# 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

• How did you go about balancing the reactions? – Avyansh Katiyar Mar 7 '18 at 9:07
• @Avnish I tried to multiply each half reactions so that they both had 24 electrons. – Ella Lewis Mar 7 '18 at 9:10
• The atoms of an element need to be balanced in the half reactions as well. – Avyansh Katiyar Mar 7 '18 at 9:25
• @EllaLewis Avnish is right. Also, we need to know the medium of the reaction (acidic or basic) to be able to completely balance the half reaction. Was it given in your book? – Gaurang Tandon Mar 7 '18 at 13:25
• Without supplying the details of the reaction medium, this question seems unanswerable. – Gaurang Tandon Jul 14 '18 at 8:16

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.

• How can you balance the reaction without knowing the medium of the reaction (acidic or basic)? – Gaurang Tandon Mar 7 '18 at 13:25
• If it's not mentioned, this is the best we can do. – strawberry-sunshine Mar 7 '18 at 13:27
• Yeah, but still it is a point worth noting. The answer will differ for both mediums. – Gaurang Tandon Mar 7 '18 at 13:29
• Yeah, you're right! – strawberry-sunshine Mar 7 '18 at 13:31