Three electrolytic cells are connected in series; the current passes through all

Question

Three electrolytic cells are connected in series; the current passes through all three, one after another. In the first cell, 1.20 g of Cd is oxidized to Cd2+; in the second, Ag+ is reduced to Ag, in the third, Fe2+ is oxidized to Fe3+. Find the number of faradays passed through the circuit. What mass of Ag is deposited at the cathode in the second cell? What mass of Fe(NO3)3 might be recovered from the third cell solution?

Explanation / Answer

1] current = amps x seconds = 316/1000 x 50.2 x 60 coulombs = 1087.332 = 1.09E+3 C 2] from Cu2+ + 2e- ---> Cu , you known that 2 x 96500 coulombs are needed to get 1 mole = 63.5g Cu. Now just use simple proportion. 2 x 96500 coulombs gives 63.5g Cu 1087.332 coulombs gives 63.5/ [2 x 96500] x 1087.332 = 0.357749g = 358mg

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