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

Jan 26, 2017

$\text{Hypochlorite}$ is reduced to $\text{chloride}$..........

#### Explanation:

And $\text{iodide}$ is oxidized to $\text{triiodide}$.............

$\text{Reduction:}$

$C l {O}^{-} + 2 {H}^{+} + 2 {e}^{-} \rightarrow C {l}^{-} + {H}_{2} O$ $\left(i\right)$, yes I know you asked for basic solution, but attend........

$\text{Oxidation:}$

$3 {I}^{-} \rightarrow {I}_{3}^{-} + 2 {e}^{-}$ $\left(i i\right)$

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

We add $\left(i\right)$ and $\left(i i\right)$ to remove the electrons.

$C l {O}^{-} + 2 {H}^{+} + 3 {I}^{-} \rightarrow {I}_{3}^{-} + C {l}^{-} + {H}_{2} O$

But you specified basic solution. So all we have to do is add $2 \times H {O}^{-}$ to each side:

$C l {O}^{-} + \cancel{2} {H}_{2} O \left(l\right) + 3 {I}^{-} \rightarrow {I}_{3}^{-} + + C {l}^{-} + 2 H {O}^{-} + \cancel{{H}_{2} O}$

To give me finally,

$C l {O}^{-} + {H}_{2} O \left(l\right) + 3 {I}^{-} \rightarrow {I}_{3}^{-} + C {l}^{-} + 2 H {O}^{-}$

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