For decolourisation of `1 "mol of" KMnO_(4)`, the moles of `H_(2)O_(2)` required is

Options

A. `1//2`
B. `3//21`
C. `5//2`
D. `7//2`

Answer

c

Explanation

Red half reaction
`MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
Oxid half reaction
`H_(2)O_(2)rarrO_(2)+2H^(+)+2e^(-)`
since 1 mole of `H_(2)O_(2)` requires `5e^(-)` therefore reduction of 1 mole of `KMnO_(4)` will require `5//2` molesof `H_(2)O_(2)`