Question

If 245 milliamps were passed through a solution containing lead(ll) ions for 1800 seconds, with (1pts) lead strips seving as electrodes, how many grams of lead would be deposited? 0.474 0.494 0.419 D 0.149

Answer 1

Answer 2

given: Current (2) = 245 mA = 245 8103 Amp Time (© = 1800 sec. Moleculan Weight (Pb) = 207.2 g/mol Pb2+ + 2e- - Pb since amol = 96485c 2457103 Aup cursent is 245x183 e Poted change g = 1800 set x 2457109 = 441 C. F = Faradey Constant= 96485 mol Since nag 2=anu, of electron [llene=2] n = no. of moles deposit 245x cure 248210 ZF ५५ (2) (96485) 2m = 2.285 x 10^9 moll deposite meiss (g) = mole x molecular weight = 2.285 x 103 mot x 2042 So Jelipide mus (p) = 0,474 gull du Scanned with CamScanner