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Write an equation for the disproportionation of chlorate(I) ions and show the oxidati

Write an equation for the disproportionation of
chlorate(i) ions and show the
oxidation states of chlorine in the products. Thanks :biggrin:
Reply 1
Avatar for krisshP
krisshP
14
Original post by Rhubarb96
Write an equation for the disproportionation of
chlorate(i) ions and show the
oxidation states of chlorine in the products. Thanks :biggrin:


See question 1bii in

http://www.ocr.org.uk/Images/60448-question-paper-unit-f332-chemistry-of-natural-resources.pdf
Original post by krisshP


Forgive me, but did the question not seem to imply that ClO- was a reactant?
Reply 3
Avatar for krisshP
krisshP
14
Original post by PythianLegume
Forgive me, but did the question not seem to imply that ClO- was a reactant?


Sorry
Original post by krisshP
Sorry


That's fine. I'm not sure how to answer it, because I don't know what the products would be. At a guess, I'd say Cl2 and ClO3, but it would be a guess.
Original post by PythianLegume
That's fine. I'm not sure how to answer it, because I don't know what the products would be. At a guess, I'd say Cl2 and ClO3, but it would be a guess.


The chlorate(I) disproportionates (simultaneously oxidises and reduces) to chloride ions and chlorate(V) ions ...
Reply 6
Avatar for MattRay94
MattRay94
1
Hi Rhubarb :smile:

The easiest way to write any disproportionation is to write single electron reactions for the reduction and the oxidation and then add the two equations to each other.

Assuming it disproportionates to chloride ions and chlorate ions, you have the following half equations.

Reduction of ClO(-) into Cl(-)

0.5 ClO(-) + H(+) + e(-) --> 0.5 Cl(-) +0.5H2O

Oxidation of ClO(-) into ClO3(-)

0.25ClO(-) + 0.5 H2O --> 0.25ClO3(-) + H(+) + e(-)

Now add the two equations, and everything simplifies nicely with the hydrogen ions, electrons and waters cancelling out.

0.75 ClO(-) --> 0.5 Cl(-) + 0.25 ClO3(-)

This can also be written with integer coefficients of 3, 2 and 1 respectively.

EDIT: Just saw you asked for oxidation states too. ClO(-) has chlorine in oxidation state +1, Cl(-) has chlorine in oxidation state -1, ClO3(-) has chlorine in oxidation state +5. You can work this out as the oxidation state of oxygen in all non-peroxide compounds is (-2), and the sum of...

oxidation state of chlorine + ( (-2) * number of oxygen atoms) = overall charge on the ion.

This is why the reaction is a disproportionation, because 0.75 moles of chlorine at oxidation state 1 is simultaneously reduced to 0.5 moles of chloride ions at oxidation state -1 and oxidised to 0.25 moles of chlorate (V) ions at oxidation state +5.

Hope this helped!
Matt


Posted from TSR Mobile
(edited 10 years ago)
Original post by Rhubarb96
Write an equation for the disproportionation of
chlorate(i) ions and show the
oxidation states of chlorine in the products. Thanks :biggrin:


3ClO- -> ClO3 - + 2Cl-
+1------> +5 + -1
Original post by Coral Reafs
3ClO- -> ClO3 - + 2Cl-
+1------> +5 + -1


yeah nice one ...

.. you have successfully re-answered a question.

value much not
Original post by charco
yeah nice one ...

.. you have successfully re-answered a question.

value much not


oops..didn't check entire page :biggrin:
Brilliant, thanks! Eve of Kent

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