A direct current of 0.125 ampere was passed through 200 mL of a 0.25M solution o

Question

A direct current of 0.125 ampere was passed through 200 mL of a 0.25M solution of Fe_2(SO_4)_3 between platinum electrodes for a period of 1.100 hours. Oxygen gas was produced at the anode. The only change at the cathode was a slight change in the color of the solution. At the end of the electrolysis, the electrolyte was acidified with sulfuric acid and was titrated with an aqueous solution of potassium permanganate. The volume of the KMnO_4 solution required to reach the end point was 24.65 mL. (a) How many faradays were passed through the system? (b) Write a balanced half-reaction for the process that occurred at the cathode during the electrolysis. (c) Write a balanced net ionic equation for the reaction that occurred during the titration with potassium permanganate. (d) Calculate the molarity of the KMnO_4.

Explanation / Answer

(a.) Amperes x time = Coulombs

     Time = 1.100 hr x (3600 sec / hr) = 3960 sec

     0.125 amperes x 3960 sec = 495 coulombs

     495 coulombs x (1 faraday / 96500 coulombs) = 5.13 x 10-3 F

(b) A balanced half-reaction for the process that occurred at the cathode during the electrolysis:

          Fe3+ + e- -------> Fe2+

(c) A balanced net ionic equation for the reaction that occurred during the titration with potassium permanganate:

         MnO4- + 5Fe2+ + 8H+ ------> 5Fe3+ + Mn2+ + 4H2O

d) Calculate the molarity of the KMnO4 solution.

     5.13 x 10-3 faraday x (1 mole Fe2+/ 1 faraday) = 5.13 x 10-3 mole Fe2+

     5.13 x 10-3mole Fe x (1 mole MnO4-/ 5 mole Fe2+) = 1.03 x 10-3 mole MnO4-

     1.03 x 10-3mole MnO4- / 24.65*0.001 liter = 0.0416 M MnO4- (or KMnO4)

     1.03 x 10-3mole MnO4- / 0.02465 liter = 0.0416 M KMnO4

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