In aqueous alkaline solution two electron reduction of `HO_(2)^(-)` gives

Options

A. `HO^(-)`
B. `H_(2)O`
C. `O_(2)`
D. `O_(2)^(-)`

Answer

a

Explanation

`HO_(2)^(-)rarrOH^(-)`
O.N of each O atom in `HO_(2)^(-)` is -1 while that in `OH^(-)` is -2 in other words total O.N of two O atoms in `HO_(2)^(-)` is -2 while that of two O atoms in `OH^(-)` -4 therefore to balance O atoms on either side add `2e^(-)` to lL.H.S and one doing so we have
`HO_(2)^(-)+2e^(-)rarr 2OH^(-)`
Since the reaction is occuring in basic medium therefore to balance H atoms add one `H_(2)O` molecules to L.H.S and one `KO^(-)` ion to R.H.S of eqn (ii) Thus the balanced equation is
`HO_(2)^(-)+H_(2)O+2e^(-)rarr3OH^(-)`