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NaiO3+ NaI+ H2O

(Alkaline Medium)

-->

Balance the following Redox Reaction by ion Electron Method :->

I2(iodine)+ NaOH-----> NaiO3+ NaI+ H2O

(Alkaline Medium)

Asked by Anish 10th August 2018, 2:20 PM
Answered by Expert
Answer:
Step 1: Write down the skeleton equation:
 
straight I subscript 2 space plus space NaOH space rightwards arrow over leftwards arrow with blank on top space NaIO subscript 3 space plus space NaI space plus space straight H subscript 2 straight O
Step 2:
a)Assign the oxidation numbers for each atom in the equation.
 
begin mathsize 14px style blank subscript 0 space space space plus 1 space minus 2 space plus 1 space space space space space space space space space plus 1 space plus 5 space minus 2 space space space space space plus 1 minus 1 space space space space plus 1 space minus 2 end subscript
straight I subscript 2 space plus space N a O H space space rightwards arrow over leftwards arrow space space space space space NaIO subscript 3 space plus space NaI space plus space straight H subscript 2 straight O end style
 
b) Identify which reactants are oxidised and which are reduced.
 
begin mathsize 14px style O x i d a t i o n colon space blank subscript 0 space space space space space space space space space space space space plus 1 space plus 5 space minus 2 space space space end subscript
straight I subscript 2 space space rightwards arrow space space space space NaIO subscript 3 space
end style
 
begin mathsize 14px style R e d u t i o n colon space blank subscript 0 space space space space space space space space space space space space plus 1 space minus 1 space space space end subscript
straight I subscript 2 space space rightwards arrow space space space space NaI space
end style
 
Step 3: Balanced the atoms in each half-cell reaction.
a) Balance all other atoms other than hydrogen and oxygen
 
begin mathsize 14px style Oxidation colon
space space space
straight I subscript 2 space plus space 2 NaOH space rightwards arrow space space space 2 NaIO subscript 3 space
end style
 
begin mathsize 14px style Redution colon

straight I subscript 2 space space plus space 2 NaOH rightwards arrow space space 2 NaI space
end style
 
b) Balance the oxygen atoms 
 
begin mathsize 14px style Oxidation colon
space space space
straight I subscript 2 space plus space 2 NaOH space plus space 4 straight H subscript 2 straight O rightwards arrow space space space 2 NaIO subscript 3 space
end style
 
begin mathsize 14px style Redution colon

straight I subscript 2 space space plus space 2 NaOH rightwards arrow space space 2 NaI space plus space straight H subscript 2 straight O
end style
c) Balance the hydrogen atoms
 
begin mathsize 14px style Oxidation colon
space space space
straight I subscript 2 space plus space 2 NaOH space plus space 4 straight H subscript 2 straight O rightwards arrow space space space 2 NaIO subscript 3 space plus space 10 straight H to the power of plus
end style
 
begin mathsize 14px style Redution colon

straight I subscript 2 space space plus space 2 NaOH space plus space 2 straight H to the power of plus rightwards arrow space space 2 NaI space plus space 2 straight H subscript 2 straight O
end style
d) For reactions in alkaline medium, add one OH- ion to each side for every H+ ion present in the equation.
 
begin mathsize 14px style Oxidation colon
space space space
straight I subscript 2 space plus space 2 NaOH space plus space 4 straight H subscript 2 straight O space plus space 10 OH to the power of minus rightwards arrow space space space 2 NaIO subscript 3 space plus space 10 straight H subscript 2 straight O
end style
 
begin mathsize 14px style Redution colon

straight I subscript 2 space space plus space 2 NaOH space plus space 2 straight H subscript 2 straight O rightwards arrow space space 2 NaI space plus space 2 straight H subscript 2 straight O space plus space 2 OH to the power of minus
end style
Step 4: Now, balance the charge.
 
begin mathsize 14px style Oxidation colon
space space space
straight I subscript 2 space plus space 2 NaOH space plus space 4 straight H subscript 2 straight O space plus space 10 OH to the power of minus rightwards arrow space space space 2 NaIO subscript 3 space plus space 10 straight H subscript 2 straight O space plus space 10 space straight e to the power of minus
end style
 
 
begin mathsize 14px style Redution colon

straight I subscript 2 space space plus space 2 NaOH space plus space 2 straight H subscript 2 straight O space plus space 2 straight e to the power of minus rightwards arrow space space 2 NaI space plus space 2 straight H subscript 2 straight O space plus space 2 OH to the power of minus
end style
Step 5: The electron lost in oxidation half-cell reaction must be equal to the electron gain in the reduction half-cell reaction. to make equal multiply the coefficients of all species by integers producing the lowest common multiple.
 
begin mathsize 14px style Oxidation colon
space space space
straight I subscript 2 space plus space 2 NaOH space plus space 4 straight H subscript 2 straight O space plus space 10 OH to the power of minus rightwards arrow space space space 2 NaIO subscript 3 space plus space 10 straight H subscript 2 straight O space plus space 10 space straight e to the power of minus space space space space space m u l t i p l y space b y space 1
end style
 
 
begin mathsize 14px style Redution colon

straight I subscript 2 space space plus space 2 NaOH space plus space 2 straight H subscript 2 straight O space plus space 2 straight e to the power of minus rightwards arrow space space 2 NaI space plus space 2 straight H subscript 2 straight O space plus space 2 OH to the power of minus space space space space space space m u l t i p l y space b y space 5
end style
 
 We get
 
begin mathsize 14px style Oxidation colon
space space space
straight I subscript 2 space plus space 2 NaOH space plus space 4 straight H subscript 2 straight O space plus space 10 OH to the power of minus rightwards arrow space space space 2 NaIO subscript 3 space plus space 10 straight H subscript 2 straight O space plus space 10 space straight e to the power of minus space space space space space
end style
 
 
begin mathsize 14px style Redution colon

5 straight I subscript 2 space space plus space 10 NaOH space plus space 10 straight H subscript 2 straight O space plus space 10 straight e to the power of minus rightwards arrow space space 10 NaI space plus space 10 straight H subscript 2 straight O space plus space 10 OH to the power of minus space space space space space space
end style
Step 6: Add both half-reactions together, we get,

 
6 straight I subscript 2 space plus 12 NaOH space plus space 14 straight H subscript 2 straight O space plus space 10 OH to the power of minus space plus space 10 straight e to the power of negative space end exponent space rightwards arrow over leftwards arrow with blank on top space 2 NaIO subscript 3 space plus 10 space NaI space plus space 20 straight H subscript 2 straight O space plus space 10 space straight e to the power of minus space plus space 10 OH to the power of minus
 
Step 7: Simplify the equation:

 
3 straight I subscript 2 space plus space 6 NaOH space rightwards arrow over leftwards arrow with blank on top space NaIO subscript 3 space plus space 5 NaI space plus space 3 straight H subscript 2 straight O
Answered by Expert 11th August 2018, 10:05 PM
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