An electric current is passed through electrolytic cells in series one containing Ag(NO₃)(aq.) and other H₂SO₄(aq.). What volume of O₂ measured at 25^oC and 750mm Hg pressure would be liberated from H₂SO₄ if

(a) one mole ion of Ag⁺ is deposited fro AgNO₃ solution

(b) 8 x 10²² ions of Ag⁺ are deposited from AgNO₃ solution .

As the cell is connected in series then 

Equivalents of silver deposited = Equivalent of O2 liberated (second law Faraday)

Ag+ + e- → Ag

According the reaction transfer of one mole of electron requires for the deposition of one mole silver. 

equivalent wt. of Ag = Molecular wt. / 1 = 108 g

 a) Equivalent of silver deposited = amt. of silver deposited / eqivalent wt. of silver 

Amt of silver deposited = 1mole(given) = 108 g.

Equivalent of silver deposited = 108/108 = 1

∴Equivalents of oxygen liberated = 1 (second law faraday as mentioned above).

O2 is liberated at anode according to following reaction. 2 H2O→ O2 + 4 H(+) + 4 e(-) 

Since the reaction involves the transfer of 4 electrons then moles of O2 liberated will Equivalents of O2 / 4 = 1/4 = 0.25 moles.

Calculate the volume of oxygen measured produced. Use the ideal gas equation rewritten as V = nRT / P where n = 0.25 mol R = 8.314 JK-1mol-1 T = 273 + 25 = 298 K P = 750 mmHg = 750 / 760 atm = (750 / 760) x 101.3 kPa = 100.1 kPa ∴ V = (0.25 x 8.314 x 298) / 100.1 =6.18  L

b) Amt of silver deposited = 8 x 1022 /6.023 x 10 23 * 108 

Equivalent of silver deposited = 8 x 1022 /6.023 x 10 23 * 108 /108 = 0.13

∴Equivalents of oxygen liberated = 0.13 

moles of O2 liberated will Equivalents of O2 / 4 = 0.13/4 = 0.03

Putting the value in ideal gas equation we get (similar as calculated above)

V = (0.03 x 8.314 x 298) / 100.1 =0.74  L