At a potential of -1.0 V (versus SCE), CCl₄ in methanol is reduced to CHCl₃ at a mercury cathode:

Several different 0.750-g samples containing CCl₄, CHCl₃, and inert organic species were dissolved in methanol and electrolyzed at -1.0 V until the current approached zero. A coulometer indicated the charge required to complete the reaction as given in the middle column of the following table. The potential of the cathode was then adjusted to -1.8 V. The additional charge given in the last column of the table was required at this potential.

Calculate the percentage of CCl₄ and CHCl₃ in each mixture.

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2CCI, + 2H* + 2e + 2Hg(/) → 2CHCI, + Hg,C2(s) At -1.80 V, the CHCI, further reacts to give CH4: 2CHCI; + 6H* + 6c + 6Hg(/) → 2CH, + 3Hg,Cl,(s)