Writing disproportionation half reactions


> 
> **Question**  
> 
>  Write the half equations for the following reaction:  
> 
>  $$\ce{3OCl- -> 2Cl- + ClO3-}$$
> 
> 
> 


I know that $\ce{Cl}$ is reduced from an oxidation state of 1+ to 1- (in $\ce{Cl-}$) and is oxidized from an oxidation state of 1+ to 5+ (in $\ce{ ClO3-}$) 


Below is as far as I have gotten: 


$$\ce{3OCl- -> ClO3- + 4e}$$
$$\ce{3OCl- + 6e -> 2Cl- }$$


I am sure this isn't correct, as I can't balance it and together they do not add up to the full equation




Hint:


$$\ce{2OCl- + 4e -> 2Cl- + 2[O] }$$
$$\ce{2[O] + OCl- -> ClO3- + 4e}$$


P.S. 


$$\ce{2[O] -> O2}$$ is likely to happen, if [O] (nascent oxygen) is not used up in some reaction. 
Here, as soon as it is released in the first reaction (reduction), it is used up in the second (oxidation).


Hope that helps.


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