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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.

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.

Grade:12

1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
Cr(OH)3(s) +IO3(-)(aq) -->CrO4(-2)(aq) + I(-)(aq)
remember in basic medium firstbalancing of oxidation number then balancing of O atom.Balancing of O atom is made by using H2O and OH- ions
add desired molecules of H2o on the side rich in O atoms and double OH- on oppsite side

Cr+3 ---CrO42-
Cr+3 + 8OH- ----------CrO42- +3e- +4H2O
3H2O +6E- +IO3- ---------I- +6OH-
Balance
2Cr(OH)3 + IO3- + 4 OH- --> 2 CrO4(2)- + I- + 5H2O

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