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

(a)

Using the equation:

PH2 = PBarometric – VP – Pheight

Putting values we get:

PH2 = 758.8 – 19.24 – Pheight

Pheight = d*g*h = 1000*9.81*0.082 = 804.42 Pa = 6.03 mm Hg

So,

PH2 = 758.8 – 19.24 – 6.03 = 733.53 mm Hg

(b)

Using equation:

PV = nRT

Putting values we get:

733.53*42.51 = n*62364*294.5

Solving we get:

n = 0.00169 moles

(c)

Number of faradays of electrons passed = 2*moles of H2 produced = 2*0.00169 = 0.00338 moles Faradays

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