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The amount of copper deposited by one Faraday current will be maximum in an acidic solution of one litre of
Options
(a) 1 M Cu₂Cl₂
(b) 2 M Cu(NO₃)₂
(c) 5 M CuSO₄
(d) 5 M Cu₃(PO₄)₂
Correct Answer:
1 M Cu₂Cl₂
Explanation:
In 1 M Cu₂Cl₂. Cu⁺ (one faraday = 96500C) deposited = 63.5 g of Cu. In 2 M Cu(NO₃)₂. Cu²⁺ (2F = 2⨯96500C) deposited 63.5 of Cu. Therefore 1F deposited 63.5/2 g of Cu. In 5 M CuF₂. Cu²⁺ (2F = 2⨯96500 C) deposited 63.5 g of Cu. Therefore 1F = 96500 C deposited 63.5/2 g of Cu. so Cu deposited by 1 Faraday will be maximum in case of 1M Cu₂Cl₂.
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