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+)^0=+1.51V`
`:. E_(OP Mn^(2+)//Mn^(7+)^0=-1.51V`
`E_(RP CO_2//C_2O_4^(2-)^0=-0.49V`
`E_(OP C_2O_4^(2-)//CO_2)^0=+0.49V`
More is `E_(OP)^0` more is the tendency to get oxidise
`C_2O_4^(2-) to 2CO_2+2e,E_(OP)^0=+0.49v`
`MnO_4^- +8H^- +5e to Mn^(2+)+4H_2O`,
`E_(RP)^0=+1.51V`
`2MnO_4^- +5C_2O_4^(2-) +16H^- to `
`10CO_2+2Mn^+ 8H_2O`, n=10
10 electrons are used in redox change.
`:. E_("Cell")^0=E_(OP)^0+E_(RP)^0=0.49+1.51=2.0V` Also
`E^0=(0.059)/n log K_c`
`:. 2=(0.059)/(10)log K_c" ":. K_c=(10^(338.98))`