The time is 1046.15 taken o deposit 1. 36 g of copper from an aqueous solution of copper(ii) sulfate by passing a current of two amperes through the solution.
Electric current is a stream of charged particles going through an electric conductor.
By law of Faraday
[tex]\rm W = Zit \dfrac{E}{F} it[/tex]
Where, W= Deposited amount is 1.36 grams
E = Equivalent weight is the Molar mass of Cu = 63.5 g mol−1
F = Faraday = 96500 C mol−1
i = Electricity passed 2 ampere.
t = Time =?
Putting the values in the equation
[tex]\rm 1.36 =\dfrac{63.5 }{96500} \times 2 \times t \\\\t =[/tex]
Thus, the time taken will be calculated as 1046.15.
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