Question

Dal lake has water 8.2 x 1012 litre approximately. A power reactor produces electricity at the rate of 1.5 x 108 coulomb per second at an appropriate voltage.How many years would it take to electrolyse the lake?

Answer 1

Answer:

Given data:

Dal lake has water approximately 8.2 * 10¹² litres

Rate of electricity produced by the power reactor = 1.5 * 10⁸ C/s

To find: no. of years to electrolyse the lake

Solution:

The reactions at anode and cathode are as follows:

At anode: 2H₂O → 4H⁺ + O₂ + 4e⁻

At cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻

In order to balance the above two reactions, we will multiply the reaction at cathode by 2 throughout.

By doing so we can see that,

4 moles of electrons are required to electrolyse 2 moles of H2O which is

= 18 * 2 = 36 g of H2O ….. [∵ the molecular weight of water is 18 g/mol]

We know the density of water is 1000 gm/litre

∴ The volume of water = 36/1000 = 36 * 10⁻³ litre

So, 36 * 10⁻³ litre of H2O we require 4 moles of electrons

∴ No. of eletrons required to electrolyse 8.2 * 10¹² litres of H2O

= [4/36 * 10⁻³] * 8.2 * 10¹²

Now,

The total charge required to electrolyse is given as,  

Q = (no. of electrons) * (Faraday’s constant)  

⇒ [I * t] =  [{4/36 * 10⁻³}* 8.2 * 10¹²] * 96500

⇒ 1.5 * 10⁸ * t = [{4/36 * 10⁻³}* 8.2 * 10¹²] * 96500

⇒ t = [791300/13.5] * 10⁷

⇒ t = 58614.81 * 10₇ sec

⇒ t = 5.8 * 10¹¹ sec  

⇒ t = 1.83 * 10⁴ years

Thus, the no. of years required to electrolyse the lake is  1.83 * 10⁴ years