In Electrolysis and Electrolytic Cells , a DC power supply was used for applied
ID: 835217 • Letter: I
Question
In Electrolysis and Electrolytic Cells, a DC power supply was used for applied voltage in an electrolytic cell. The average voltage difference across two 3.9-ohm resistors in series was 0.978 V for 26.0 minutes while collecting 27.7 mL H2(g) at atmospheric pressure at the cathode.
a. If the barometric pressure was 755.1 mmHg, and the temperature was 21.9 oC, calculate the number of moles of H2 (gas) produced, given the vapor pressure of water at 20.0 oC is 17.5 mmHg. (1 point)
Hint: Use ideal gas (R = 0.0821 L atm deg-1 mol-1) and P(total) = P(H2) + P(H2O).
b. Calculate the total charge, in Coulombs, that flowed through the circuit while the 27.7 mL H2(g) was produced. (1 point)
Hint: According to Ohm
Explanation / Answer
(a) P(H2) = P(total) - P(H2O)
= 755.1 - 17.5 = 737.6 mmHg
= 737.6/760 = 0.97053 atm
Ideal gas equation: PV = nRT
Moles of H2 = n = PV/RT
= 0.97053 x (27.7/1000)/(0.0821 x (273.15 + 21.9))
= 0.0011098 = 0.00111 mol = 1.11 x 10^(-3) mol
(b) Current = voltage/resistance
= 0.978/(2 x 3.9) = 0.1254 A
Charge = current x time in seconds
= 0.1254 x 26.0 x 60
= 195.6 C (approximately 196 C)
(c) 2 H+ + 2 e- => H2
Moles of electrons = 2 x moles of H2
= 2 x 0.0011098 = 0.0022196 mol
Faraday constant = charge/moles of electrons
= 195.6/0.0022196
= 88124 C/mol = 8.81 x 10^4 C/mol
Percent error = ((9.65 x 10^4 - 88124)/9.65 x 10^4) x 100%
= 8.68%