Question

A 0.1002 g sample containing only CCl4 and CHCl3 was dissolved in methanol and electrolyzed at the surface of a mercury electrode at –1.0 V, which required 50.20 s at a constant current of 0.200 A. The potential of the cathode was then adjusted and held constant at –1.80 V. Completion of the titration at this potential required 382.35 s at a constant current of 0.200 A. Calculate the respective percentages of CCl4 and CHCl3 in the mixture. The electrolysis reactions are given below. 2CCl4 + 2H+ + 2e- + 2Hg(l) → 2CHCl3 + Hg2Cl2(s) [occurs at –1.0V] 2CHCl3 + 6H+ + 6e- + 6Hg(l) → 2CH4 + 3Hg2Cl2(s) [occurs at –1.80V]

Answer 1

Formula: The charge (Q) = It, where 'I' is the current in Amp. and 't' is the time in sec.

Therefore, the charge required by CCl₄ = 0.2*50.2, i.e. 10.04 Coulombs

According to Faraday's laws, mass of CCl₄ deposited at the electrode = (E_CCl4/F)*Q, where E_CCl4 = equivalent weight of CCl₄, i.e. 154 g.eq^-1

= (154/96485)*10.04, i.e. 0.016 g.

Therefore, the percent weight of CCl4 in the given mass of mixture = (0.016/0.1002)*100, i.e. ~16%

The charge required by CHCl₃ = 0.2*382.35, i.e. 76.47 Coulombs

Here, the voltage needed = 1.8 V, i.e. the charge required by CHCl₃ = 76.47/1.8, i.e. 42.48 Coulombs

According to Faraday's laws, mass of CHCl₃ deposited at the electrode = (E_CHCl3/F)*Q, where E_CHCl3 = equivalent weight of CHCl₃, i.e. 119.5 g.eq^-1

= (119.5/96485)*42.48, i.e. 0.053 g.

Therefore, the percent weight of CCl4 in the given mass of mixture = (0.053/0.1002)*100, i.e. ~53%