The following data was obtained for Part B of the lab: Mass of Cu before electro
ID: 565287 • Letter: T
Question
The following data was obtained for Part B of the lab: Mass of Cu before electrolysis = 21.1301 g Mass of Cu after electrolysis = 20.9894 g Time of electrolysis = 464 sec Average current = 0.92 Amp Volume of gas collected = 54.1 mL Atmospheric pressure (corrected to standard conditions) = 760.3 mmHg Room temperature = 18.7 °C Vapour pressure of water at 18.7 °C = 16.1 mmHg Using this data determine: Moles of Cu consumed:___moles Moles of H2 produced:___moles Experimental value of Faraday’s constant based on Cu consumed:___C/mole e- Experimental value of Avogadro’s number using the charge of one electron (1.602 x 10-19 C)
Explanation / Answer
From the data above
Given,
mass of Cu electrolyzed = 21.1301 - 20.9894 = 0.1407 g
moles of Cu produced = 0.1407 g/63.55 g/mol = 0.0022 mol
2 mole e- produce 1 mole Cu
So,
mole e- = 2 x 0.0022 mol = 0.0044 mole e-
using I (current) and t (time),
charge (Q) = I x t
= 0.92 amp x 464 sec
= 426.88 C
Faraday constant = Q/mole e-
= 426.88 C/0.0044 mole e-
= 97018.2 C/mole e-
---
Volume of gas collected = 54.1 ml = 0.0541 L
Pressure of H2 gas = 760.3 mmHg - 16.1 mmHg = 744.2 mmHg
= 744.2 mmgHg/760 = 0.98 atm
Temperature = 18.7 oC + 273 = 291.7 K
moles H2 produced = PV/RT = 0.98 x 0.0541/0.0821 x 291.7 = 0.0022 mol
moles of electron = 2 x 0.0022 mol = 0.0044 mole e-
number of electron used = Q/1.602 x 10^-19 C
= 426.88 C/1.602 x 10^-19 C
= 2.66 x 10^21
Avogadro's number = number of electrons/moles of electron
= 2.66 x 10^21/0.0044
= 6.04 x 10^23 electron/mole
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