# How do you balance the following redox reaction? "MnO"_4^(-) + "I"^(-) -> "MnO"_2 + "IO"_3^(-) ?

Apr 25, 2015

$2 M n {O}_{4}^{-} + {H}_{2} O + {I}^{-} \rightarrow 2 M n {O}_{2} + I {O}_{3}^{-} + 2 O {H}^{-}$

Set up the 1/2 equations:

$M n {O}_{4}^{-} \rightarrow M n {O}_{2}$

Assign oxidation numbers to Mn

$+ 7 \rightarrow + 4$

So this will need to take in 3 electrons. Add H+ to the left and water to the right to balance the oxygens$\Rightarrow$

$M n {O}_{4}^{-} + 4 {H}^{+} + 3 e \rightarrow M n {O}_{2} + 2 {H}_{2} O$ $\textcolor{red}{\left(1\right)}$

and:

${I}^{-} \rightarrow I {O}_{3}^{-}$

Assign oxidation numbers to $I$:

$- 1 \rightarrow + 5$

So this will need to give up 6 electrons. Add water to the left and H+ to the right to balance the oxygens:

${I}^{-} + 3 {H}_{2} O \rightarrow I {O}_{3}^{-} + 6 {H}^{+} + 6 e$ $\textcolor{red}{\left(2\right)}$

To balance the electrons we need to x $\textcolor{red}{\left(1\right)}$ by 2 then add both sides of $\textcolor{red}{\left(1\right)}$ and $\textcolor{red}{\left(2\right)}$ together:

$2 M n {O}_{4}^{-} + \cancel{8 {H}^{+}} + \cancel{6 e} + {I}^{-} + \cancel{3 {H}_{2} O} \rightarrow M n {O}_{2} + \cancel{4 {H}_{2} O} + I {O}_{3}^{-} + \cancel{6 {H}^{+}} + \cancel{6 e}$

This cancels down to:

$2 M n {O}_{4}^{-} + 2 {H}^{+} + {I}^{-} \rightarrow 2 M n {O}_{2} + {H}_{2} O + I {O}_{3}^{-}$ $\textcolor{red}{\left(3\right)}$

For basic conditions add 2OH- to both sides of $\textcolor{red}{\left(3\right)}$ to remove the 2H+ ions on the left, making water:

$2 M n {O}_{4}^{-} + \cancel{2 {H}_{2} O} + {I}^{-} \rightarrow M n {O}_{2} + \cancel{{H}_{2} O} + I {O}_{3}^{-} + 2 O {H}^{-}$

Cancelling out the water gives:

$2 M n {O}_{4}^{-} + {H}_{2} O + {I}^{-} \rightarrow 2 M n {O}_{2} + I {O}_{3}^{-} + 2 O {H}^{-}$