6. At a potential of -1.00 V vs. SCE, carbon tetrachloride (CCl) in methanol und

Question

6. At a potential of -1.00 V vs. SCE, carbon tetrachloride (CCl) in methanol undergoes a two- electron reduction to chloroform (CHCl5) at a mercury cathode: 2CCl4 + 2H+ + 2 Hg + 2 e 2 CHCl3 HigpCi2(s) + At a potential of -1.80 V vs. SCE, chloroform (CHCls) in methanol undergoes a six-electron reduction to methane (CH4) at a mercury cathode: A 0.1248-gram sample, consisting of CCl4, CHCl, and some inert material, was dissolved in methanol and the resulting solution was subjected to a controlled-potential electrolysis at a was 12.68 SCE, and the quantity of electricity passed during this second electrolysis was 59.15 cou- lombs. Calculate the weight percentages of CCl4 and CHCl, in the original sample. mercury cathode held at-1,00 V vs. SCE; the quantity of electricity passed in this elcctrolysis passed in this electrolysis coulombs. Then the potential of the mercury cathode was adjusted to -1.80 V vs.

Explanation / Answer

From the reduction reaction , we can say that

to convert 2 moles of CCl4 to CHCl3 , 2 Faraday of electricity is required

so, 1 mole CCl4 = 154 g

2*96500 coulombs of electricity is required to convert 2*154 = 308 g of CCl4  

12.68 coulombs of electricity is required to convert 308*12.68/2*96500 = 0.020 g

0.020 g = 1.29*10-4 mole CCl4

hence amount of CHCl3 produced = 1.29*10-4 mole = 1.29*10-4 *119.5 = 0.015 g

In the 2nd reduction reaction

to convert 2 moles of CHCl3 to CH4,  6 Faraday of electricity is required

1 mole CHCl3 = 119.5 g

or

6*96500 coulombs of electricity is required to convert 2*119.5 g CHCl3

59.15 coulombs of electricity is required to convert 2*119.5*59.15/6*96500 = 0.0244 g

therefore total amount of CHCl3 = 0.0244 g

and mass of CHCl3 taken initially  = total mass of CHCl3 - mass of CHCl3 produced in the first reduction process

= 0.0244 - 0.015 = 0.0094 g

now, weight percentage of CCl4 = 0.020*100/0.1248 = 16.02 %

weight percentage of CHCl3 = 0.0094*100/0.1248 = 7.53 %

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