Question

16. A) How many grams of cadmium metal will be deposited from a solution that contains Cd²⁺ions if a current of 1.08 A is applied for 45.6 minutes _____ grams

B) How many grams of cadmium metal will be deposited from a solution that contains Cd²⁺ions if a current of 0.535 A is applied for 35.7 minutes ____ grams

C) How many grams of iron metal will be deposited from a solution that contains Fe²⁺ions if a current of 0.680 A is applied for 65.5 minutes _____ grams

Answer 1

A)

Electrolysis equation is:

Cd2+ + 2e- ------> Cd

1 mol of Cd requires 2 mol of electron

1 mol of electron = 96485 C

So,1 mol of Cd requires 192970 C

let us calculate the charge passed:

t = 45.6 min = 45.6*60 s = 2.736*10^3 s

time, t = 2.736*10^3s

Q = I*t

= 1.08A * 2.736*10^3s

= 2.955*10^3 C

mol of Cd plated = 2.955*10^3/192970 = 1.531*10^-2 mol

Molar mass of Cd = 1.124*10^2 g/mol

mass of Cd = number of mol * molar mass

= 1.531*10^-2 * 1.124*10^2

= 1.721 g

Answer: 1.72 g

B)

Electrolysis equation is:

Cd2+ + 2e- ------> Cd

1 mol of Cd requires 2 mol of electron

1 mol of electron = 96485 C

So,1 mol of Cd requires 192970 C

let us calculate the charge passed:

t = 35.7 min = 35.7*60 s = 2.142*10^3 s

time, t = 2.142*10^3s

Q = I*t

= 0.535A * 2.142*10^3s

= 1.146*10^3 C

mol of Cd plated = 1.146*10^3/192970 = 5.939*10^-3 mol

Molar mass of Cd = 1.124*10^2 g/mol

mass of Cd = number of mol * molar mass

= 5.939*10^-3 * 1.124*10^2

= 0.6675 g

Answer: 0.668 g

C)

Electrolysis equation is:

Fe2+ + 2e- ------> Fe

1 mol of Fe requires 2 mol of electron

1 mol of electron = 96485 C

So,1 mol of Fe requires 192970 C

let us calculate the charge passed:

t = 65.5 min = 65.5*60 s = 3.93*10^3 s

time, t = 3.93*10^3s

Q = I*t

= 0.68A * 3.93*10^3s

= 2.672*10^3 C

mol of Fe plated = 2.672*10^3/192970 = 1.385*10^-2 mol

Molar mass of Fe = 55.85 g/mol

mass of Fe = number of mol * molar mass

= 1.385*10^-2 * 55.85

= 0.7735 g

Answer: 0.774 g