First slide

Faraday's Laws

Question

When 0.1 mol $M n O_{4}^{2 -}$ is oxidised, the quantity of electricity required to completely oxidise $M n O_{4}^{2 -}$ to $M n O_{4}^{1 -}$ is

Easy

Solution

The oxidation reaction is
$\underset{0.1 m o l}{\overset{+ 6}{M n} O_{4}^{2 -}} \underset{}{\to} \underset{}{\overset{+ 7}{M n} O_{4}^{-}} + \underset{0.1 m o l}{e^{-}}$
$Q = 0.1 \times F = 0.1 \times 96500 C = 9650 C$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App