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

Anode reaction `: C_(2)O_(4)^(2-) rarr 2CO_(2)+cancel(2e^(-))]xx5`
Cathode reaction `:`
`cancel(5e^(-))+MnO_(4)^(c-)+8H^(o+) rarrMn^(2+)+4H_(2)O]xx2`
Cell reaction `:`
`ulbar(5C_(2)O_(4)^(2-)+2MnO_(4)^(c-)+16H^(o+)rarr10CO_(2)+2Mn^(2+)+8H_(2)O)`
`E^(c-)._(cell)=(E^(c-)._(reduction))_(c)-(E^(c-)._(reduction))_(a)`
`=1.51-(-0.49)=2.0V`
Cell representation `:`
`Pt,CO_(2)(g)|H_(2)C_(2)O_(4)H^(o+)||H^(o+),MnO_(4)^(c-)|Mn^(2+)|Pt`
`Delta_(r)G^(c-)=-nFe^(c-)._(cell)(n_(cell)=10)`
`=-10xx96500Cxx2.0V`
`=-193xx10^(4)J mol ^(-1)`
`Delta_(r)G^(c-)=-2.303RT log K_(c)`
`-193xx10^(4)J mol^(-1)=-2.303xx8.134 J K^(-1)mol ^(-1)xx298Kxxlog K_(c)`
`:. log K_(c)=338.08`
`K_(c)=Antil og(338.08)`
`:. K_(c)=1.2 xx 10^(338)`