In alkaline medium, `ClO_(2)` oxidises `H_(2)O_(2) "to" O_(2)` and is itself reduced to `Cl^(ө)`. How many moles of `H_(2)O_(2)` are oxidised by `1 "mol of " ClO_(2)` ?

Options

A. `1.0`
B. `1.5`
C. `2.5`
D. `3.5`

Answer

C

Explanation

`overset(+4)(ClO_2) rarroverset(-1)(Cl^(-))` (Reduction)
Step 1 : `2H_2O+ClO_2 rarrCl^(-)+4OH^(-)`
Step 2 : `ClO_(2)+2H_2O + 5e^(-)rarrCl^(-)+4OH^(-) " "...(i)`
`H_2overset(-1)(O_2) rarroverset(0)O_2` (Oxidation)
Step 1: `2OH^(-)+H_2O_2rarrO_2+2H_2O`
Step 2 : `H_2O_2 +2OH^(-) rarrO_2 +2H_2O +2e^(-)" " …(ii)`
By applying eq. (i) `xx2 +` eq. (ii) `xx5` :
`2ClO_2 + 4H_2O + 5H_2O_2 + 10OH^(-) rarr 2Cl^(-) +8OH^(-) +5O_2 +10H_2O`
`2ClO_2 + 5H_2O_2 + 2OH^(-) rarr2Cl^(-)+ 5O_2 + 6H_2O`
5 moles of `H_2O_2` oxidize by 2 mole of `ClO_2`
5/2 moles of `H_2O_2` will oxidize by 1 mole of `ClO_2 .`