Respuesta :
Answer:
The correct option is: B.) 10.0 x 3600 x (1/96500) x (1/2.00) x 63.5
Explanation:
Given reduction half-reaction: Cu²⁺ + 2e⁻ → Cu
Given: Electrons transferred: n = 2, Current: I = 10.0 A,
Time: t = 60.0 min = 60 × 60 sec = 3600 sec (∵ 1 min = 60 sec)
Since, Electric charge (Q) = current × time
∴ Q = I × t
⇒ Q = 10.0 A × 3600 sec = (10.0 × 3600) C
Since, one faraday charge is equal to the charge of one mole electrons.
∴ One mole electron = 1 Faraday (F) = 96,500 coulombs (C)
⇒ 1 C = 1 ÷ 96,500 mole electron
∴ (10.0 × 3600) C = [(10.0 × 3600) × (1 ÷ 96,500)] mole electron
Now since 2 mole electrons reduces 1 mole Cu²⁺ to Cu.
So, 1 mole electrons reduces (1/2) mole Cu²⁺
Therefore, moles of Cu²⁺ reduced by [(10.0 × 3600) × (1 ÷ 96,500)] mole electrons = [(10.0 × 3600) × (1 ÷ 96,500)] mole electron × (1 mole Cu²⁺ ÷ 2 mole electron)
= [(10.0 × 3600) × (1 ÷ 96,500) × (1 ÷ 2)] moles of Cu²⁺
As number of moles = mass taken ÷ molar mass
⇒ mass of copper plated = number of moles × molar mass
As the molar mass of copper = 63.5 g/ mol
∴ mass of copper plated = [(10.0 × 3600) × (1 ÷ 96,500) × (1 ÷ 2)] moles × 63.5 g/mol = [(10.0 × 3600) × (1 ÷ 96,500) × (1 ÷ 2) moles × 63.5] g
Therefore, the mass of copper plated = [(10.0 × 3600) × (1 ÷ 96,500) × (1 ÷ 2) moles × 63.5] g
