[xshadow]

Hi

Can I balance this redox reaction with the half reaction method?

PbSO₄(s) + H₂O₂ ---->  PbO₂ + H₂SO₄
OX:  PbSO4 ---> PbO₂ + H2SO₄
RID:  H₂O₂ ---> PbO₂

the iussue here is that I can't find the right two half reaction...
In the reduction half reaction I don't have "Pb" in the reagent so I can't balance correctly that


Some help??
thanks

[chenbeier]

Get rid of spectator ions,.use Water as oxygen transfer.

Pb2+ + 2 H2O => PbO2 + 4 H+ + 2 e-  Oxidation

H2O2 + 2H+ + 2e- =>  2 H2O Reduction.

Please continue by yourself

[xshadow]

Get rid of spectator ions,.use Water as oxygen transfer.

Pb2+ + 2 H2O => PbO2 + 4 H+ + 2 e-  Oxidation

H2O2 + 2H+ + 2e- =>  2 H2O Reduction.

Please continue by yourself

But PbSO4 is a solid

Shouldn't keep it in a molecular form (very weak electrolyte)
Thanks

[chenbeier]

If the reaction should take place a little bit has to be dissolved. You can also develop if you use PbSO4 solid.

[xshadow]

If the reaction should take place a little bit has to be dissolved. You can also develop if you use PbSO4 solid.

So I don't understand if there is a way to use half reaction method  for this reaction using PbSO4 (s)  (in the molecular form) in the  oxidation halfway reaction.

Because I can't
Our teacher said that weak electrolyte should be in a molecular form alsoconsidering the ionic equation (and then the net one) so I'm confused when you have considered PbSO4 in a dissociated form

: )

[chenbeier]

What about
PbSO₄ + 2 H₂O => PbO₂ + 4 H⁺ + SO₄ ^2- + 2 e^-

[xshadow]

What about
PbSO₄ + 2 H₂O => PbO₂ + 4 H⁺ + SO₄ ^2- + 2 e^-

Thanks!!
So I also  have to put SO4-2 even if is a spectator ion  (but I need it in order to balance Sulfur)

[chenbeier]

If you use it, then you have to put on both sides.