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Lead(IV) oxide, PbO2, reacts with concentrated hydrochloric acid to produce chlorine, lead(II) ions, Pb2+,and water.
(i) Write a half-equation for the formation of Pb2+ and water from PbO2 in the presence of H+ ions.
(ii)Write a half-equation for the formation of chlorine from chloride ions.
(iii) Hence deduce an equation for the reaction which occurs when concentrated hydrochloric acid is added to lead(IV) oxide, PbO2
Please help
(i) Write a half-equation for the formation of Pb2+ and water from PbO2 in the presence of H+ ions.
(ii)Write a half-equation for the formation of chlorine from chloride ions.
(iii) Hence deduce an equation for the reaction which occurs when concentrated hydrochloric acid is added to lead(IV) oxide, PbO2
Please help
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For part (i), I got
PbO2 + 4H+ → Pb2+ + 2H2O + 2e-
The mark scheme says:
PbO2 + 4H+ + 2e− → Pb2+ + 2H2O
I don't understand why the electrons are on the other side. Can someone please explain?
PbO2 + 4H+ → Pb2+ + 2H2O + 2e-
The mark scheme says:
PbO2 + 4H+ + 2e− → Pb2+ + 2H2O
I don't understand why the electrons are on the other side. Can someone please explain?
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#3
(Original post by 010534)
For part (i), I got
PbO2 + 4H+ → Pb2+ + 2H2O + 2e-
The mark scheme says:
PbO2 + 4H+ + 2e− → Pb2+ + 2H2O
I don't understand why the electrons are on the other side. Can someone please explain?
For part (i), I got
PbO2 + 4H+ → Pb2+ + 2H2O + 2e-
The mark scheme says:
PbO2 + 4H+ + 2e− → Pb2+ + 2H2O
I don't understand why the electrons are on the other side. Can someone please explain?
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#4
Tips:
to balance a redox equation firstly u balance the oxygen with water, then balance the hydrogens in water with H+ then balance the charges on both sides
or u can balance the charges on Pb first with electrons then only oxygen with water then hydrogens in water with H+
to balance a redox equation firstly u balance the oxygen with water, then balance the hydrogens in water with H+ then balance the charges on both sides
or u can balance the charges on Pb first with electrons then only oxygen with water then hydrogens in water with H+
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#5
(Original post by 010534)
For part (i), I got
PbO2 + 4H+ → Pb2+ + 2H2O + 2e-
The mark scheme says:
PbO2 + 4H+ + 2e− → Pb2+ + 2H2O
I don't understand why the electrons are on the other side. Can someone please explain?
For part (i), I got
PbO2 + 4H+ → Pb2+ + 2H2O + 2e-
The mark scheme says:
PbO2 + 4H+ + 2e− → Pb2+ + 2H2O
I don't understand why the electrons are on the other side. Can someone please explain?
In PbO2, the oxidation state of the lead is +4, because the overall oxidation state of PbO2 is 0 because charges cancel out, leaving a compound neutral. The oxidation state of O is always -2, and there are two oxygen atoms here, so the charge on the O2 is -4. Therefore to make PbO2 0, the oxidation number of Pb must be +4.
If something has a charge of +4, this means that it has lost 4 electrons, right? (think about sodium as an ion - Na+) So surely in order to become Pb2+, it has to have lost 2 electrons. This means that it has to have gained 2 electrons in order to become Pb2+ on the other side of the equation.
And it's the PbO2 that has to have gained 2 electrons in order to become Pb2+, because if you think about it, if the + 2e- was on the other side of the equation, that would be equivalent to the PbO2 losing two electrons, which would not make Pb2+, but Pb6+, which isn't what you want as a product.
Hope that helps - please reply if that explanation was ok
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(Original post by sue99)
I am not sure if this will make sense but here goes:
In PbO2, the oxidation state of the lead is +4, because the overall oxidation state of PbO2 is 0 because charges cancel out, leaving a compound neutral. The oxidation state of O is always -2, and there are two oxygen atoms here, so the charge on the O2 is -4. Therefore to make PbO2 0, the oxidation number of Pb must be +4.
If something has a charge of +4, this means that it has lost 4 electrons, right? (think about sodium as an ion - Na+) So surely in order to become Pb2+, it has to have lost 2 electrons. This means that it has to have gained 2 electrons in order to become Pb2+ on the other side of the equation.
And it's the PbO2 that has to have gained 2 electrons in order to become Pb2+, because if you think about it, if the + 2e- was on the other side of the equation, that would be equivalent to the PbO2 losing two electrons, which would not make Pb2+, but Pb6+, which isn't what you want as a product.
Hope that helps - please reply if that explanation was ok
I am not sure if this will make sense but here goes:
In PbO2, the oxidation state of the lead is +4, because the overall oxidation state of PbO2 is 0 because charges cancel out, leaving a compound neutral. The oxidation state of O is always -2, and there are two oxygen atoms here, so the charge on the O2 is -4. Therefore to make PbO2 0, the oxidation number of Pb must be +4.
If something has a charge of +4, this means that it has lost 4 electrons, right? (think about sodium as an ion - Na+) So surely in order to become Pb2+, it has to have lost 2 electrons. This means that it has to have gained 2 electrons in order to become Pb2+ on the other side of the equation.
And it's the PbO2 that has to have gained 2 electrons in order to become Pb2+, because if you think about it, if the + 2e- was on the other side of the equation, that would be equivalent to the PbO2 losing two electrons, which would not make Pb2+, but Pb6+, which isn't what you want as a product.
Hope that helps - please reply if that explanation was ok
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It helped a lot^^
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#9
(Original post by Gracie jewel)
Does anyone know what part three of this question is??? I'm so stuck 😥
Does anyone know what part three of this question is??? I'm so stuck 😥
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