Question

35. In the electrolysis of a solution containing Ni2+(aq), metallic Ni(s) deposits on the cathode. Using a current of 0.150 A for 12.2 min, what mass of nickel will form? (a) 0.0580 g (b) 5.56 x 104 g (c) 0.116 g (d) 1.86 x 10-5 (e) 0.025 g

Answer 1

Step 1 Textual Explanation : According to finst law of Farraday - during electrolysis, the mass of the metal diposited in the electrode is directly propora tional to the amount of electron city passes throough the dectro. lytes. so w do a w=2x8 Here, 2 = proportionality Constant] Again, if I ampere current is passes in t second time, then d=Ixt : Coulomb = amperex second o W = 2xIxt

Here, w = mass of metal diposited I > Electric current in ampere z – Electrochemical equivalent. t = time of electrolysis in S. Step 2 : Calculation of I : Here, the reaction occured is Ni(aq) +2e - Ni (s) 58.7 g 0 24 Z = Electrochemical equivalent - Equivalent weight of Ni Faraday's constant Atomic weight of Ni Valency of Ni Faraday's constant Now, Farraday's constant = 96500 C

Atomic weight of Ni = 58.79 valency of Ni=2. 58:7/2 : Z = = 96500 g. C-1 – 3.04 x 10-4 9.C-1 Step 3: Datas : We have the given datas as below ( Electric current I = 0.150 A Time, t = 12.2 min =12.2 x60 s. a t =732 si Electroochemical Equivalent, 2= 3.04 x 10-4 9.6-1 [Calculated in step-2]

- 1 Step-4 Final Calculation : Now plugging all the datas. from step-3 in the equation in step-1, we get . . W = 3.04 x 1079 9.2 x 0.150 A x 732 s = 3.04 x 109 g. 5 x 109.8 & : C = 4 x 5 = 0.03 g [~ 0.025 3] Ans So, the correct option i.e. ans. wer is - |(e): 0.025 g Aus