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How much copper is deposited on the cathode if a current of 3A is passed through aqueous CuSO_(4) solution for 15 minutes? |
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Answer» Solution :Quantity of electricity passed = Current in amperes x TIME in SECONDS. `Q = 3 xx 15 xx 60 = 2700` C `Cu^(++) + 2e^(-) to Cu` Two mole electron or 2F charge can DEPOSIT 1. Mole copper i.e. 63.5 g and so 2700 C will deposit `=(2700 xx 63.5)/(2 xx 96500) = 0.889` g |
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