Copper may be used for electroplating with a half reaction of cu2+ + 2e- > cu. if a current of 10.0a is applied to a cu2+ solution for 60.0 min, the mass of copper plated out can be calculated as
a.) 10.0 x 3600 x (1/96500) x (2.00/1) x 63.5
b.) 10.0 x 3600 x (1/96500) x (1/2.00) x 63.5
c.) 96500 x (1/10.0) x (1/2.00) x (1/3600) x 63.5
d.) 96500 x (1/10.0) x (1/2.00) x (1/3600) x (1/63.5)

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

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.