Question

Canada produces about 1.5 million tonnes of chlorine per year by the electrolysis of brine (NaCl(aq)). If the required electrolytic cells are operated at a potential difference of 4.5 V, calculate how many total kilowatt hours (kWh) of electrical power are required. 1 kWh = 3.6 × 10⁶ J.

Answer 1

2Cl^- -----> Cl₂

Therefore n = 2, 2 electrons

We know that 1F – 96500 Q (Q=C mol⁻¹ e^-1)

Therfore 1 mole of Cl₂ (70.9 g molar mass) require 2 x 96500 Q of electricity.

Then 1.5 million tonnes (1.5 x 10¹² g ) requires=? (X) (1 million ton=1x10¹² g)

X = (2 x 96500 Q x 1.5 x 10^12 g)/ 70.9 g = 4.08 x 10¹⁵ Q
Now per 1 hour (3600s): charge Q = It; where I=current, t=time

I = Q / t = 4.08 x 10¹⁵ Q / 3600 s = 1.13 x 10¹² A

The power: P = IU

Where U= potential difference,

P= 1.13 x 10¹² A x 4.5 V = 5.1 x 10¹² W = 5.1 x 10⁹ kW. (1kW=1000 W)

For 1 year process: 365 days = 8760 hours

5.1 x 10⁹ kW x 8760 h = 4.47 x 10¹³ kWh electric power are required.