Sir please help me balance this equation-

Cr(oh)3+IO3^-1-------->I-+CrO4^2-(in basic medium) by ion electron method.

Sir it is written in my book to first balance all the atoms other that o and h by seeing increase and decrease in oxidation number. After that balance o by adding h20 to excess side and adding double of that to other side. But in this case the ans is the other way round.

I am getting

2CR(oh)3+io3- >>>>I-+2crO4 2-

After that I feel that there is one extra oxygen in lhs but in the ans there are 5h20 on the other side. Pls help me.

Step-I: Write the two halves ; one the reduction-half and the other oxidation-half.
Cr(OH)_3.............>CrO₄^2-...........................(i) oxidation-half
IO₃^-.......... ...>I^-…...............................(ii): Reduction-half
Step-II: Balenced in each of the half reaction the atoms of elements whose Oxidation number is changing other than O and H
In this case both the half reaction have atoms already balenced.
Step-III: Observe the change in oxidation number per molecule/ionin both thehalf reaction.
Here Cr goes from +3 to +6 so change is 6-3=3 per Cr(OH)3 molecule
Similarly I goes from +5 to -1; so change is 6 per one ion of IO₃^-
Step-IV:Now multiply both the half reaction with suitable numbers so that the gain in Oxidation number in one half reaction would be equal to loss in other half
Here, in (i), gain=3
in (ii), loss=6
so multiply (i) with 2 fand multiply (ii) with1 and thenadd both the equation.
2Cr(OH)3....+.....IO3-.....> 2CrO₄^2-…...+ ….I-
Step-V: Balence O atom by adding appropriate number of water molecules and then to balence H, add H+ ion
2Cr(OH)₃....+.....IO₃^-.......> 2CrO₄^2-…...+ ….I-...+H2O.+4H+
....................If Medium is Basic.......................
Step-VI:If medium is acidic then you are done but if medium is basic then add OH- ion on both the side and the number of OH-ion=H= ion added in step-V.
Now if H+ and OH- ion are in same side, combine H+ and OH- ion and make H2O and then remove the number of water molecules common on both the side.
2Cr(OH)3....+.....IO3-....+4OH-...> 2CrO42-…...+ ….I-...+H2O.+4H++4OH-(basic)
Now combine the h+ & OH- to make H2O