On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 C mol⁻¹)

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Accepted Answer

0.02 M. The charge passed is calculated as Q = I × t = 1 × (16 × 60 + 5) = 965 C. Using Faraday's laws, moles of Cu deposited = Q / (n × F), where n = 2 for Cu²⁺. This gives moles of Cu = 965 / (2 × 96500) = 0.005 mol. Since the solution is 1 L, molarity = 0.005 × 2 = 0.02 M.

On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu = 63.5; Faraday constant = 96,500 C mol⁻¹)

Solution

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