用惰性电极电解2L 1mol·L -1 CuSO 4 溶液,在电路通过0.5mol电子后,计算:(1)阴极上析出金属质量。

2025年04月07日 05:38
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网友(1):

解:2L 1mol·L -1 的CuSO 4 溶液中含CuSO 4 2mol,转移0.5mol电子,CuSO 4 未完全电解。设生成Cu的质量为x,生成O 2 的体积为y,同时溶液中增加H + 的物质的量为z
2Cu 2+ + 2H 2 O =2Cu + O 2 ↑ + 4H +   4e -
       2×64g 22.4L 4mol 4mol
        x   y   z 0.5mol
(1)X= =16g
(2)Y= =2.8L
(3)Z= =0.5mol
c(H + )= =0.25mol·L -1

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