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
B. 1.5
C. 2.5
D. 3.5

Answer

C

Explanation

`ClO_(2)to Cl^(-)`
`ClO_(2)+2H_(2)O+5e^(-)toCl^(-)+4OH^(-)`
`H_(2)O_(2) to O_(2)`
`H_(2)O_(2)+2OH^(-)toO_(2)+2H_(2)O+2e^(-)`
`[ClO_(2)+2H_(2)O+5e^(-)to`
`Cl^(-)+4OH^(-)]xx2`
`([H_(2)O_(2)+2OH^(-)toO_(2)+2H_(2)O+2e^(-)]xx5)/(2ClO_(2)+5H_(2)O+2OHto2Cl^(-)+5O_(2)+6H_(2))`
`2ClO_(2)=5H_(2)O_(2)`
`ClO_(2)=2.5H_(2)O_(2)`
Hence on mole of `ClO_(2)` oxidises
2.5 moles of `H_(2)O_(2)`.