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In a classroom, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday's laws of electrolysis. Calculate the quantity of electricity required to deposit 0.09 g of aluminium during the following electrode reaction: Al^3 + 3e^(-) to Al ( Al=27u) |
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Answer» Solution :Quantity of electricity required to deposit 27 G (1 mole ) of Al = 3F = `3xx96500` coulombs `THEREFORE .Quantity of electricity required to deposit (0.09g of Al `3xx96500xx0.09)/27`=965 coulombs |
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