How do I determine which is the reduction reaction in a hydrogen peroxide and potassium permanganate solution?
Answers
Answered by
0
First off, in your question, you wrote hydrogen peroxide incorrectly. it should be H2O2, not 2H2O. Using superscripts and subscripts help. That confused me for a bit until I read the word question you graciously included. The other issue is that your question does not give the final products, it just tells you about the reactants so you will have to do a bit of investigating.
To determine which reactant is oxidized and which is reduced, you need to determine which element’s oxidation state changed. If the oxidation state went up (more positive) then it was oxidized. If the oxidation state of an element went down (less positive) then it was reduced.
To determine the oxidation state, there are myriad methods used. The simplest is the actual rule of what the oxidation state (OS) measures. the OS is the charge that would exist on any atom in a molecule or ion, if all the bonds (a bond joins two atoms) were ionic ones with both electrons in the bond going to the more electronegative atom of the two. The other way I use is to recognise certain elements in compounds almost always tend to a particular oxidation state, and then I do some simple adding and subtracting to make sure that the sum of the oxidation states adds up to the charge on the species.
Take MnO−4MnO4−. The sum of the oxidation states must add up to -1 because that is the charge of the ion itself. Each oxygen tends to have a -2 OS so four oxygens have a total “charge” of -8. The Mn, must be a +7 charge so that
+7 + (-8) = -1.
OK, so the permanganate has a Mn with a +7 OS.
I looked in the table of standard reduction potentials and found a reaction for the reduction of permanganate. The Mn-containing product in this reaction is MnO2.MnO2. It is a neutral molecule so the sum of the OS must be zero. Two oxygens each -2 means the Mn must be +4 so that
+4 + (-4) = 0
So the manganese changed from +7 to +4. It was reduced. If this is the reduction then the peroxide must have been oxidized.
Peroxide can actually go either way. Hydrogen peroxide can be reduced to water or oxidized to oxygen gas, depending on what it is reacting with. It can even oxidize and reduce itself in a disproportionation reaction, giving oxygen gas and water. That’s why hydrogen peroxide is always stored in dark opaque (brown) bottles since light helps speed up this self-destruction of the hydrogen peroxide.
Since the permanganate is clearly being reduced, the peroxide must be oxidized and the product is O2(g)O2(g).
so, the net ionic reduction reaction is (ignore the potassium as a spectator ion)
MnO−4+4H++3e−→MnO2+2H2OMnO4−+4H++3e−→MnO2+2H2O
The net ionic oxidation reaction is
H2O2→O2+2H++2e−H2O2→O2+2H++2e−
Since the oxidation creates two electrons but the reduction uses up three electrons, we multiply the reduction by 2 (total 6 electrons used) and the oxidation by 3 (6 electrons produced) so that the electrons cancel and the overall balanced equation is
2MnO−4+2H++3H2O2→2MnO2+4H2O+3O22MnO4−+2H++3H2O2→2MnO2+4H2O+3O2
To determine which reactant is oxidized and which is reduced, you need to determine which element’s oxidation state changed. If the oxidation state went up (more positive) then it was oxidized. If the oxidation state of an element went down (less positive) then it was reduced.
To determine the oxidation state, there are myriad methods used. The simplest is the actual rule of what the oxidation state (OS) measures. the OS is the charge that would exist on any atom in a molecule or ion, if all the bonds (a bond joins two atoms) were ionic ones with both electrons in the bond going to the more electronegative atom of the two. The other way I use is to recognise certain elements in compounds almost always tend to a particular oxidation state, and then I do some simple adding and subtracting to make sure that the sum of the oxidation states adds up to the charge on the species.
Take MnO−4MnO4−. The sum of the oxidation states must add up to -1 because that is the charge of the ion itself. Each oxygen tends to have a -2 OS so four oxygens have a total “charge” of -8. The Mn, must be a +7 charge so that
+7 + (-8) = -1.
OK, so the permanganate has a Mn with a +7 OS.
I looked in the table of standard reduction potentials and found a reaction for the reduction of permanganate. The Mn-containing product in this reaction is MnO2.MnO2. It is a neutral molecule so the sum of the OS must be zero. Two oxygens each -2 means the Mn must be +4 so that
+4 + (-4) = 0
So the manganese changed from +7 to +4. It was reduced. If this is the reduction then the peroxide must have been oxidized.
Peroxide can actually go either way. Hydrogen peroxide can be reduced to water or oxidized to oxygen gas, depending on what it is reacting with. It can even oxidize and reduce itself in a disproportionation reaction, giving oxygen gas and water. That’s why hydrogen peroxide is always stored in dark opaque (brown) bottles since light helps speed up this self-destruction of the hydrogen peroxide.
Since the permanganate is clearly being reduced, the peroxide must be oxidized and the product is O2(g)O2(g).
so, the net ionic reduction reaction is (ignore the potassium as a spectator ion)
MnO−4+4H++3e−→MnO2+2H2OMnO4−+4H++3e−→MnO2+2H2O
The net ionic oxidation reaction is
H2O2→O2+2H++2e−H2O2→O2+2H++2e−
Since the oxidation creates two electrons but the reduction uses up three electrons, we multiply the reduction by 2 (total 6 electrons used) and the oxidation by 3 (6 electrons produced) so that the electrons cancel and the overall balanced equation is
2MnO−4+2H++3H2O2→2MnO2+4H2O+3O22MnO4−+2H++3H2O2→2MnO2+4H2O+3O2
Similar questions