In the first cell, 42.51 mL of hydrogen gas were formed at the cathode.The tempe

Question

In the first cell, 42.51 mL of hydrogen gas were formed at the cathode.The temperature was 21.5C and the vapor pressure of water at this temperature is 19.240 torr. The Barometric pressure was 758.8 mm Hg and the height of water in the buret above the level of water in the beaker was 8.2 cm. An electrolysis apparatus was set up connecting three cells in a series. How long will it take to form this much Hydrogen (H2) (and copper and X) if an ammeter, connected in series with the cells, shows a constant reading of 0.250 amps? (an amp is equal to one Coulomb per second, and there are 96,485 Coulombs per mole of electrons)

Explanation / Answer

moles H2 formed = PV/RT

with,

P = (758.8 - 19.240)/760 = 0.973 atm

V = 0.04251 L

R = gas constant

T = 21.5 + 273 = 294.5 K                   

So,

moles of H2 formed = 0.973 x 0.04251/0.08205 x 294.5 = 0.0017 mol       

moles e- needed = 0.0017 x 2 = 0.0034 mol e-

time required to deposit = 0.0034 x 96485/0.250 = 1312.2 s

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