Transcript Slide 1

ElectroChemistry
GHS Honors Chem
Electrochemistry
GHS Honors
Chem
• 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
Chem
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.
GHS Honors
Chem
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).
GHS Honors
Chem
Redox Reactions
• The substance being reduced is
called the oxidizing agent.
• The substance being oxidized is
called the reducing agent.
GHS Honors
Chem
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.
GHS Honors
Chem
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
GHS Honors
Chem
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)
GHS Honors
Chem
– 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.
GHS Honors
Chem
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
GHS Honors
Chem
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
GHS Honors
Chem
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?
GHS Honors
Chem
Redox Reactions
1. Identify the substance oxidized, substance reduced,
oxidizing agent, reducing agent, and write oxidation
and reduction half reactions
2H2
Oxidized
(RA)
GHS Honors
Chem
+ 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 +
GHS Honors
Chem
→
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 +
GHS Honors
Chem
2e-  Sn+2
Oxidation Half reaction
Reduction Half Reaction
Redox Reactions
Redox Reactions
Worksheet
GHS Honors
Chem
Electrochemical Cells
GHS Honors
Chem
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
+
GHS Honors
Chem
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)
GHS Honors
Chem
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
GHS Honors
Chem
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
GHS Honors
Chem
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
GHS Honors
Chem