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 0.26 A is drawn for 1.5 hrs?

mg Ni

b) What is [Cr³⁺] when [Ni²⁺] has dropped to 10^-4 M?

[Cr³⁺] =  M

c) What is the cell potential at the conditions described in Part b?

E =  V

Answer 1

a)

mg of Ni plated in

t = 1.5 h = 1.5*3600 = 5400 s

Ctotal = I*t = 0.26*5400

C = 1404 C

1 mol of e- = 96500 C

mol of e- = 1404/96500 = 0.01454 mol of e-

2 mol of e- required = 1 mol of Ni

0.01454 mol --> 1/2*0.01454 = 0.00727 mol of Ni

mass = mol*MW = 0.00727*58.6934 = 0.42670 g = 426.7 mg of Niickel

b)

initally

[Ni+2] = 1

[Cr+3] = 1

after reaction

Cr + Ni2+ = Cr+3 + Ni

balance

2Cr + 3Ni2+ = 2Cr+3 + 3Ni

[Ni+2] = 1 - 3x

[Cr+3] = 1 + 2x

and we know

[Ni+2] = 1 - 3x = 10^-4

x = ((10^-4)-1)/(-3) = 0.3333

[Cr+3] = 1 -+2*0.3333 = 1.6666 M

c)

Ni2+ + 2 e− ⇌ Ni(s) −0.25; Cr3+ + 3 e− ⇌ Cr(s) −0.74

E°cell = -0.25 --0.74 = +0.49 V

apply NErnst equation

Q = [Cr+3]^2 / [Ni+2]^3

n = 6 e-

E = E° - 0.0592/n*log((1.6666 ^2) / (0.3333^3))

E = 0.49 - 0.0592/6*log((1.6666 ^2) / (0.3333^3))

E = 0.4714V