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 - Vapor Pressure(H_2O) = ____mmHg = ____atm V(H_2) = ___mL = ____ L T = ___ K n(H_2) = ____ moles n_H_2 = PV/RT (where P = P_H_2) mole H_2 requires passage of ___ faradays Number of faradays passed = ___ Loss of mass of metal electrode = ___g Number of grams 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. MMFe = g. The charge n on the Fe ion is therefore ____.(see Eq.3)

Explanation / Answer

P = 748 - 23.8 = 724.2 m Hg = 724.2/760 = 0.9528 atm

V = 96.30 mL

T = 25°C = 298 K

mol = from n = PV/(RT) = 0.9528 *96.30/1000 / (0.082*298) = 0.00375 mol of H2

1 mol of H2 --> 2H+ + 2e- --> H2

2 Faradays per mol of H2

No. Farady passed = 0.00375 *2 = 0.0075 Faradays

Charge = 723.75

loss of metal electrode = 0.208 g

mol of metal = 0.208 / 0.0075 = 27.733 g = EM

then

MM Fe = 55.8 g

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