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

Question

In the first cell, 42.51 mL of hydrogen gas were formed at the cathode. The temperature was 21.5 C 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. Calculate:

a. Partial Pressure of Hydrogen (H2) in buret (PH2 = Barometric – VP – Pheight) (The pressure due to the height of water still in the buret is a conversion from mmH2O to mmHg. Since the density of water is 1.00 g/mL and the density of mercury is 13.6 g/mL, Pheight in mmHg = (Height in mm)*(1.00/13.6))

b. Moles of Hydrogen (H2) gas formed (PV = nRT) (R = 0.08206 (L atm)/(K mol) = 62364 (mL torr)/(K mol))

c. Number of Faradays (= moles of electrons) that passed through the cell using the conversion (2 mol e- per 1 mol H2)

Explanation / Answer

since Pressure = density*g*height

presssre of water needs to be converted into pressure expressed in terms of Hg.

hecne pressure of 8.2 cm of water column= 8.2*1/13.6 =0.6029 mm Hg

partial pressure of water = barometric pressure (mm Hg)- Vapor pressure of water (mm Hg)- hieght of water column

=758.8- 19.240-0.6029 =738.96 mmHg=738.96/760 atm =0.9723 atm

from PV= nRT, n= PV/RT= 0.9723*31.5/1000L / ((0.0821*(21.5+273))=0.001267 moles

H2 --------->2H+2e-

0.001267 moles of hydrogen generates 2*0.001267 moles of electrons= 0.002534 moles of electrons

hence Faradays= 0.002534moles of electron* 1 Farday/ mole of electron = 0.002534 Faradays.

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