Question

(A) A current of 4.62 A is passed through a Ni(NO3)2 solution for 1.70 hours. How much nickel is plated out of the solution?

(B) A current of 4.09 A is passed through a Cr(NO3)2 solution. How long (in hours) would this current have to be applied to plate out 7.70 g of chromium?

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 = 1.7 hr = 1.7*3600 s = 6.12*10^3 s

time, t = 6.12*10^3s

Q = I*t

= 4.62A * 6.12*10^3s

= 2.827*10^4 C

mol of Ni plated = 2.827*10^4/192970 = 0.1465 mol

Molar mass of Ni = 58.69 g/mol

mass of Ni = number of mol * molar mass

= 0.1465 * 58.69

= 8.60 g

Answer: 8.60 g

B)

Electrolysis equation is:

Cr2+ + 2e- ------> Cr

1 mol of Cr requires 2 mol of electron

1 mol of electron = 96485 C

So,1 mol of Cr requires 192970 C

let us calculate mol of element deposited:

use:

number of mol, n = mass/molar mass

= 7.7/52

= 0.1481 mol

total charge = mol of element deposited * charge required for 1 mol

= 0.1481*1.93*10^5

= 2.857*10^4 C

use:

time = Q/i

= 2.857*10^4/4.09

= 6.986*10^3 seconds

= 6.986*10^3/3600 hr

= 1.941 hr

Answer: 1.94 hr