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

#2MnO_4^(-)+H_2O+I^(-)rarr2MnO_2+IO_3^(-)+2OH^(-)#

Set up the 1/2 equations:

#MnO_4^(-)rarrMnO_2#

Assign oxidation numbers to Mn

#+7rarr+4#

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

#MnO_4^(-)+4H^(+)+3erarrMnO_2+2H_2O# #color(red)((1))#

and:

#I^(-)rarrIO_3^(-)#

Assign oxidation numbers to #I#:

#-1rarr+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^(-)+3H_2OrarrIO_3^(-)+6H^(+)+6e# #color(red)((2))#

To balance the electrons we need to x #color(red)((1))# by 2 then add both sides of #color(red)((1))# and #color(red)((2))# together:

#2MnO_4^(-)+cancel(8H^(+))+cancel(6e)+I^(-)+cancel(3H_2O)rarrMnO_2+cancel(4H_2O)+IO_3^(-)+cancel(6H^(+))+cancel(6e)#

This cancels down to:

#2MnO_4^(-)+2H^(+)+I^(-)rarr2MnO_2+H_2O+IO_3^(-)# #color(red)((3))#

For basic conditions add 2OH- to both sides of #color(red)((3))# to remove the 2H+ ions on the left, making water:

#2MnO_4^(-)+cancel(2H_2O)+I^(-)rarrMnO_2+cancel(H_2O)+IO_3^(-)+2OH^(-)#

Cancelling out the water gives:

#2MnO_4^(-)+H_2O+I^(-)rarr2MnO_2+IO_3^(-)+2OH^(-)#