Question

How do you solve `I^- + ClO^- rarr I_3^- +Cl` using the redox reaction method in a base solution?

Answer 1

`"Hypochlorite"` is reduced to `"chloride"`..........

Explanation:

And `"iodide"` is oxidized to `"triiodide"`.............

`"Reduction:"`

`ClO^(-) +2H^(+) + 2e^(-) rarrCl^(-) + H_2O` `(i)`, yes I know you asked for basic solution, but attend........

`"Oxidation:"`

`3I^(-) rarrI_3^(-) + 2e^(-)` `(ii)`

For each half equation, mass and charge are conserved as is required. Am I right?

We add `(i)` and `(ii)` to remove the electrons.

`ClO^(-) +2H^(+) + 3I^(-) rarrI_3^(-) + Cl^(-) +H_2O`

But you specified basic solution. So all we have to do is add `2xxHO^-` to each side:

`ClO^(-) +cancel2H_2O(l) + 3I^(-) rarrI_3^(-) + + Cl^(-) +2HO^(-)+cancel(H_2O)`

To give me finally,

`ClO^(-) +H_2O(l) + 3I^(-)rarrI_3^(-) + Cl^(-) + 2HO^(-)`

Which I think is balanced with respect to mass and charge.