Question

Calculate the amount in grams of metal that is deposited at the cathode by running a current of 3.15 A for 78 min in an electrolysis reaction for an aqueous solution containing a)Ni2+,b) Cr3+, c)Cu2+.

Answer 1

a)

Electrolysis equation is:

Ni2+ + 2e- ------> Ni

1 mol of Ni requires 2 mol of electron

1 mol of electron = 96485 C

So,1 mol of Ni requires 192970 C

let us calculate the charge passed:

t = 78.0 min = 78.0*60 s = 4.68*10^3 s

time, t = 4.68*10^3s

Q = I*t

= 3.15A * 4.68*10^3s

= 1.474*10^4 C

mol of Ni plated = 1.474*10^4/192970 = 7.64*10^-2 mol

Molar mass of Ni = 58.69 g/mol

mass of Ni = number of mol * molar mass

= 7.64*10^-2 * 58.69

= 4.484 g

Answer: 4.48 g

b)

Electrolysis equation is:

Cr3+ + 3e- ------> Cr

1 mol of Cr requires 3 mol of electron

1 mol of electron = 96485 C

So,1 mol of Cr requires 289455 C

let us calculate the charge passed:

t = 78.0 min = 78.0*60 s = 4.68*10^3 s

time, t = 4.68*10^3s

Q = I*t

= 3.15A * 4.68*10^3s

= 1.474*10^4 C

mol of Cr plated = 1.474*10^4/289455 = 5.093*10^-2 mol

Molar mass of Cr = 52 g/mol

mass of Cr = number of mol * molar mass

= 5.093*10^-2 * 52

= 2.648 g

Answer: 2.65 g

c)

Electrolysis equation is:

Cu2+ + 2e- ------> Cu

1 mol of Cu requires 2 mol of electron

1 mol of electron = 96485 C

So,1 mol of Cu requires 192970 C

let us calculate the charge passed:

t = 78.0 min = 78.0*60 s = 4.68*10^3 s

time, t = 4.68*10^3s

Q = I*t

= 3.15A * 4.68*10^3s

= 1.474*10^4 C

mol of Cu plated = 1.474*10^4/192970 = 7.64*10^-2 mol

Molar mass of Cu = 63.55 g/mol

mass of Cu = number of mol * molar mass

= 7.64*10^-2 * 63.55

= 4.855 g

Answer: 4.86 g