Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110

General Chemistry

Fall 2003

Lecture Notes::Lec 15_29 September

© R. Paselk 2003
 
     
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Guest lecturer, Dr. Robert Zoellner

Batteries

Batteries are simply galvanic cells, or stacks of galvanic cells. I want to look at a couple of examples. Today we'll focus on the lead acid battery, one of the most important and common batteries.

Lead storage batteries are important and hard to replace because they have a high current density, are tough, durable, handle a wide range of environmental conditions, and are rechargeable.

The half reactions in these batteries can be written as:

anode: Pb(s) + HSO4-(aq) Æ PbSO4(s) + H+ + 2e -

cathode: PbO2(s) + HSO4-(aq) + 3 H+ + 2e - Æ PbSO4(s) + 2 H2O

The anode of the battery is a lead grid coated in "spongy lead" to provide a high surface area and thus high current capacity, while the cathode is a lead grid coated with spongy lead(IV) oxide. (overhead, figure in text)

The rechargeability of this battery arises from the fact that the reactants and reaction products of both half cells are deposited on the electrodes and so are available for both the forward and reverse reactions.

You should read material on other batteries such as dry cells. Note why they are not rechargeable. Also read text on fuel cells, and compare to guest lecture.

Today I want to begin with discussion on electrolytic cells.

Consider a cell composed of Sn2+/Sn and Cu2+/Cu half cells.

Sn2+ + 2e- Æ Sn; E° = - 0.14V

Cu2+ + 2e- Æ Cu; E° = + 0.34V

Sn(s) | Sn2+(aq) || Cu2+(aq) | Cu(s)

anode: Sn Æ Sn2+ + e-; E° = + 0.14V

cathode: Cu2+ + e- Æ Cu; E° = + 0.34V


Cu2+ + Sn Æ Cu + Sn2+; E° = 0.48V

 

Now let's substitute a power supply, (e.g. a battery) with a controller to provide a variable voltage. If we supply just 0.48V with + on + and - on - what will happen?

Reaction stops - DE = 0 so DG = 0!

If we now increase, the voltage, the reverses!

That is electrons are now entering the tin metal electrode, leaving the copper metal electrode. Thus the tin now becomes the cathode, and the copper becomes the anode.

For the electrolytic cell the reactions of the galvanic cell are reversed:

Tin = cathode: Sn2+ + e- Æ Sn

Copper = anode: Cu Æ Cu2+ + e-

The cell is driven backwards by the electromotive force delivered by the power supply.

In principles any galvanic cell can be reversed. Not always true in practice, however. Kinetics, other competing reactions can prevent reversal.

 

Electrolysis

Electrolysis is the process whereby a thermodynamically normally non-spontaneous reaction is forced by the application of external energy. Lets look at the example of the electrolysis of HCl. Chemically the favored reaction would be:

H2 + Cl2 Æ 2 HCl

So if we bubbled hydrogen and chlorine gases over the two electrodes we'd expect to make hydrogen chloride (electrolytic cells just enable us to do chemistry in a controlled and partitioned fashion). But if we set up a cell with a power supply we should be able to force the reverse reaction:

Pt | H2 (g) | H+, Cl-(aq) | Cl2 (g) | Pt ; E° = 1.36V

H2 (g) Æ 2 H+(aq) + 2 e-

Cl2 (g) + 2 e- Æ 2 Cl-(aq)


H2 + Cl2 Æ 2 HCl; E° = 1.36V

 

If we apply a reversed voltage greater than 1.36 V we will form H2 + Cl2 from HCl.

Why don't we need a salt bridge?? (Half reactions are isolated by the fact that streams of gas phase bubbles are separated.)

The external voltage needed to begin electrolysis = decomposition potential. This is often > cell voltage, sometimes by as much as a volt, due to kinetic and or surface effects. The extra voltage is called the overvoltage or overpotential.

Sometimes its not possible to reverse a reaction because of side reactions, e.g.:

Zn | Zn2+ || Cu2+ | Cu

If done in aqueous solution we cannot get Zn to plate out because H2O breaks down first to give H2 rather than Zn.

 

Commercial examples

Molten NaCl electrolysis (>800 °C) see figure in text.

anode: 2 Cl- Æ Cl2 (g) + 2 e-

cathode: (Na+ + e- Æ Na(l)) x 2

What happens with NaCl in H2O? Note species present H2O (H+, O-) Na+, Cl-. Predict outcome from your chemical background.

Aluminum Production: Can't produce aluminum in aqueous solution for the same reason we can't electrolyse NaCl in water - hydrogen gas is released instead of metal. We also can't electrolyse the natural form of aluminum (aka alumina, corundum, sapphire etc.), Al2O3 since its melting point exceeds 2,000°C (too hot for stable, practical containment). however, mixed in proper proportion with Na3AlF6 the melting point is reduced to about 1,000°C which is doable. See figure in text.

anode: 2 Al2OF62- + 12 F- + C Æ 4 AlF63- + CO2 + 4 e-

cathode: AlF63- + 3 e- Æ Al + 6 F-

To give an overall reaction of:

2 Al2O3 + 3 C Æ 4 Al + 3 CO2

Note that the carbon electrodes decompose to give the carbon dioxide.


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