Question

8.2 × 10¹² litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 × 10⁶ Cs⁻¹ at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake? Assume that there is no loss of water except due to electrolysis.

Answer 1

Electrolysis of water

At anode:

\[\ce{2H2O -> 4H^+ + O2 + 4e^-}\] .......(1)

At cathode:

\[\ce{2H2O + 2e^- -> H2 + 2OH^-}\] ............(2)

Overall reaction:

\[\ce{6H2O -> 4H^+ + 4OH^- + 2H2 + O2}\]

(or) Equation (1) + (2) × 2

\[\ce{2H2O -> 2H2 + O2}\]

∴ According to faradays Law of electrolysis, to electrolyse two mole of Water (36 g ≃ 36 mL of H₂O), 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated = 4 × 96500 C.

∴ When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to

= `(4 xx 96500  "C")/(36  "mL") xx 9 xx 10^12  "L"`

= `(4 xx 96500 xx 9 xx 10^12)/(36 xx 10^-3) "C"`

= 96500 × 10¹⁵ C

∴ Given that in 1 second, 2 × 10⁶ C is generated therefore, the time required to generate 96500 × 10¹⁵ C is

= `(1  "S")/(2 xx 10^6  "C") xx 96500 xx 10^15  "C"`

= 48250 × 10⁹ S

∴ Number of years = `(48250 xx 10^9)/(265 xx 24 xx 60 xx 60)`

= 1.5299 × 10⁶ years

1 year = 365 days

= 365 × 24 hours

= 365 × 24 × 60 min

= 365 × 24 × 60 × 60 sec.