Question

Calculate the current at the cathode of a Hofmann Electrolytic cell with an Initial pH of the electrolyte solution of 5.56. A total time of 40min ,Cathode volume of 52mL and a final pH of 11.18 at the cathode.

Answer 1

In Hoff Man's Electrolytic cell We have given that :PH 5.56 Volume of Cathode = 52 hode & Some time requered during PHlinn = 11-18 electrolysis = uomin As we know moles of particle x n-factor & F = Q where, Q = charge & Current (1) x Time (Total) t F = Faraday = 96500c moles = amount of particle PH) : 5 56 =) - lo [HP] 2 pH), 2.11.18 -) - logpht] = 11:18 [H + 1 ; [HI] = = 9.75 x 10-6 6.6 x 10-12 10-6 X 52 mole moles of (H+]; = 19.75 X mol, 1000 =1.43010-7 mole = 16.6x10-12 x 52 mole moleof (H - 3x10-13 more moles of Htion tranferred : .432 X 10 -

: moles of particles = 1.432x 10-t molt 1-factor i. Ixt = moles x n-factor & F C for Ht = 1) I = 1.432x 10-7 x 1 x 96500 40x60 = 5.629 x 10-6 A = 5.630 mA