Question

Number of coulombs required to liberate 0.5 mol of $O_2$ is:

• A. 19300
• B. 193000
• C. 96500
• D. 9650

Answer 1

The correct option is B 193000

Formation of $O_2$ is ${4e}^{-}$ transfer.

i.e. $4\times 96500C\to 1mol$ of $O_2$

$\therefore 0.5mol{O}_2\to \frac{1}{2}\times 4\times 96500$

$=193000C$