96,485 coulombs will deposit 27/3 = 9 grams Al. You have 18 g; therefore,
96,485 coulombs x (18 g/9 g) = 96,485 x 2 = 192,970 coulombs.
96,485 coulombs will deposit 107.9 g Ag so how much Ag will be deosited with 192,970 coulombs? That's 107.9 g Ag x (192,970/96,495) = ?
Two cells are connected in series. One contain AlCl3, and the other contains AgNO3 as the electrolytes .what mass of Ag is deposited when 18g of Al is deposited at cathode?
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