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標題:
[10分] Electrolysis!!急
發問:
What mass of copper is deposited by the passage of 2.0A for 482s through a 2.0 M solution of CuSo4? Please answer it step by step... Thank you for your help 更新: but the answer is 0.317 g... it needs to consider the Faraday law...i think=) thanks
最佳解答:
1 ampere = 1 coulomb of electrons per second 1 coulomb = 6.24151e18 electrons in 482 seconds, no. of coulomb of electrons passes through the circuit = 482 x 1 = 482 C = 3.008e21 electrons = 0.005 mole of electrons according to half-equation: Cu(2+) + 2e- ------> Cu one copper atom is produced for every 2 electrons. therefore, no. of mole of copper metal produced = 0.005/2 = 0.0025 mole mass of copper formed = 0.0025 x 63.5 = 0.159 g concentration of solution doesn't matter -- unless the question asks you about the error and related improvement: if solution is too diluted, side reactions like electrolysis of water may occur. this consumes some of electrons and reduce the actual amount of copper produced. 2010-09-25 02:14:18 補充： sorry i missed faraday's law. no. of mole = (1/no. of electrons needed in equation) x (total charge / 96485) = (1/2) x (482x2/96485) = 0.005 mole mass = 0.005*63.5 = 0.3175 g 2010-09-25 02:15:44 補充： in fact my original calculation is ALMOST correct: i just missed multiplying current (2A) to time. after correction, the answer should be EXACTLY the same. of course faraday's law is more convinient, but principle is the same.
其他解答:
[10分] Electrolysis!!急
發問:
What mass of copper is deposited by the passage of 2.0A for 482s through a 2.0 M solution of CuSo4? Please answer it step by step... Thank you for your help 更新: but the answer is 0.317 g... it needs to consider the Faraday law...i think=) thanks
最佳解答:
1 ampere = 1 coulomb of electrons per second 1 coulomb = 6.24151e18 electrons in 482 seconds, no. of coulomb of electrons passes through the circuit = 482 x 1 = 482 C = 3.008e21 electrons = 0.005 mole of electrons according to half-equation: Cu(2+) + 2e- ------> Cu one copper atom is produced for every 2 electrons. therefore, no. of mole of copper metal produced = 0.005/2 = 0.0025 mole mass of copper formed = 0.0025 x 63.5 = 0.159 g concentration of solution doesn't matter -- unless the question asks you about the error and related improvement: if solution is too diluted, side reactions like electrolysis of water may occur. this consumes some of electrons and reduce the actual amount of copper produced. 2010-09-25 02:14:18 補充： sorry i missed faraday's law. no. of mole = (1/no. of electrons needed in equation) x (total charge / 96485) = (1/2) x (482x2/96485) = 0.005 mole mass = 0.005*63.5 = 0.3175 g 2010-09-25 02:15:44 補充： in fact my original calculation is ALMOST correct: i just missed multiplying current (2A) to time. after correction, the answer should be EXACTLY the same. of course faraday's law is more convinient, but principle is the same.
其他解答:
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