NEET Class neet Answered
In most situations of balancing an equation, you are not told whether the reaction is redox or not.
In these circumstances, you can use a procedure called the oxidation number method.
The skeleton equation is:
Cr(OH)3 + IO3- → CrO42- + I-
Step 2
(a) The oxidation number of various atoms involved in the reaction.
+3 -2 +1 +5 -2 +6 -2 -1
Cr(OH)3 + IO3- → CrO42- + I-
(b) Identify and write out all the redox couple in reaction.
+3 -2 +1 +6 -2
Oxidation: Cr(OH)3 → CrO42- + 3e-
+5 -2 -1
Reduction: IO3- + 6e- → I-
Step 3
Balance the atoms in each half reaction
a) Balance all other atoms except hydrogen an oxygen
+3 -2 +1 +6 -2
Oxidation: Cr(OH)3 → CrO42- + 3e-
+5 -2 -1
Reduction: IO3- + 6e- → I-
b) Balance the charge:
+3 -2 +1 +6 -2
Oxidation: Cr(OH)3 + 5OH- → CrO42- + 3e-
+5 -2 -1
Reduction: IO3- + 6e- → I- + 6OH-
c) Balance the oxygen atoms
+3 -2 +1 +6 -2
Oxidation: Cr(OH)3 + 5OH- → CrO42- + 3e- + 4H2O
+5 -2 -1
Reduction: IO3- + 6e- + 3H2O → I- + 6OH-
Step 4: Make electron gain equivalent to electron lost.
+3 -2 +1 +6 -2
Oxidation: Cr(OH)3 + 5OH- → CrO42- + 3e- + 4H2O ................ multiply by 2
+5 -2 -1
Reduction: IO3- + 6e- + 3H2O → I- + 6OH- ................ multiply by 1
We get,
+3 -2 +1 +6 -2
Oxidation: 2Cr(OH)3 + 10OH- → 2CrO42- + 6e- + 8H2O
+5 -2 -1
Reduction: IO3- + 6e- + 3H2O → I- + 6OH-
Step 5: Add the half-reactions together.
+3 -2 +1 +5 -2 +6 -2 -1
2Cr(OH)3 + IO3- + 10OH- + 6e- + 3H2O → 2CrO42- + 6e- + 8H2O + I- + 6OH-
Step 6 : Simplify the equation:
+3 -2 +1 +5 -2 +6 -2 -1
2Cr(OH)3 + IO3- + 4OH- → 2CrO42- + I- + 5H2O
and we will get the balanced equation,
2Cr(OH)3 + IO3- + 4OH- → 2CrO42- + I- + 5H2O