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_H2 = P_bar - VP_H_2O = ___________ mm Hg = ____________ at, 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 meal 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. 3.) In ordinary units, the faraday is equal to 96.480 coulombs. A coulomb is the amount of electricity passed w hen 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. From the given data

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

V(H2) = 96.30 ml = 0.0963 L

T = 25 oC = 25 + 273 = 298 K

n(H2) = PV/RT = 0.004 mols

1 mole of H2 require a passage of 96485 faraday

Number of faradays passed = 0.0038 x 96485 = 366.643 faraday

Loss of mass of metal anode = 0.208 g

Number of grams of metal lost per faraday = 0.208/366.643 = 0.00057 g/F = EM

When the metal is Fe

molar mass = 55.845 g/mol

The charge n on Fe ion is therefore = 55.845 x 0.00057 = 0.03

2. Value of avogadro's number = 96480/1.6022 x 10^-19 = 6.022 x 10^23

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