# What is the Balanced redox reaction for the following equation: I2 + Ca(ClO)2➡ Ca(IO3)2 +CaCl2 ?

Apr 15, 2017

Elemental iodine is oxidized to iodate anion, ""^(+V)IO_3^-......

#### Explanation:

And hypochlorite ion, ""^(+I)ClO^-, is REDUCED to ""^(-I)Cl^-

$\text{Oxidation reaction:}$

$\frac{1}{2} {I}_{2} + 3 {H}_{2} O \rightarrow I {O}_{3}^{-} + 6 {H}^{+} + 5 {e}^{-}$ $\left(i\right)$

$\text{Reduction reaction:}$

""^(+I)ClO^(-) +2H^(+) + 2e^(-) rarr Cl^(-)+H_2O $\left(i i\right)$

For both reactions, charge and mass are (I think!) balanced, as indeed they must be if we purport to represent chemical reality.

The overall redox equation eliminates the presence of electrons, and so we take $2 \times \left(i\right) + 5 \times \left(i i\right) :$

${I}_{2} + 6 {H}_{2} O + 5 {\text{^(+I)ClO^(-) +10H^(+) + 10e^(-) rarr 2}}^{+ V} I {O}_{3}^{-} + 12 {H}^{+} + 10 {e}^{-} + 5 C {l}^{-} + 5 {H}_{2} O$

And now we just cancel the common reagents..........

${I}_{2} + \cancel{6} {H}_{2} O + 5 {\text{^(+I)ClO^(-) +cancel(10H^(+)) + cancel(10e^(-)) rarr 2}}^{+ V} I {O}_{3}^{-} + \cancel{12} 2 {H}^{+} + \cancel{10 {e}^{-}} + 5 C {l}^{-} + \cancel{5 {H}_{2} O}$

To get finally.............

${I}_{2} + {H}_{2} O + 5 C l {O}^{-} \rightarrow 2 I {O}_{3}^{-} + 5 C {l}^{-} + 2 {H}^{+}$

Is this balanced with respect to mass and charge?

This MIGHT seem difficult, but remember all I have done is to balance mass and charge using (fictional!) $\text{oxidation state}$, and (fictional!) electrons......You can get very good at these sorts of equations provided you know how to assign $\text{oxidation number}$.