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Grade 12Physical Chemistry

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.

Jai Mahajan

11 Years agoGrade 12

4 Answers

Sunil Kumar FP

11 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

Mrinali

7 Years ago

In acidic medium , this method can be followed

Cr(oh)3+IO3^-1--->I^-1+CrO4^-2

Cr(OH)3-->CrO4^-2

On left side 1 oxygen is less so we add H2O on left side but if we do that there are 5 hydrogen less on right side so the total reaction will be

Cr(OH)+H2O+3e-->CrO4^2- + 5H^+

+3e because left side is less by 3 electrons

In the same way we get

IO3^- +6H^+ ---> I^-1 + 3H2O +6e

Now to balance electrons we multiply 2 on both sides of first reaction and by doing so...we get

2{Cr(OH)3+H2O +3e-->CrO4^2- +5 H^+}

1{IO3^-1 +6H^+ ---> I^- +3H2O +6e}

----------------------------------------------------------------

2Cr(OH)3 +2H2O +6e---->2Cro4^-2 + 10H^+

IO3^-1 +6H^+ ------>I^- +3H2O +6e

-----------------------------------------------------------------

2Cr(OH)3 + IO3^-1 -------> 2CrO4^2- +I^-1 +4H^+1 +H2O

ankit singh

5 Years ago

Cr(OH)3 GIVES Cr+3 + 3OH-

Hear OH shows that the equation should be balanced in basic medium

Cr+3----->(Cro4)2-
Cr+3----->(Cro4)2- +4H2O
8OH- + Cr+3---->(Cro4)2- + 4H2O
8OH- + cr+3--->4H2O+(Cro4)2-+ 3e- this is our first equation

Now

IO3 - ----->I-
IO3 - +3H2O ----> I- + 6OH-
IO3- + 3H2O + 6e- -----> I- +6OH- Tthis is our second equation

Multiplying first euation with three and second equation with six

Indian Sketcher Official

3 Years ago

Reduction Half-Reaftion →

IO₃^- →I^-

6e^- + 3H₂O + IO₃^- →I^- + 6OH^-

Oxidation Half-reaction →

Cr(OH)₃→(CrO₄)^2-

5OH^- + Cr(OH)₃→(CrO₄)^2- + 4H₂O + 3e^-

Multiplting by 2,

10OH- + 2Cr(OH)3 →2(CrO4)2- + 8H2O + 6e-

Add both half reactions to get final result

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