Answer:
a. 135 g
b. 60.6 min
Step-by-step 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
