Oxidation and ReductionChapters 20 & 21

Oxidation vs Reduction

• Oxidation= A substance loses electrons

• Reduction A substance gains electrons• 2Al(s)

0 + 3CuCl2(aq) → 2AlCl3(aq) + 3Cu(s)0

• Al(s)0 → AlAl(aq)(aq)

+3+3 aluminum oxno increases

• Cu(aq)+2 → Cu(s)

0 copper oxno decreases

What’s really happening…

• 2Al(s)0 + 3CuCl2(aq) → 2AlCl3(aq) + 3Cu(s)

0

• 2Al(s)0 → 2AlAl(aq)(aq)

+3+3 + 6e- Al oxno increases

• 3Cu(aq)+2 + 6e- → Cu(s)

0 Cu oxno decreases

• These are called half reactions. Notice the number of electrons lost is the same as the number gained.

What????• Oxidation= increase in oxygen atoms, increase

in oxno, loss in electrons• Reduction= loss of oxygen atoms, decrease in

oxno, gain in electrons

• Remember:• “L.E.O. says G.E.R.” • Loss of electrons oxidation• Gain electrons reduction

Agents

• The oxidizing agent causes oxidation of another substance. Example: copper

• Cu(aq)+2 → Cu(s)

0

• The reducing agent causes reduction of another substance. Example: aluminum

• Al(s)0 → AlAl(aq)(aq)

+3+3

Activity Series of Metals (p. 668)

When the the reaction happens, electrons move from Al to Cu

• 2Al(s)0 → 2AlAl(aq)(aq)

+3+3 + 6e-

• 3Cu(aq)+2 + 6e- → Cu(s)

0

• This electron flow can be measured as voltage!• We will see how later.

Types of Redox Reactions• Direct Combination:• S + O2 → SO2• Decomposition:• HgO → 2Hg + O2• Single Replacement:• Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)• Cu(s) + 2Ag+

(aq) → Cu+(aq) + 2Ag(s) (net ionic)

• But: Cu(s) + ZnCl2(aq) → NR• Cu(s) + Zn+2

(aq) → No reaction (due to relative reactivity rank)

Balancing Redox Equations

• Some equations are are difficult to balance by inspection or trial and error that worked up until now.

• The fundamental principle is that the number of electrons lost in the oxidation process must equal the number of electrons gained in the reduction process.

Electrochemical cells

• Use redox reactions to either produce or use electricity.

Voltaic Cells

• In late 1700’s Italian physician Luigi Galvani twitched frog legs by connecting two metals. Italian scientist Alessandro Volta concluded the two metals in the presence of water produce electricity.

Voltaic Cells

• Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

• Zn(s) → Zn+2(aq) + 2e- oxidation

• Cu+2(aq) + 2e- → Cu(s) reduction

• Half Cell- Zn (anode) • Pushes e- to Cu • (cathode)

Voltaic Cell

• Electrons move spontaneously from the anode (-) to the cathode (+)

• The salt bridge allows• Electrons to move freely • Without mixing solutions.

Cell Potential

• Ability to move e- through a wire from one electrode to another is the electrical or cell potential. It is measured in volts (v)

• For Example: A Zn-Cu cell with 1 M solutions produces 1.10 volts.

• Here is how it works:

Standard Reduction Potentials (p. 693)

Standard Electrode Potentials

• Ecell = Eoxidation + Ereduction

• Ecell0= sum of the oxidation potential (Eox

0) plus reduction potential (Ered

0)• The standard state conditions are noted with the 0.• E0 are determined by measuring half cell potential

differences.• Zn(s) → Zn+2

(aq) + 2e- E0 ox = + 0.76 V

• Zn+2(aq) + 2e- → Zn(s) E0 red = - 0.76 V

Calculating Cell Potentials

• Zn(s) → Zn+2(aq) + 2e- E0

ox = + 0.76 V

• Cu+2(aq) 2e- → Cu(s) E0

red = + 0.34 V

• Total Voltage (Ecell) = + 1.10 Volts• Practice Problems #1 and 2 on P. 696.

That’s it for this

• Electrifying lecture!