A 0.1002 g sample containing only CCl4 and CHCl3 was dissolved in methanol and e

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]

Explanation / Answer

Total charge required by CCl4 = current*time in seconds = 0.2*50.2 = 10.04 Culombs

Thus, as per Faraday's law, equivalents of CCl4 present = 10.04/96500 = 1.04*10-4

Now,

Total charge required by CHCl3 = current*time in seconds = 0.2*382.35 = 76.47 Columbs

Thus, as per Faraday's law, equivalents of CHCl3present at the time of reaction = 76.47/96500 = 7.92*10-4

Thus, equivalents of CHCl3 present in the mixture = (7.92*10-4) - (1.04*10-4) = 6.88*10-4

Now, equivalent mass of CCl4 = molar mass/2 = 154/2 = 77 g/equivalent

Equivalent mass of CHCl3 = molar mass/6 = 119.5/6 = 19.92 g/equivalent

Thus, mass of CCl4 present = number of equivalents*equivalent mass = 0.008 g

Mass of CHCl3 present = number of equivalents*equivalent mass = 0.014 g

Thus, % of CCl4 = (mass of CCl4/mass of mixture)*100 = 36.36%

% of CHCl3 = 63.64 %

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