ElectroChemistry
GHS Honors Chem
Electrochemistry
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• Electrochemistry is the study of
the relationships between
electrical energy and chemical
reactions
• It’s the study of how chemical
energy is changed to electrical
energy through the exchange
(flow) of electrons
Oxidation-Reduction
Reactions
Oxidation-reduction reactions are
chemical reactions involving the
exchange of electrons between two
substances.
• During an oxidation reaction, there is a
loss of electrons.
• For example, the oxidation of Fe(s) to
Fe+2(aq) is accompanied by the loss of
two electrons.
Fe(s)  Fe+2(aq) + 2 e GHS Honors
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Oxidation-Reduction
Reactions
• During a reduction reaction, there is a gain of
electrons.
• Example: reduction of Cu+2(aq) to Cu(s) is
accompanied by the gain of two electrons.
Cu+2(aq) + 2 e -  Cu(s)
• Short-cut:
• Oxidation Is Loss (OIL) of electrons
(electrons on the right-hand side of the equation)
• Reduction Is Gain (RIG) of electrons
(electrons on the left-hand side of the equation)
• Remember OIL-RIG.
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Redox Reactions
• Oxidation and reduction occur together.
• Hence, they are called redox reactions.
• In a redox reaction, one substance
gains electrons (reduction), while the
other substance loses electrons
(oxidation).
Cu+2(aq) + Fe(s)  Fe+2(aq) + Cu(s)
• In this reaction, Cu+2(aq) is reduced to
Cu (s) and Fe(s) is simultaneously
oxidized to Fe+2(aq).
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Redox Reactions
• The substance being reduced is
called the oxidizing agent.
• The substance being oxidized is
called the reducing agent.
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Oxidation Number
• The oxidation number is a number
which tells us how oxidized or
reduced a given element of a given
substance is.
• The higher the oxidation number is,
the more oxidized the element is.
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Assigning Oxidation
Numbers
1. Rule # 1: The oxidation number for an element in
the elemental state is 0.
– the oxidation numbers for Al in Al(s) is 0.
2. Rule # 2: The oxidation number for oxygen in most
oxygen compounds is equal to - 2.
3. Rule # 3: Group IA compounds, have an oxidation
number of +1
– Na in NaCl(s) or in Na2SO4(s) has an oxidation number
of +1
4. Rule # 4: Group IIA compounds, the oxidation
number of the metal is +2.
– Ca in CaCO3 and in Ca(NO3)2 has an oxidation number
of +2
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Assigning Oxidation
Numbers
5. Rule # 5: In all HALOGEN-containing compounds,
the oxidation number for the halogen is -1.
– The oxidation number of F in NaF, CaF2 and AlF3 is
always -1.
– In NaCl, Cl has an oxidation number of -1.
6. Rule # 6: In a compound, the oxidation number of
hydrogen is +1 if H is bonded to a nonmetal.
– H in NH3, in CH4, in H2O and in HCN has the same
oxidation number of +1.
7. Rule # 7: In a compound, the oxidation number of
hydrogen is -1 if H is bonded to a metal. (Note: in
this case, H behaves like an anion and is called
hydride)
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– H in NaH (sodium hydride), in CaH2 (calcium hydride) has
the oxidation number of -1.
Assigning Oxidation
Numbers
8. Rule # 8: The sum of the oxidation numbers of all
elements in a compound is equal to 0 (the charge of
the compound).
– For example in NO2  Ox.# of N + 2 x Ox # O = 0.
9. Rule # 9: The sum of the oxidation numbers of all
elements in a polyatomic ion is equal to the charge
of the polyatomic ion.
– For example in CO3-2, Ox # C + 3 x Ox # O = -2.
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Assigning Oxidation #
Examples
1. Na2SO4:
– Oxidation number of Na = “Na” = +1
– Oxidation number of O = “O” = -2
– Oxidation number of S: = “S” to be
calculated
– 2 “Na” + “S” + 4 “O” = 0
– 2 x (+1) + “S” + 4 x (-2) = 0
– “S” – 6 = 0
– “S” = +6
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Assigning Oxidation #
Examples
1. Ox # of Pt in K2PtCl4
2. Ox # of Mn in KMnO4
3. Ox # of Pb in PbSO4
4. Ox # of Pb in PbS2
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5. Ox # of each C in C2H3O2-
Redox Reactions
Identifying:
•What is Oxidized (the Reducing Agent)
•What is Reduced (the Oxidizing Agent)
… in a Redox Reaction
How’s It Done?
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Redox Reactions
1. Identify the substance oxidized, substance reduced,
oxidizing agent, reducing agent, and write oxidation
and reduction half reactions
2H2
Oxidized
(RA)
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+ O2
Reduced
(OA)
→
2H2O
2H2 0
→ 4H+1 + 4e- Oxidation Half reaction
O2 0 + 4e-
→ 2O-2
Reduction Half Reaction
Redox Reactions
2. Identify the substance oxidized, substance reduced,
oxidizing agent, reducing agent, and write oxidation
and reduction half reactions
Mg
+
Zn+2
Oxidized
Reduced
(RA)
(OA)
Mg
Zn+2 +
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→
2e-
Mg+2 +
→ Zn
→
Mg+2
+
Zn
2e- Oxidation Half reaction
Reduction Half Reaction
Redox Reactions
3. Identify the substance oxidized, substance reduced,
oxidizing agent, reducing agent, and write oxidation
and reduction half reactions
Sn
+ SnO2 + 2H2SO4 → 2SnSO4 + 2H2O
Oxidized Reduced
(RA)
(OA)
Sn  Sn+2 + 2e-
Sn+4 +
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2e-  Sn+2
Oxidation Half reaction
Reduction Half Reaction
Redox Reactions
Redox Reactions
Worksheet
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Electrochemical Cells
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Let’s Look at the Handouts as an Intro to Electrochemical Cells
Calculating Standard Cell
Potentials
Example: Calculate the standard cell
potential for:
Fe (s)  Fe+2 (aq)  Ag+ (aq)  Ag (s)
Fe+2 + 2e-  Fe(s) Eo = -0.44V
Ag+1 + 1e-  Ag(s) Eo = +0.80V
The Ag reduction has a higher E0 (higher on the list), so it is the
reduction … the Fe reaction is turned around to an oxidation:
+2
Fe (s)  Fe
+
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Ag
-
(aq)
+2e
-
(aq)
+ e  Ag
(s)
Oxidation
E°anode = +0.44 V
Reduction
E°cathode = + 0.80 V
Calculating Standard Cell
Potentials
• In order to write the overall redox reaction, multiply
the reduction half-reaction by 2, but NOT the
potential value. Thus,
-
Fe (s)  Fe+2 (aq) + 2 e
-
2 x (Ag+ (aq) + e  Ag
(s))
Oxidation
E°anode = +0.44 V
Reduction
E°cathode = + 0.80 V
Hence,
Fe (s)  Fe+2 (aq) + 2 e
-
-
2 Ag + (aq) + 2 e  2 Ag
(s)
Oxidation
E°anode = +0.44 V
Reduction
E°cathode = + 0.80 V
Overall Redox reaction:
• Fe (s) + 2 Ag+ (aq)  Fe+2 (aq) + 2 Ag (s)
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Calculating Standard Cell
Potentials
E°cell = E°red +
E°ox
E°cell = (+ 0.80 V) + (+0.44 V)
E°cell = + 1.24 V
• Since E°cell is positive, the cell
operates spontaneously.
• Reaction will take place only if E°cell is
positive
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Calculating Standard Cell
Potentials
Example: Calculate the standard cell
potential for:
Al (s)  Al+3 (aq)  Hg+2 (aq)  Hg (s)
Al+3 +
3e-  Al(s) Eo = -1.66V
Hg+2 + 2e-  Hg(s) Eo = +0.85V
The Hg reduction has a higher E0 (higher on the list), so it is the
reduction … the Al reaction is turned around to an oxidation:
Al(s)
 Al+3 + 3e-
Eo = +1.66V
Hg+2 + 2e-  Hg(s) Eo = +0.85V
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Calculating Standard Cell
Potentials
The Hg reduction has a higher E0 (higher on the list), so it is the
reduction … the Al reaction is turned around to an oxidation:
2 x (Al(s)
 Al+3 + 3e-)
Eo = +1.66V
3 x (Hg+2 + 2e-  Hg(s)) Eo = +0.85V
2Al(s) + 3Hg+2
 2Al+3 + 3Hg
Eo Cell: Eo red + Eo ox
Eo Cell = 0.85V + 1.66V = 2.51V
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