Question

How much time in hours is required for a current of 2 amp to decompose electrolytically 18 gm of water?

• A. 96500
• B. 9650
• C. 26.8
• D. 2.68

Answer 1

Answer: (C)

26.8

18g of water$=\frac{18}{18}=1$ mole of $H_{2}O$

1 mole of water requires 2 moles of electron.

$2H_{2}O\to 2H_2+O_2$

$\therefore charge=2F=2\times 96500C$

$96500\times 2=2\times t$

$\therefore t=\frac{96500}{3600}$

$t=26.8$ hr

Option C is correct.