During the Kolbe's electrolytic decarboxylation of CH_3COONa

1.	During the Kolbe's electrolytic decarboxylation of CH_3COONa
Answer» `H_2` is LIBERATED at CATHODE because DISCHARGE potential of `H^+` is less than discharge potential of `Na^+` `CH_3COO^-` is discharged preferentially at anode than `OH^-` because discharge potential of `CH_3COO^-` is less than discharge potential of `OH^-` NaOH is discharged at cathode `CO_2` is liberated at anode along with alkane Solution :During the KOLBE's electrolytic decarboxylation of RCOONa in aqueous medium `RCOONa to RCOO^(-) +Na^(+) , H_2O to H^(+) + OH^(-)` (a) is correct because the discharge potential of `H^+` is less than the discharge potential of `Na^+` because `E_(H^+//H_2)^o=0, E_(Na^(+)//Na)^o=-Ve` HENCE, `H^+` is discharged at cathode as `H_2` (b)is correct because the discharge potential of `RCOO^-` is less than the discharge potential of `OH^-` , hence `RCOO^-` is discharged at anode. (d) is correct At anode `R-COO^(-) to R-undersetunderset(O)(\|\|)C-oversetdotO+e^-` , `R-undersetunderset(O)(\|\|)C-oversetdotO to R^(dot)+CO_2 , R^(dot)+R^(dot) to R-R`

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