Question

Hint (a) How many minutes does it take to form 10.0 L of O, measured at 91.2 kPa and 28°C from water if a current of 1.3 A passes through the electrolytic cell? Solution minutes Guided Solution (6) What mass of H, forms? inces

Answer 1

Givene . Volume of 0 (V) = 1000L Pressure (P) = 91.2K Pa = 0.9000 74 aton Temprature (I) = 28°C = 301 K Alecording to ideal gas lawo PV=nRT n=pv. 0.900074 atm x 10L RT I 0.082057 Latm mollit X 3015 n= 9.00074 mol 24.699157 n=0.364 mol | Required Reactione 1 2h01) 24 (g) + 0 (g) . Reduction (Cathode) . 24+ (aq) + 2e hl) Oxidation - (Anode) - 240(1) O₂ (g) + 4HT (age) the I mol O₂ reacts with es = . Umol. 264 mol o reacts with e's = 4x0.364 = 1.4576 mol We know that Charge of 16 = 964858 charge of 1.4576 mol e = 9648581-4576 __ =140642.275 C CS Scanned with CamScanner

No. of Colcombs = current x tine 140642-275 - 1-3xt t = 140642=275 seconds 1-3 t=108186.365 seconds to 1803.106 min Ito 1803 min / (6) According to stoichiometry I noo of moles of th = 2x no. of moles of o noo of moles of th= 2x 6-36 44mol -07288 mol I mol of H₂ = 28 0-7288 mol of H - ŽXO- = 1.4577 g (5 significant) 1.458 g (4 significant) 1.48g (3 Sigurficant) - OR OR CS Scanned with CamScanner