Question

When electricity (the flow of electrons) is passed through a solution, it causes an oxidation-reduction (redox) reaction to occur. If the solution contains a metal cation such as Ag+, the flow of electrons will reduce the silver ion, causing solid silver to plate onto the electrode. The amount of metal plated depends on the number of electrons passed. The total charge of a mole of electrons is 96, 485 coulombs (C) and 1 ampere = 1 coulomb/second (C/s).

What mass of Cu(s) is electroplated by running 28.5 A of current through a Cu2+ (aq) solution for 4.00 h? Express your answer to three significant figures and include the appropriate units.
How many minutes will it take to electroplate 37.1 g of gold by running 5.00 A of current through a solution of Au+(aq)? Express your answer to three significant figures and include the appropriate units.

Answer 1

a. 135 g

b. 60.6 min

Explanation:

a. What mass of Cu(s) is electroplated by running 28.5 A of current through a Cu2+ (aq) solution for 4.00 h? Express your answer to three significant figures and include the appropriate units.

The chemical equation for the reaction is given below

Cu²⁺(aq) + 2e⁻ → Cu(s)

We find the number of moles of Cu that are deposited from

nF = It where n = number of moles of electrons, F = Faraday's constant = 96485 C/mol, I = current = 28.5 A and t = time = 4.00 h = 4.00 × 60 min/h × 60 s/min = ‭14,400‬ s

So, n = It/F = 28.5 A × ‭14,400‬ s/96485 C/mol = ‭410,400‬ C/96485 C/mol = 4.254 mol

Since 2 moles of electrons deposits 1 mol of Cu, then 4.254 mol of electrons deposits 4.254 mol × 1 mol of Cu/2 mol = 2.127 mol of Cu

Now, number of moles of Cu = n' = m/M where m = mass of copper and M = molar mass of Cu = 63.546 g/mol

So, m = n'M

= 2.127 mol × 63.546 g/mol

= 135.15 g

≅ 135 g to 3 significant figures

b. How many minutes will it take to electroplate 37.1 g of gold by running 5.00 A of current through a solution of Au+(aq)?

The chemical equation for the reaction is given below

Au⁺(aq) + e⁻ → Au(s)

We need to find the number of moles of Au in 37.1 g

So, number of moles of Au = n = m/M where m = mass of gold = 37.1 g and M = molar mass of Au = 196.97 g/mol

So, n = m/M = 37.1 g/196.97 g/mol = 0.188 mol

Since 1 mol of Au is deposited by 1 moles of electrons, then 0.188 mol of Au deposits 0.188 mol of Au × 1 mol of electrons/1 mol of Au = 0.188 mol of electrons

We find the time it takes to deposit 0.188 mol of electrons that are deposited from

nF = It where n = number of moles of electrons, F = Faraday's constant = 96485 C/mol, I = current = 5.00 A and t = time

So, t = nF/It

= 0.188 mol × 96485 C/mol ÷ 5.00 A

= ‭18173.30‬ C/5.00 A

= 3634.66 s

= 3634.66 s × 1min/60 s

= 60.58 min

≅ 60.6 min to 3 significant figures