Voltage Needed for Electrolysis• To drive a reaction in the non-spontaneous direction, we mustapply a voltage at least as large as the voltage produced bythe spontaneous reaction.the spontaneous reaction.

Zn (s) | Zn2+ (aq) | Cu2+ (aq) | Cu (s) discharge: E0 = 1.10 V

we need to apply at least (-)1 10 V to overcome the driving forcewe need to apply at least (-)1.10 V to overcome the driving forceof the standard Daniell cell and charge it.

• Often the applied voltage must be larger than this minimum(thermodynamic) value to get an “uphill” reaction to go at areasonable rate.reasonable rate.

This excess voltage is called the overvoltage.

many multi-electron redox processes are slow (e.g. H2O oxidation) a y u e ec o edo p ocesses a e s o (e g 2O o da o )and have large overvoltages (> 0.5 V).

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Faraday’s Law of Electrolysis Faraday’s Law: The amounts of the substances consumed or produced at the electrodes of an electrolysis cell is

ti l t th h th t th h th llproportional to the charge that passes through the cell.

Zn2+ (aq) + 2 e− → Zn (s)Zn (aq) + 2 e → Zn (s)stoichiometry: 2 moles of electrons to make 1 mole of Zn

We can relate: current ↔ moles electrons ↔ amounts of products

1. Charge = current × time 1 C = 1 A × 1 s

2. Charge of 1 mole electrons = F = 96,485 C / mol293

EXAMPLE: Calculate the volume of H2 gas at 25 °C and 1 atm that will collect at the cathode when water is1 atm that will collect at the cathode when water is electrolyzed for 2.00 hours with a 10.0 A current.

2 H O → 2 H + O

Total charge passed through cell:

2 H2O → 2 H2 + O2

g p g10 Amp × 2.00 hr × 3600 s/hr = 72,000 C

Moles of electrons:Moles of electrons:72,000 C / 96,485 C mol-1 = 0.746 mol e-

Moles of H2 (2 e- per H2):Moles of H2 (2 e per H2): 0.746 / 2 = 0.373 moles H2

Volume at 1 atm:Volume at 1 atm:V = nRT/P = (0.373)(0.08206)(298)/(1) = 9.12 L294

Electroplating / ElectrodepositionThe process of coating a cathode with a film (usually a metal) using an electrolysis cell.

295 The metal ions are said to “plate out” onto the cathode.

EXAMPLE: What is the oxidation number of the osmium ion in an unknown salt if 26.7 grams of osmium metal gplates out when a current of 15 amps is passed through a solution of this salt for 1 hour?

Cathode: Osn+ + n e- → Os (s) n = ?

Total charge passed through cell:Total charge passed through cell:15 Amps × 1.00 hr × 3600 s/hr = 54,000 C

Moles of electrons:Moles of electrons:54,000 C / 96,485 C mol-1 = 0.56 mol e-

0 56 mol e- produced 26 7 grams of osmium metal how0.56 mol e produced 26.7 grams of osmium metal – how many moles of Os is that?

Moles of osmium: 26.7 g / 190.23 g mol-1 = 0.14 mol Os6 g / 90 3 g o 0 o Os

4 electrons used per Os: n = 4296

EXAMPLE: A constant current is passed through an electrolysis cell containing molten Al O at 900°C for 18electrolysis cell containing molten Al2O3 at 900 C for 18 hours. If 480,000 grams of Al are obtained, what is the current in amperes?

Moles of aluminum produced:480,000 g / 27 g mol-1 = 17,778 moles Al

(17,778 mol Al) × (3 mol e- / mol Al) = 53,333 mol e-

g g

Moles of electrons needed:(17,778 mol Al) (3 mol e / mol Al) 53,333 mol e

Current needed: (53 333 mol e-) × (96 485 C/mol) / 64 800 s = 79 411 A(53,333 mol e ) × (96,485 C/mol) / 64,800 s = 79,411 A

Aluminum production consumes roughly 5%

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Aluminum production consumes roughly 5% of electricity generated in the U.S.

Aluminum Production

aluminum production needs lots ofaluminum production needs lots of electricity, so aluminum smelters are

usually sited near cheap hydroelectric power.

In U.S.: Columbia River, WAGrand Coulee Dam

y p

A single cell may use 500 000 amps

Electrolyte is molten Al2O3 in cryolite (Na3AlF6)

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A single cell may use 500,000 amps operating at 3-5 V