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

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Rhubarb96
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Write an equation for the disproportionation of
chlorate(i) ions and show the
oxidation states of chlorine in the products.
Thanks
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krisshP
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(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
See question 1bii in

http://www.ocr.org.uk/Images/60448-q...-resources.pdf
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username878045
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Forgive me, but did the question not seem to imply that ClO- was a reactant?
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krisshP
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(Original post by PythianLegume)
Forgive me, but did the question not seem to imply that ClO- was a reactant?
Sorry
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username878045
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(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.
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charco
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(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 ...
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MattRay94
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Hi Rhubarb

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


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Coral Reafs
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(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
3ClO- -> ClO3 - + 2Cl-
+1------> +5 + -1
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charco
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(Original post by Coral Reafs)
3ClO- -> ClO3 - + 2Cl-
+1------> +5 + -1
yeah nice one ...

.. you have successfully re-answered a question.

value much not
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Coral Reafs
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(Original post by charco)
yeah nice one ...

.. you have successfully re-answered a question.

value much not
oops..didn't check entire page
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Eve of Kent
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Brilliant, thanks! Eve of Kent
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