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)` ?

Answer

C

Explanation

`ClO_(2) to Cl^(-)`
`ClO_(2) + 2H_(2)O + 5e^(-) to Cl^(-) + 4OH^(-)`
`H_(2)O_(2) to O_(2)`
`H_(2)O_(2) + 2OH^(-) to O_(2) + 2H_(2)O + 2e^(-)`
`ClO_(2) + 2H_(2)O + 5e^(-) to Cl^(-) + 4OH^(-) xx 2`
`H_(2)O_(2) + 2OH^(-) to O_(2) + 2H_(2)O + 2e^(-) xx 5`
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`2ClO_(2) + 5H_(2)O_(2) + 2OH^(-) to 2Cl^(-) + 5O_(2) -6H_(2)O`
`2ClO_(2) equiv 5H_(2)O_(2)`
`therefore ClO_(2) = 2.5H_(2)O_(2)`