Electrolysis (Cr/Cr^3+//Ni^2+/Ni)?

Question

Use the following cell to answer the questions:

Cr / Cr3+ (1.00 M) // Ni2+ (1.00 M) / Ni.

a) How many milligrams of Ni would be plated on the cathode if a current of 0.37 A is drawn for 3.5 hrs?
b) What is [Cr3+] when [Ni2+] has dropped to 10-4 M?
c) What is the cell potential at the conditions described in Part b?

Answer 1

The related standard reduction potentials are:
Ni(2+)(aq) + 2e- ==> Ni(s), E° = -0.28V
Cr(3+)(aq) + 3e- ==> Cr(s), E° = -0.74V
Thus the emf of the cell at standard condition is:
2Cr(s) + 3Ni(2+)(aq) ==> 2Cr(3+)(aq) + 3Ni(s), E° = 0.46V

(a) The molar mass of Ni is: 58.7g/mol.
(0.37C/s)*(3.5*3600s)*(58.7g/mol)/(2*96485C/mol)
= 1.4 g = 1.4x10^3 milligrams.

(b) Insufficient data to calculate, such as the volume of Ni(2+) and Cr(3+).

(c) Need to know [Cr3+] first.