oxidation and reduction
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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!
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