borek]

Yes there is - if you can't see it, assign oxidation numbers to all elements on both sides.

 

[veganazi]

Um 2H2O + 2Cl(-) → O2 + 4H(+) + 6e(-)? Is this right?

 

[Borek]

Um 2H2O + 2Cl(-) → O2 + 4H(+) + 6e(-)? Is this right?

No.

Have you tried to assign oxidation numbers? As long as you will ignore hints you will not move forward.

 

[DeeNos]

The balanced equation for this redox reaction can be achieved by breaking it down into two half-reactions. The first half-reaction involves the reduction of PbO2 to Pb(2+), which you have correctly identified. The second half-reaction involves the oxidation of Cl(-) to form O2. The balanced equation for this half-reaction is 2Cl(-) → O2 + 2e(-).

To balance the overall equation, we can multiply the first half-reaction by 2 to balance the number of electrons on both sides. This gives us 2Pb(4+) + 4e(-) → 2Pb(2+). We can then combine this with the second half-reaction to get the final balanced equation: 2Pb(4+) + 4e(-) + 4Cl(-) → 2Pb(2+) + 4Cl(-) + O2.

It is important to remember that the number of electrons transferred in the oxidation and reduction half-reactions must be equal in order for the overall equation to be balanced. I hope this helps with your understanding of balancing redox equations using the 1/2 reaction method.