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Ionic Stoichiometry help please

(d) A solution of potassium manganate(VII), KMnO4, can be standardised by titration
with arsenic(III) oxide, As2O3. In this reaction, 5 mol of arsenic(III) oxide are
oxidised to arsenic(V) oxide, As2O5, by 4 mol of manganate(VII) ions, MnO4
–.
Calculate the final oxidation number of the manganese.
Arsenic changes by III to V that is a loss of two electrons.
5 moles x 2 electrons = 10 electrons
this interacts with 4 moles of manganate(VII) ions
If 4 moles of manganate VII ions absorb 10 electrons then the change is by 10/4 = 2.5
So the final oxidation state = 7 - 2.5 = 4.5 !!

seems pretty unlikely to me...

However, if only 2 moles of manganate (VII) were involved then the change would be to Mn2+ ... far more satisfactory.
Reply 2
Well, the answer is Mn2+
Artemidoros.
Well, the answer is Mn2+


so there is a typo in the question...
Reply 4
charco
so there is a typo in the question...


Thats exactly what i copy pasted from the past paper, so..
2MnO4- + 2H2O + 5As2O3 -- > 2Mn2+ + 5As2O5 + 4H+

Half equation 1:
MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

As2O3 + 2H2O --> As2O5 + 4e + 4H+

To equalise the electrons multiply the top equation by 4 and the bottom equation by 5

4MnO4- + 32H+ + 20e --> 4Mn2+ + 16H2O
5As2O3 + 10H2O --> 5As2O5 + 20e + 20H+

and add together (cancelling out where appropriate)

4MnO4- + 12H+ + 5As2O3 --> 4Mn2+ + 5As2O5 + 6H2O

OK that's the correct equation, so where was my logic wrong?


Arsenic changes by III to V that is a loss of two electrons.
5 moles x 2 electrons = 10 electrons
this interacts with 4 moles of manganate(VII) ions
If 4 moles of manganate VII ions absorb 10 electrons then the change is by 10/4 = 2.5


Arsenic changes by III to V that is a loss of two electrons. BUT per mole of As2O3 but there are two atoms of As therefore the transition As2O3 --> As2O5 involves 4 electrons per mole5 moles x 4 electrons = 20 electrons
this interacts with 4 moles of manganate(VII) ions
If 4 moles of manganate VII ions absorb 20 electrons then the change is by 10/2 = 5

typo? I'll get me coat... :getmecoat:
Reply 6
Original post by charco
2MnO4- + 2H2O + 5As2O3 -- > 2Mn2+ + 5As2O5 + 4H+

Half equation 1:
MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

As2O3 + 2H2O --> As2O5 + 4e + 4H+

To equalise the electrons multiply the top equation by 4 and the bottom equation by 5

4MnO4- + 32H+ + 20e --> 4Mn2+ + 16H2O
5As2O3 + 10H2O --> 5As2O5 + 20e + 20H+

and add together (cancelling out where appropriate)

4MnO4- + 12H+ + 5As2O3 --> 4Mn2+ + 5As2O5 + 6H2O

OK that's the correct equation, so where was my logic wrong?



Arsenic changes by III to V that is a loss of two electrons. BUT per mole of As2O3 but there are two atoms of As therefore the transition As2O3 --> As2O5 involves 4 electrons per mole5 moles x 4 electrons = 20 electrons
this interacts with 4 moles of manganate(VII) ions
If 4 moles of manganate VII ions absorb 20 electrons then the change is by 10/2 = 5

typo? I'll get me coat... :getmecoat:

4Mn gains 20 electrons , mean 1Mn gains 5 electrons. Since the oxidation number is plus 7 in reactants side after gaining 5 electrons, it goes to +2 charge

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