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HETAV PATEL

Grade 11,

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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

4 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
Mrinali
13 Points
5 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
askIITians Faculty 614 Points
3 years ago

Cr(OH)3 GIVES Cr+3 + 3OH- 

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

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

Now 

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

 

Multiplying first euation with three and second equation with six

Indian Sketcher Official
26 Points
one year 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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