Question

At a potential of 21.0 V (versus SCE), carbon tetrachloride in methanol is reduced to chloroform at a Hg cathode:

$2{\mathrm{CCl}}_4+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}+2\mathrm{Hg}\left(\mathrm{l}\right)\to 2{\mathrm{CHCl}}_3+{\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(\mathrm{s}\right)$

At 21.80 V, the chloroform further reacts to give methane:

$2{\mathrm{CHCl}}_3+6{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}+6\mathrm{Hg}\left(\mathrm{l}\right)\to 2{\mathrm{CH}}_4+3{\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(\mathrm{s}\right)$

Several 0.750-g samples containing CCl₄, CHCl₃, and inert organic species were dissolved in methanol and electrolyzed at 21.0 V until the current approached zero. A coulometer indicated the charge required to complete the reaction, as given in the second column of the following table. The potential of the cathode was then adjusted to 21.80 V. The additional charge required to complete the reaction at this potential is given in the third column of the table. Calculate the percent CCl₄ and CHCl₃ in each mixture.

Short answer

Sample	% CCl4	%CHCl3
1	2.30	2.02
2	4.36	1.48
3	1.32	1.50
4	2.47	1.86

Step-by-step solution

Given Information

At – 1.0 V (versus SCE)

$2{\mathrm{CCl}}_4+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}+2\mathrm{Hg}\left(\mathrm{l}\right)\to 2{\mathrm{CHCl}}_3+{\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(\mathrm{s}\right)$

At – 1.80 V,

$2{\mathrm{CHCl}}_3+6{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}+6\mathrm{Hg}\left(\mathrm{l}\right)\to 2{\mathrm{CH}}_4+3{\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(\mathrm{s}\right)$

Samples of 0.750 g containing CCl₄,CHCl₃ and other inert organic species. They were dissolved in methanol and electrolyzed at – 1.0 V. The potential of the cathode is then adjusted to -1.80 V.

Sample No:	Charge required at – 1.0 V, C	Charge required at – 1.8 V, C
1	10.84	69.20
2	20.52	88.50
3	6.22	45.98
4	11.60	68.62

Explanation

Sample 1

At – 1.0 V,

$10.84\mathrm{C}\times \frac{1\mathrm{mole}}{96485\mathrm{C}}\times \frac{2\mathrm{mol}{\mathrm{CCl}}_4}{2\mathrm{mole}}=1.12\times {10}^{-4}\mathrm{mol}$

At – 1.80 V,

role="math" localid="1649393612805" $69.20\mathrm{C}\times \frac{1\mathrm{mole}}{96485\mathrm{C}}\times \frac{2\mathrm{mol}C{\mathrm{HCl}}_3}{6\mathrm{mole}}=2.39\times {10}^{-4}\mathrm{mol}$

Initial number of moles of CHCl₃ present in the sample

$2.39\times {10}^{-4}\mathrm{mol}-1.12\times {10}^{-4}\mathrm{mol}=1.27\times {10}^{-4}\mathrm{mol}$

$\%{\mathrm{CCl}}_4=\frac{1.12\times {10}^{-4}\mathrm{mol}\times 153.82\mathrm{g}/\mathrm{mol}}{0.750\mathrm{g}}\times 100\%=2.297\%$

$\%C{\mathrm{HCl}}_3=\frac{1.27\times {10}^{-4}\mathrm{mol}\times 119.37\mathrm{g}/\mathrm{mol}}{0.750\mathrm{g}}\times 100\%=2.02\%$

Likewise, we can calculate the % CCl₄ and % CHCl₃ in the sample using a spread sheet.

Sample	charge at -1.0 V	charge at -1.8 V	mol CCl4	mol CHCl3	Initial moles of CHCl3	% CCl4	%CHCl3
1	10.84	69.2	0.00011235	0.00023907	0.000126721	2.30	2.02
2	20.52	88.5	0.00021268	0.00030575	9.30715E-05	4.36	1.48
3	6.22	45.98	6.4466E-05	0.00015885	9.43843E-05	1.32	1.50
4	11.6	68.62	0.00012023	0.00023707	0.00011684	2.47	1.86