In an electrolysis cell similar to the one employed in this experiment, a studen

Question

In an electrolysis cell similar to the one employed in this experiment, a student observed that his unknown metal anode lost 0.208 g while a total volume of 96.30 ml. of H_2 was being produced The temperature in the laboratory was 25 degree C and the barometric pressure was 748 mm Hg. At 25 degree C the vapor pressure of water is 23.8 mm Hg. To find the equivalent mass of his metal, he filled in the blanks below. Fill in the blanks as he did. P_H_2 = P_bar - VP_H_2O = mm Hg = atm V_H_2 = mL = L T = K n_H_2 = moles n_H_2 = PV/RT (where P = P_H_2) 1 mole H_2 requires passage of faradays No. of faradays passed = Loss of mass of metal anode = g No. grams of metal lost per faraday passed = no. grams lost/no. faradays passed = g = EM The student was told that his metal anode was made of iron. MM Fe = g. The charge n on the Fe ion is therefore. (See Eq. 31.) In ordinary units, the faraday is equal to 96,480 coulombs. A coulomb is the amount of electricity passed when a current of one ampere flows for one second. Given the charge on an electron, 1.6022 Times 10^-19 coulombs. calculate a value for Avogadro's number.

Explanation / Answer

1. for the experiment

P(H2) = 748 - 23.8 = 724.2 mmHg = 0.953 atm

V(H2) = 96.30 ml = 0.0963 L

T = 25 + 273 = 298 K

n(H2) = PV/RT = 0.953 x 0.0963/0.08205 x 298 = 3.75 x 10^-3 mol

1 mol of H2 requirepassage of = 2 F

No. of Faraday passed = 2 x 3.75 x 10^-3 = 7.50 x 10^-3 F

Loss of metal mass = 0.208 g

grams of metal lost per faraday = 0.208/7.50 x 10^-3 = 27.71 g

MM of Fe = 55.85 g/mol

The charge of Fe = 55.85/27.71 = 2

2. Value of avogadro's number = (1 e-/1.6022 x 10^-19 C)(96480 C/1 Faraday)(1 Faraday/1 mole)

                                                  = 6.022 x 10^23 e-/mole

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