± Electrolysis and Current?
2016-05-06 4:06 am
A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for 3.00 min , 11.8 mL of water-saturated H2(g) is collected. The temperature and pressure of the system are 25 ∘C and 715 torr.
回答 (1)
2016-05-06 4:28 am
Consider the H₂ gas produced :
P = 715/760 atm
V = 11.8 mL = 0.0118 L
R = 0.0821 atm L / (mol K)
T = (273 + 25) K = 298 K
n = ? mol
PV = nRT
n = PV/(RT)
No. of moles of H₂, n = (715/760) × 0.0118 / (0.0821 × 298) = 4.54 × 10⁻⁴ mol
2H⁺(aq) + 2e⁻ → H₂(g)
OR: 2H₃O⁺(aq) + 2e⁻ → 2H₂O(l) + H₂(g)
According to the equation above, the production of 1 mole of H₂ needs 2 moles of e⁻.
Number of moles of e⁻ = (4.54 × 10⁻⁴) × 2 = 9.08 × 10⁻⁴ mol
Each molecules of e⁻ carries 96500 C of electricity.
Amount of electricity, Q = 96500 × (9.08 × 10⁻⁴) = 87.6 C
Q = It
Current, I = Q/t = 84.7 / (3 × 60) = 0.487 A
P = 715/760 atm
V = 11.8 mL = 0.0118 L
R = 0.0821 atm L / (mol K)
T = (273 + 25) K = 298 K
n = ? mol
PV = nRT
n = PV/(RT)
No. of moles of H₂, n = (715/760) × 0.0118 / (0.0821 × 298) = 4.54 × 10⁻⁴ mol
2H⁺(aq) + 2e⁻ → H₂(g)
OR: 2H₃O⁺(aq) + 2e⁻ → 2H₂O(l) + H₂(g)
According to the equation above, the production of 1 mole of H₂ needs 2 moles of e⁻.
Number of moles of e⁻ = (4.54 × 10⁻⁴) × 2 = 9.08 × 10⁻⁴ mol
Each molecules of e⁻ carries 96500 C of electricity.
Amount of electricity, Q = 96500 × (9.08 × 10⁻⁴) = 87.6 C
Q = It
Current, I = Q/t = 84.7 / (3 × 60) = 0.487 A
收錄日期: 2021-04-18 14:48:24
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