In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is :

Question

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is : (A) 10 (B) 2 (C) 1 (D) 5

Tardigrade Admin · Accepted Answer

2 overset text+ 7M nO4 + 5C2 O2-4 + 16H+ arrow 2 overset text+ 2 M n2+ +10 CO2 + 8H2 O 10 e- trans for 10 molecules of CO2 so per molecule of CO2 transfer of e- is '1'

Q. In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of $C O_{2}$ is :

10

2

1

5

$2 M + 7 n O_{4} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M + 2 n^{2 +} + 10 C O_{2} + 8 H_{2} O$
$10 e^{-}$ trans for $10$ molecules of $C O_{2}$ so per molecule of $C O_{2}$ transfer of $e^{-}$ is $^{'} 1^{'}$

Q. In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2​ is :

10

2

1

5

2M+ 7nO4​+5C2​O42−​+16H+→2M+ 2 n2++10CO2​+8H2​O 10e− trans for 10 molecules of CO2​ so per molecule of CO2​ transfer of e− is ′1′

Q. In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of $C O_{2}$ is :

$2 M + 7 n O_{4} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M + 2 n^{2 +} + 10 C O_{2} + 8 H_{2} O$
$10 e^{-}$ trans for $10$ molecules of $C O_{2}$ so per molecule of $C O_{2}$ transfer of $e^{-}$ is $^{'} 1^{'}$