Honors Chemistry

Chapter 21 Notes – Electrochemistry

(Teacher edition)

Suggested Chapter 21 problem set: 52, 58, 68, 74, 72

21.1Electrochemical Cells

Two Branches of Electrochemistry:

Electrochemistry involves the relation of flow of electrons (current) to chemical changes.

Electrochemistry studies the conversion between chemical and electrical energy.

The 2 main branches are: (both are tied to redox reactions)

Electrolysis: converting electrical energy to chemical energy (electroplating,

charging battery).

Electrochemical cell: converting chemical energy to electrical energy (using a

battery).

Half Reactions and Half Reaction Equations:

Example:

2Na + Cl2  2NaCl

half reaction equations:

2Na  2Na+ + 2e-(oxidation: loss of electrons)

Cl2 + 2e-  2Cl-1(reduction: gain of electrons)

Electric Current: The flow of electric charge (movement of electrons).

in metals – metallic conduction

in solution – ionic conduction – positive and negative ions move along a path

Electrons move away from each other (repulsion) and are attracted to a positive charge.

Direct current: electrons move in one direction only (as opposed to alternating current).

Electrolysis – Current Through an Electrolyte:

Diagram:

Current passes into 1 plate: e- are added from this plate to a chemical in solution.

This is reduction.

Molecules give up electrons at the other plate. This is oxidation.

Electrolysis: electric current causes a redox reaction in a water solution of electrolytes or a

liquid electrolyte.

Cathode: negative plate where reduction occurs.

Anode: positive plate where oxidation occurs.

21.3Electrolytic Cells

Electrolysis – Molten Sodium Chloride:2NaCl(l)  2Na(l) + Cl2(g)

Half reactions:2Na+ + 2 e-  2Na(l)

2Cl-  Cl2(g) + 2e-

Electrolysis of Water:2H2O(l)  2H2(g) + O2(g)

cathode reaction:

electrons added

4H2O + 4e-  2H2 + 4OH-1 (basic)

anode reaction:2H2O  O2 + 4H+1 + 4e- (acidic)

Add the two reactions above and cancel appropriately:

6H2O  2H2 + O2 + 4H+1 + 4OH-1

2H2O(l)  2H2(g) + O2(g

H2SO4 is added to increase the pace of the reaction (electrolyte). HCl cannot be added as Cl-1 could be oxidized to form Cl2 gas. A ternary acid must be used (HNO3, HClO4, etc.).

Electrolysis of Salt Water:

2H2O + 2e-  H2 + 2OH-1

2Cl-1  Cl2 + 2e-

Na is not as easily reduced as water (really Hydrogen).

Water is not as easily oxidized as Cl-1 (really Oxygen)think electronegativity

ionic equation:

2H2O + 2Cl-1  H2 + Cl2 + 2OH-1

molecular equation:

2H2O + 2NaCl(aq)  H2 + Cl2 + 2NaOH(aq)

Na+ is spectator ion

Electroplating – coating a cheap metal with an expensive metal

Show diagram: see previous diagram on page 2

Imagine that the anode is a piece of silver metal and that the cathode is a spoon

made up of some other cheap metal.

anode reaction (silver metal plate):AgAg++e-

cathode reaction (spoon) :Ag++e-Ag

21.2 Half-Cells and Cell Potentials - We will skip everything about types of cells – this

information is not necessary to do problems.

To figure out which happens, oxidation or reduction, for two sets of chemicals – choose

the higher, more positive value.

Ex1:2H+ + 2e-  H2 (0.0 V)

Or…

Ag+ + e-  Ag (+0.80 V)

So…

Ag+ + e-  Ag

And…

H2  2H+ + 2e-

Ex2: Cu and Hydrogen (Cu reduction potential = +.52, H = 0.0)

So… Cu gets reduced, Hydrogen gets oxidized.

To find the total Eo (electromotive force) for the reaction: just add the Eo for each

reaction, but switch sign if it’s an oxidation.

Ex3:

Pb+2+2 e-Pb(-0.13 V)

Sn+4+2 e-Sn+2(+0.15 V)

So…

Sn+4+2 e-Sn+2(+0.15 V)

PbPb+2+2 e-(+0.13 V)

+0.28 V

a positive value means the reaction happens

Ex4:

Br2+2 e-2 Br-1(+1.09 V)

Cl2+2 e-2 Cl-1(+1.36 V)

So…

Cl2+2 e-2 Cl-1(+1.36 V)

2 Br-1Br2+2 e-(-1.09 V)

+0.25 Vdoes happen

This makes sense:

2 NaBr + Cl2  2NaCl(Cl is more active than Br)

2 NaCl + Br2  NR(Br can’t replace Cl)

Ex5:

Al + K+1  does this happen?

Al+3+3 e-  Al (-1.66 V)

K+1+ e-K(-2.93 V)

So…

Because Al is more easily reduced, this reaction doesn’t happen.

Al+KCl NR

This makes sense. Remember that K is higher on the activity series than Al.

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