If the oxidation of oxalic acid by acidic `MnO_(4)^(c-)` solution is carried out in a reversible cell, then what is the electrode reaction and equilibirum constant of the cell reaction.
Given `:`
`E^(c-)._((MnO_(4)^(c-)|Mn^(2+)))=1.51V`
`E^(c-)._((CO_(2)|C_(2)O_(4)^(2-)))-0.49V`

Explanation

`E_(RP_(Mn^(7+)//Mn^(2+)))^(@) = 1.51V`
`E_(OP_(Mn^(2+)//Mn^(7+)))^(@) = 1.51V`
`E_(RP_(CO_(2)//C_(2)O_(4)^(2-)))^(@) = 1.51V`
`E_(RP_(C_(2)O_(4)^(2-).//CO_(2)))^(@) = 0.49V`
More is `E_(OP)^(@)`, more is the tendency to get oxidise
`{:(C_(2)O_(4)^(2-)rarr2CO_(2)+2e E_(OP)^(@) = +0.49V),(MnO_(4)^(-)+8H^(+)+5erarrMn^(2+)+4H_(2)O),(" "E_(RP)^(@
) = +1.51V):}/(2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 10CO_(2)+2Mn^(+)+8H_(2)O,)`
`n = 10 because 10` electrons are used in redox change.
`:. E_(Cell)^(@) = E_(OP)^(@)+E_(RP)^(@) = 0.49+1.51 = 2.0V`
Also `E^(@) = (0.059)/(n)logK_(c)`
`:. 2 = (0.059)/(10)"logK_(c)`
`:. K_(c) = 10^(338.98)`