Two voltameters, one containing a solution of silver salt and the other a solution of a trivalent-metal salt, are connected in series. A current of 2 A is maintained for 1.50 hours. It is observed that 1.00 g of the trivalent metal is deposited. Determine the amount of silver deposited during this period, given that the atomic weight of silver is 107.9 g mol ⁻¹ .

Step-by-step solution

According to Faraday's first law of electrolysis, the mass of a substance deposited at an electrode is given by: m = E I t F For silver, the valency is 1 , so its equivalent weight E_ Ag is equal to its atomic weight: E_ Ag = 107.9 g eq ⁻¹ The total charge passed through the series circuit is: Q = I t = 2 A (1.50 3600) s = 10800 C Substituting the values into the formula: m_ Ag = 107.9 10800 96500 m_ Ag = 1165320 96500 12.075 g Rounding off to one decimal place, the mass of silver deposited is 12.1 g . The data reg