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write the balanced half reaction for the conversion of HOBr(aq) to Br2(aq) in acidic solution

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  1. You need to learn how to balance half reactions that are oxidation-reduction. I will balance this one for you and provide you with a link to tell you the procedure for doing this. You can add the (aq) to the materials and I won't type them in.
    HOBr ==> Br2
    Br goes from +1 on the left to zero on the right. So Br is the element changing oxidation state. I start by balancing the Br so we can compare two Br with Br2 (the same number of atoms).
    2HOBr ==> Br2
    From +2 on the left to 0 on the right is a gain of 2 e.
    2HOBr + 2e ==> Br2
    Now we count up the charge on the left (it is -2) and the charge on the right (it is 0) and add H^+ to the appropriate side to balance the charge. That gives us
    2HOBr + 2e + 2H^+ ==> Br2
    Now add H2O to the other side to balance the H atoms. That will be
    2HOBr + 2e + 2H^+ ==> Br2 + 2H2O
    The oxygen should balance.
    Notice that the equation balances three ways.
    1. The atoms balance.
    I have 4H on the left and 4 on the right.
    I have 2 O on the left and 2 O the right.
    I have 2Br on the left and 2 Br ;on the right.
    2. The charge balances.
    I have zero charge on left and right.
    3. The change in electrons balance.
    The number of electrons has changed from +2 to zero and that is a gain of 2e.
    I will make another post for the site I mentioned above.

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