Question

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Answer 1

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Answer 2

To deposit 1 mole of Cu at cathode from Cu²⁺SO⁴⁻ solution = 2 moles of electrons are required

That is

To deposit 63.5 g = 63.5 g / 63.5 g × 2 = 1 × 2 = 2 moles

Thus total no. of electrons = 2 × 6.022 × 10²³ = 12.044 × 10²³

∴ The answer is option (3) 12.044 × 10²³