1 e– =

electroplating: using electrolysis to deposit a thinlayer of one metal onto another

For electrolysis, currentin amperes (1 A = 1 C/s)is passed through theliquid. One mole oftransferred e– carrieswith it 96,500 C of charge.

1.60 x 10–19 C

Fe cathode

to be plated with Ni

Ni anode

Ni2+(aq)

Vext

(Faraday’s constant)

6.02 x 1023 e– = 96,500 C~

For how long must a 50.0 A current be passed throughmolten BaBr2 in order to produce 500. g of barium?

500. g Ba

Ba g 137.3Ba mol 1

Ba mol 1e mol 2 -

-e mol 1C 500,96

C 50s 1

= 14,100 s

= 3.90 h

Hall-Heroult process for purifying Alfrom Al2O3 and Na3AlF6 (1886)Paul Heroult

(1863–1914)Charles Martin Hall

(1863–1914)

**To solve these problems, use unit cancellation.

For a voltaic cell, the maximumwork the system can do on thesurroundings is given by:

wmax = DG = –nFE

where E is the reduction potential ofthe system at the conditions specified

For an electrolytic cell, the work that the surroundingsdo (i.e., that the external “oomph” does) on the systemis given by:

w = nFEext

from externalvoltage source

(E = Eext)

(i.e., it could be Eo, but maybe not).

wmax WILL BE (–), and w WILL BE (+).

(1 h)

The unit for electric power is the watt (1 W = 1 J/s).Electric companies often measure electrical energyin the kilowatt-hour (kWh).

1 kWh =

h 1s 36001000 W

W1J/s 1

How many J is 1 kWh?

= 3.6 x 106 J

= 3.6 MJ

To read an electric meter: 1. Read left to right. 2. Always round down.

8 4 1 6

(Calc. w = nFEext in Joules first, then convert to kWh.)

Calculate the ideal number of kWh required to produce10.0 kg of calcium from the electrolysis of moltencalcium chloride if the applied emf is 75 V. Assume100% efficiency.

n = ?

Ca g 40.1Ca mol 110,000 g Ca

Ca mol 1e mol 2 -

= 498.75 mol e–

w = nFEext = 498.75 (96,500)(75) = 3.61 x 109 J

3.61 x 109 J

J/s 1 W1

W1000kW 1

s 3600h 1

= 1.00 x 103 kWh

In reality, electrolysisis only about 50%efficient, which meansthat half of the energythat is put in is givenoff as exhaust heat.