Question

Electric current passed through two coulometers connected in series.The first one contained a solution of a metal-nitrate, the second one a solution of H₂SO₄. After some time 0,675 g of the unknown metal could be found on a Pt electrode of the first coulometer and 73,1 cm³ of hydrogen (16°C, 770 Torr) were created on the Pt electrode of the second coulometer. What was the unknown metal?

Answer 1

Hi this problem is based on Farday's second law, which states that when the same quantity of electricity is passed through the different electrolytes in series, the amount of various chemical substances deposited or liberated are directly proportional to their chemical equivalents (equivalent weights).

w1 / w2 = E1 / E2

w1 = amount of metal in grams = 0.675 g

w2 = amount of hydrogen gas ---> to be calculated indirectly

P = 770 torr = 1.01316 atm; T = 16 C = 289 K; R = 0.082057 L.at/ mol.K; V = 73.1 cm3 = 73.1 mL = 0.0731 L

PV = nRT ==> n = PV / RT = [1.01316 atm x 0.0731 L] / [0.082057 L.at/ mol.K x 289 K] = 0.00312307155 mol

convert moles to mass = 0.00312307155 mol x 2.0158 g /mol = 0.00629548763 g

w2 = 0.0063 g

E1 = to be calculated;

E2 = each hydrogen molecule supplies two moles of hydrogen ions, so its equivalent weight is 2.0158 g /mol / 2 eq/ mol = 1.0079 g/ eq

so E2 = 1.0079 g

E1 = (w1 / w2) x E2 = (0.675 g / 0.0063 g) x 1.0079 g = 107.989285714 g ~ 108 g

from periodic table this atomic mass represents silver Ag.

The unknown metal is Argentum - Ag (Silver common name)

Hope this helped you!

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