Question

. A student carried out the electrolysis of a copper electrode in an electrolysis cell like the one employed in this experiment. The student weighed the electrode before the electrolysis and found its mass to be 35.251 g. During the electrolysis, a total volume of 96.30 mL of H2 was produced. Following the electrolysis the electrode had a mass of 35.008 g. The temperature in the laboratory was 25.2C and the barometric pressure was 975 mbar. 1 mol of H2 requires the passage of ________ faradays of charge. Loss in mass by anode _________________ g Equivalent mass of metal ___________________ g Molar mass of copper ___________________ g Charge, n, on cation ______________ Barometric pressure (1 atm = 1013.25 mbar) __________ mbar = __________ atm Temperature, T ___________ C = ____________ K Vapor pressure of H2O at T __________ mm Hg = __________ atm Partial Pressure of dry H2 ______________ atm Total volume of H2 produced, V __________ mL = __________ L

Answer 1

1 mol of H2 requires the passage of 2 mol faradays of charge.

Loss in mass by anode 35.251 g. - 35.008 = 0.243 g

Equivalent mass of metal = 32 g

Molar mass of copper = 64 g

Charge, n, on cation = 2

Barometric pressure = 975m bar = 975/1013.25 = 0.962 atm

Temperature, T 25.2⁰ C = 25.2 +273 = 298.2 K

Vapor pressure of H2O at T 23.8   mm Hg = 0.0313atm

Partial Pressure of dry H2 0.962 atm - 0.0313atm = 0.9307

Total volume of H2 produced, V 96.30 mL

Use PV=nRT to calculate moles of H2

R =0.082 L atm K−1 mol−1

n = 0.962atm x 0.0 963L/0.082 x 298.2 = 0.00378 moles

1 mole H2 requires passage of 2 faradays

No. of faradays passed = .00378 moles = 2faradays/1mol H2 = 0.00756 faraday

Loss of mass of metal anode = 0.243 g

No. grams of metal lost per faraday passed = 0.243g/.00756 faradays = 32 g = EM