Question

A Cr3+(aq) solution is electrolyzed, using a current of 7.00 A .

a. What mass of Cr(s) is plated out after 2.00 days?

b. What amperage is required to plate out 0.230 mol Cr from a Cr3+ solution in a period of 8.00 h ?

Answer 1

a)

mass of Cr(s) could be plated in d = 2 days = 2*24 h = 48 h = 48 h * 3600 s/h = 172800 seconds

I = 7 A= 7 C/s

total charge = I*t = (172,800)(7) = 1,209,600 C

recall that, according to Faraday

1 mol of e- = 96500 C

x mol of e- = 1,209,600 C

then

x = 1,209,600/96500 = 12.534 mol of e-

for electrolysis:

Cr+3 + 3e- = Cr(s)

1 mol of Cr(s) = 3 mol of e-

x mol fo Cr(s) = 12.534 mol of e-

x = 12.534/3 = 4.178 mol of Cr(s)

mass = mol*M;W

MW of Cr = 51.99610

mass = 51.99610*4.178 = 217.23 g of Cr(s)

b)

Find I (Amp) for

n = 0.23 mol of Cr in t = 7 h

time = 8 h = 8*3600 s = 28,800 s

calculate moles of e- required

1 mol fo Cr = 3 mol of e-

0.23 mol of Cr = x mol of e-

x = 0.23*3 = 0.69 mol of e-

1 mol of e- = 96500 C

0.69 mol of e- = (0.69)(96500) = 66,585 C

then

I = C/t = (66,585)/28,800 = 2.31197 C/s

I = 2.31197 A