Redox Reaction: solve the following equation by ion electron method in acidic medium

 

Asked by kkdmmdsd | 9th Jun, 2019, 04:27: PM

Expert Answer:

In ion electron method, divide the equation in two part 
 
Mn to the power of plus 2 space end exponent plus space ClO subscript 3 to the power of minus rightwards arrow space MnO subscript 2 plus Cl to the power of minus
plus 2 space space space space space space space space plus 5 space space space space space space space space space space space plus 4 space space space space space minus 1

Oxidation minus
Mn to the power of plus 2 end exponent rightwards arrow MnO subscript 2

left parenthesis 1 right parenthesis space Add space straight H subscript 2 straight O space to space balance space straight O space atoms
Mn to the power of plus 2 end exponent plus 2 straight H subscript 2 straight O rightwards arrow MnO subscript 2

left parenthesis 2 right parenthesis space Add space straight H to the power of plus space space end exponent ions space to space balance space straight H space atoms
Mn to the power of plus 2 end exponent plus 2 straight H subscript 2 straight O rightwards arrow MnO subscript 2 plus 4 straight H to the power of plus

left parenthesis 3 right parenthesis Add space electrons space to space balance space charge space on space both space side
Mn to the power of plus 2 end exponent plus 2 straight H subscript 2 straight O rightwards arrow MnO subscript 2 plus 4 straight H to the power of plus plus 2 straight e to the power of minus space space space space space space space space space space space space space space space space............... left parenthesis 1 right parenthesis

Reuduction minus
ClO subscript 3 to the power of minus rightwards arrow Cl to the power of minus

left parenthesis 1 right parenthesis space Add space straight H subscript 2 straight O space to space balance space straight O space atoms
ClO subscript 3 to the power of minus rightwards arrow Cl to the power of minus plus 3 straight H subscript 2 straight O

left parenthesis 2 right parenthesis Add space straight H to the power of plus space space end exponent ions space to space balance space straight H space atoms
ClO subscript 3 to the power of minus plus 6 straight H to the power of plus rightwards arrow Cl to the power of minus plus 3 straight H subscript 2 straight O

left parenthesis 3 right parenthesis space Add space electrons space to space balane space charge space on space both space side
ClO subscript 3 to the power of minus plus 6 straight H to the power of plus plus 7 straight e to the power of minus rightwards arrow Cl to the power of minus plus 3 straight H subscript 2 straight O space space space space space space space space space space space space space space................... left parenthesis 2 right parenthesis


After space adding space equation space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis

We space will space get comma

2 ClO subscript 3 to the power of minus plus 7 Mn to the power of plus 2 end exponent plus 8 straight H subscript 2 straight O rightwards arrow 7 MnO subscript 2 space plus space 2 Cl to the power of minus plus space 16 straight H to the power of plus

Answered by Ravi | 10th Jun, 2019, 12:43: PM