A 212 mL sample of a 0.500 M NaCl solution with an initial pH of 7.00 is subject

Question

A 212 mL sample of a 0.500 M NaCl solution with an initial pH of 7.00 is subjected to electrolysis. After 17 minutes, a 10.0-mL portion (or aliquot) of the solution was removed from the cell and titrated with 0.100 M HCl solution. The endpoint in the titration was reached upon addition of 22.2 mL of HCl.

Part A

Assuming constant current, what was the current (in A) running through the cell?

A 212 mL sample of a 0.500 M NaCl solution with an initial pH of 7.00 is subjected to electrolysis. After 17 minutes, a 10.0-mL portion (or aliquot) of the solution was removed from the cell and titrated with 0.100 M HCl solution. The endpoint in the titration was reached upon addition of 22.2 mL of HCl.

Part A

Assuming constant current, what was the current (in A) running through the cell?

Explanation / Answer

0.1 M HCl means 0.1 moles in 1000 mL

so in 22.2 mL, moles of HCl = (0.1/1000) X 22.2 = 0.00222 moles

that means 0.00222 moles of NaOH was formed during the electrolysis in 10 mL solution

so in 212 mL solution no. of moles of NaOH produced = 0.00222 x 212/10 = 0.047 mols

so 0.047 mols of electrons have passed in 17 X 60 = 1020 seconds

1 mol of electrons = 96500 C

so 0.047 mols of electrons = 96500 X 0.047 = 4542 C

q = I X t

therefore 4542 C = I X 1020 sec

therefore I = 4542 C /1020 s = 4.45 A

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