Question

3. How many grams of Ag(s) could be plated out on the cathode of an electrolytic cell by the passage of 5.0 Amps of current through AgNO3 solution for 1 hour? Faraday's number = 96485 Coul/mol. a) 5.5 x 10-ºg b) 0.19 g c) 5.2 x 105 g d) 2.0 g e) 20 g

Answer 1

ve out on the cathode an electro of cell day the passage current of AgNO3 Ihr ? number How many grams of Ag (8) could plated letic passage of of 5.0 Amps solution through for = 96 4 8 5 coul/mol Faraday the question from Electro chemistry more preciesly from faraday' ammount of Substanie deposited at an electrode directly proportional to ammout of This is and law. praday's first low : - "The is charge passed! py wap We amount of Substance a charge 2Q

where zos electrochemical equivalent As hee Know' 25 wlu E- Equivalent weight Faraday f 1 for WE EQ F as we Know li I current t time E It f

as we know, Equivalent weight Atomic weight n- factor comes to this question deta's are given dela's I= 5A If 96 4 8 5 t = I h 3600 see (as thr = 3600 see Atomic mass of Ag = 1088 at anode reaction is Ag + 1 + e- Ag NO: of electron lost L so) factor = 1

Now, equivalent weight of Ag molecular Ag weight of I / so Equivalent hecight of Ag 108 g as We Know € It + put all the deta in this given get, equation (108) (5) (8600 96 485 so / 011 201 g 209 or W 20 laprox)