Determine whether `O_(2)(g)` can oxidize sulphate `(SO_(4)^(2-))` ion to peroxodisulphate `(S_(2)O_(8)^(2-))` ion or not in an acidic solution with `O_(2)(g)` being reduced to water.
Given `: E^(c-)._(O_(2)|H_(2)O)=1.20V` and `E^(C-)._(S_(2)O_(8)^(2-)|2SO_(4)^(2-))=2.0V`

Explanation

The oxidation reaction of `SO_(4)^(2-)` to `S_(2)O_(8)^(2-)` and reduction reaction of `O_(2)` to `H_(2)O` are as below `:`
`(` Reduction `) O_(2)(g)+4H^(o+)(aq)+4e^(-) rarr 2H_(2)O," "E^(c-)._(red)=1.20V`
`(` Oxidation `)4SO_(4)^(2-) rarr 2S_(2)O_(8)^(2-)+4e^(-)" "E^(e-)._(o x i d )=-2.0V` `ulbar(Net reaction:O_(2)(g)+4SO_(4)^(2-)+4H^(o+) rarr2S_(2)O_(8)^(2-)+2H_(2)O)`
`E^(c-)._(cell)=E^(c-)._(red)+E^(c-)._(o x i d)`
`=E^(c-)._(O_(2)|2H_(2)O)+E^(c-)._(SO_(4)^(2-)|S_(2)O_(8)^(2-))`
`=1.20V-2.0V=-0.8V`
Since `E^(c-)._(cell)` is negative, thus, `O_(2)` will not oxidize `SO_(4)^(2-)` to `S_(2)O_(8)^(2-)`.