A dilute aqueous solution of Na2SO4 is electrolyzed between Pt electrodes for 3.75 h with a current of 2.85 A .?

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2016-07-25 7:21 am

更新1:

Part A What volume of gas, saturated with water vapor at 25∘C and at a total pressure of 744 mmHg , would be collected at the anode? The vapor pressure of water at 25∘C is 23.8 mmHg.

更新2:

I know that I have to use PV= nRT, but How do I get the number of moles?

回答 (1)

戇戇居士

2016-07-25 8:10 am

Anode reaction :
4OH⁻(aq) → 2H₂O(aq) + O₂(g) + 4e⁻
OR: Mole ratio O₂ : e⁻ = 1 : 4

Quantity of negative charges passed = (2.85 A) × (3.75 × 3600 s) = 38475 C

1 mole of electrons carries 96500 C of negative charges.
No. of moles of e⁻ passed = (38475 C) / (96500 C/mol)
No. of moles of O₂ formed = (38475/96500 mol) × (1/4) = 0.0997 mol

For the oxygen formed :
P = (744 - 23.8) mmHg = 720.2 mmHg
V = ? L
n = 0.0997 mol
R = 62.36 mmHg L / (mol K)
T = (273 + 25) = 298 K

PV = nRT
Then, V = nRT/P

Volume of O₂ formed = 0.0997 × 62.36 × 298 / 720.2 L = 2.57 L

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