The proposed mechanism for the reaction ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq) is
1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq) FAST
2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq) FAST
3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq) SLOW
What is the overall equation? (Type your answer using the format [NH4]+ for NH4+. Use the lowest possible coefficients. Enter 0 if necessary. Do not leave any box blank.)
(aq) + I -(aq) Cl -(aq) + (aq) (b) Identify the intermediates, if any.

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq).

Explanation:

1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq)

2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq)

3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq)

Adding all the 3 equations together gives and it gives:

ClO-(aq) + H2O(l) + I-(aq) + HClO(aq) + OH -(aq) + HIO(aq)

---> HClO(aq) + OH-(aq) + HIO(aq) + Cl -(aq) + H2O(l) + IO-(aq)

Deleting the same species on both sides of the equation gives:

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)

The overall equation:

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)