1 electrochemistry. 2 electron transfer reactions electron transfer reactions are oxidation-...
TRANSCRIPT
1ELECTROCHEMISTRY
2
Electron Transfer ReactionsElectron Transfer Reactions
• Electron transfer reactions are oxidation-
reduction or redox reactions.
• Results in the generation of an electric
current (electricity) or be caused by
imposing an electric current.
• Therefore, this field of chemistry is often
called ELECTROCHEMISTRY.
3
Terminology for Redox Reactions
Terminology for Redox Reactions
• OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen.
• REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.
• OXIDIZING AGENT—electron acceptor; species is reduced.
• REDUCING AGENT—electron donor; species is oxidized.
• OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen.
• REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.
• OXIDIZING AGENT—electron acceptor; species is reduced.
• REDUCING AGENT—electron donor; species is oxidized.
©
Oxidation and reduction happen at the same time
There is no net gain or loss of electrons.
e- You can’t just create them or destroy them!
Example:
Fe2O3 + Al Fe + Al2O3
Fe3+ Al0 Fe0 Al3+
Fe3+ + 3e- Fe Reduction
Al Al2O3 + 3e- Oxidation
Ionic half-equations
OIL - Oxidation Is Loss of electrons
RIG - Reduction Is Gain of electrons
Remember:
And often... (as a quick and simple way to tell):
Oxidation is gain in oxygen or loss of hydrogen
Reduction is loss of oxygen or gain of hydrogen
+7+6+5+4+3+2+10-1-2-3-4-5-6-7
Reduction
Oxi
datio
n
Oxidation state
Oxidation and reduction can be seen as movement up or down a scale of oxidation states
8OXIDATION & REDUCTION
• OXIDATION:
Zn(s) Zn+2(aq) + 2 e-
Metallic zinc is oxidized to zinc ion. Metallic zinc is serving as a reducing agent.(electron loser)
• REDUCTION:
Cu+2(aq) + 2 e- Cu(s)
Copper ion is reduced to copper metal. Copper ion is serving as an oxidizing agent (electron gainer)
• In the overall reaction two electrons are transferred from the zinc metal to the copper ion.
Zn(s) + Cu+2(aq) Zn+2
(aq) + Cu(s)
9You can’t have one… without the other!
• Reduction (gaining electrons) can’t happen without an oxidation to provide the electrons.
• You can’t have 2 oxidations or 2 reductions in the same equation. Reduction has to occur at the cost of oxidation
10OXIDATION NUMBERS
• Before discussing the mechanics of oxidation numbers it is important to realize that:
• (1) oxidation numbers have no physical significance. They are merely a way to tell which substances gain or lose electrons and how many.
• (2) oxidation numbers are assigned to atoms never molecules. Molecules contain atoms with oxidation numbers but they themselves cannot be assigned oxidation numbers.
• (3) there a several rules used to assign oxidation numbers. They must be observed carefully.
11OXIDATION NUMBERS RULES
• (1) Elements have oxidation numbers of zero. For example Cu has an oxidation number of zero. In Cl2 each atom of chlorine has an oxidation number of zero.
• (2) The oxidation number of a monatomic ion (consisting of one atom) is the charge on the ion. In Cu+2 the oxidation number is +2. In Cl- the oxidation number is –1.
• (3) The oxidation number of combined oxygen is always –2 except in peroxides such as hydrogen peroxide, H2O2. In CO2 each oxygen atom has an oxidation number of –2. In SO3 each oxygen is –2.
12FINDING UNKNOWN OXIDATION NUMBERS• Problem: What is the oxidation number of Mn in MnO4
-, the permanganate ion?
• Solution: Mn is unknown Each O is –2 All of the oxidation numbers must add up to –1 Mn + 4(-2) = -1, Mn = +7 in permanganate• Problem: What is the oxidation number of Cr in K2Cr2O7,
potassium dichromate?• Solution: Cr is unknown Each K from Column I is +1 Each O is –2 All the oxidation numbers add up to 0 (no charge is shown for
K2Cr2O7)
2(+1) + 2 Cr + 7(-2) = 0, 2 Cr = 12, Cr = 12 / 2 = +6 Cr = +6 in potassium dichromate
13USING OXIDATION NUMBERS TO DETERMINE ELECTRON TRANSFERS
• Using oxidation numbers we will find the oxidizing agent, the reducing agent, the number of electrons lost and gained, the oxidation half-reaction, the reduction half-reaction and the final balanced equation for:
H+ + MnO4- + Cl- Mn+2 + Cl2 + H2O
• Step I – Find the oxidation number of each atom. H+ + MnO4
- + Cl- Mn+2 + Cl2 + H2O
+1 (+7, -2) -1 +2 0 (+1, -2)• Step II – Determine which oxidation numbers change
MnO4- Mn+2 2 Cl- Cl2
+7 +2 2(-1) 0
5 electrons gained 2 electrons lost
(MnO4- is the oxidizing agent. It gained electrons from Cl- )
(Cl- is the reducing agent. It lost electrons to MnO4- )
If oxidization # s decrease, Reduction occurs
If oxidization # s increase,Oxidation occurs
14USING OXIDATION NUMBERS TO DETERMINE
ELECTRON TRANSFERS• Step III - Electrons lost by a reducing agent must always equal electrons
gained by the oxidizing agent!
• Therefore: 2 (MnO4- + 5e- Mn+2 )
2MnO4- + 10e- 2Mn+2 (reduction half-reaction*)
• And 5(2Cl- Cl2 + 2e-)
10 Cl- 5Cl2 + 10e- (oxidation half-reaction)
• Step IV -Completing the first half-reaction with H+ ions and H2O molecules: 16 H+ + 2MnO4
- + 10e- 2Mn+2 + 8 H2O
• Step V - Adding the oxidation half-reaction and reduction half-reaction the 10 electrons gained and lost cancel to give the overall reaction:
16 H+ + 2MnO4- + 10 Cl- 2 Mn+2 + 5 Cl2 + 8 H2O
• * Half-reaction refers to the reaction showing either the electron gain (reduction) or the electron loss (oxidation) step of the reaction.
15
OXIDATION-REDUCTION REACTIONS
OXIDATION-REDUCTION REACTIONS
Direct Redox ReactionOxidizing and
reducing agents in direct contact.
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
16
OXIDATION-REDUCTION REACTIONS
OXIDATION-REDUCTION REACTIONS
Indirect Redox Reaction
A battery functions by transferring electrons through an external wire from the reducing
agent to the oxidizing agent.
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Why Study Electrochemistry?Why Study Electrochemistry?
• Batteries• Corrosion• Industrial production
of chemicals such as Cl2, NaOH, F2 and Al
• Biological redox reactions
The heme group
18
Balancing Equations for Redox Reactions
Some redox reactions have equations that must be balanced by special techniques.
MnO4- + 5 Fe2+ + 8 H+
---> Mn2+ + 5 Fe3+ + 4 H2O
Mn = +7 Fe = +2Fe = +3Mn = +2
19
Balancing Equations
Consider the reduction of Ag+ ions with copper metal.
Cu + Ag+ --give--> Cu2+ + Ag
20Balancing Equations
Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction.
Ox Cu ---> Cu2+
Red Ag+ ---> AgStep 2: Balance each element for mass. Already done
in this case.Step 3: Balance each half-reaction for charge by
adding electrons.Ox Cu ---> Cu2+ + 2e-Red Ag+ + e- ---> Ag
21Balancing Equations
Step 4: Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the oxidizing agent requires.
Reducing agent Cu ---> Cu2+ + 2e-Oxidizing agent 2 Ag+ + 2 e- ---> 2 AgStep 5: Add half-reactions to give the overall equation.Cu + 2 Ag+ ---> Cu2+ + 2Ag
The equation is now balanced for both charge and mass.
Redox in presence of acid
a) Identify, as oxidation or reduction, the formation of NO2 from NO3
- in the presence of H+ and deduce the half-equation for the reaction.
NO3- NO2
+5 +4Reduction
Steps to take1. Write a balanced equation for the species
2. Work out “before and after” oxidation states
3. Balance oxidation states with electrons
NO3- NO2
+5 +4
NO3- + e- NO2
NO3- + e- + 2H+ NO2
5. If the equation still doesn’t balance, add enough water to one side so it balances
NO3- + e- + 2H+ NO2 + H2O
4. If all the charges don’t balance, add H+ ions to one of the sides to balance them
Beautiful!
Give these a goNa Na+
Fe2+ Fe3+
I2 I¯
C2O42- CO2
H2O2 O2
H2O2 H2O
NO3- NO
NO2 NO3-
SO42- SO2
Na Na+
Fe2+ Fe3+
I2 I¯
C2O42- CO2
H2O2 O2
H2O2 H2O
NO3- NO
NO2 NO3-
SO42- SO2
Balance the half equations
All good?
Na Na+ + e-
Fe2+ Fe3+ + e-
I2 + 2e- 2I¯
C2O42- 2CO2 + 2e-
H2O2 O2 + 2H+ + 2e-
H2O2 + 2H+ + 2e- 2H2O
NO3- + 4H+ + 3e- NO + 2H2O
NO2 + H2O NO3- + 2H+ + e-
SO42- + 4H+ + 2e SO2 + 2H2O
Na Na+ + e-
Fe2+ Fe3+ + e-
I2 + 2e- 2I¯
C2O42- 2CO2 + 2e-
H2O2 O2 + 2H+ + 2e-
H2O2 + 2H+ + 2e- 2H2O
NO3- + 4H+ + 3e- NO + 2H2O
NO2 + H2O NO3- + 2H+ + e-
SO42- + 4H+ + 2e SO2 + 2H2O
Combining half equations
Just a mashing together of two half-equations!
... followed by some satisfying cancelling-out.
NO3- + e- + 2H+ NO2 + H2O
b) Deduce the overall equation for the reaction of copper with NO3
- in acidic conditions to give Cu2+.
Reduction
Cu Cu2+ + 2e- Oxidation
Now what?
1. Multiply the equations so that the number of electrons in each is the same
2. Add the two equations and cancel out the electrons on either side of the equation
3. If necessary, cancel out any other species which appear on both sides of the equation
2NO3- + 2e- + 4H+ 2NO2 + 2H2O
Cu Cu2+ + 2e-
Cu + 2NO3- + 2e- + 4H+ Cu2+
+ 2e- + 2NO2 + 2H2O
Cu + 2NO3- + 4H+ Cu2+
+ 2NO2 + 2H2O
X 2
Steps to take to combine equations
Give these a go1. Fe2+ ions are oxidised to Fe3+ ions by ClO3
- ions in acidic conditions. The ClO3
- ions are reduced to Cl- ions. Write the overall reaction.
2. Write an overall reaction for MnO4- reducing H2O2
to O2 and creating Mn2+ .
Fe2+ Fe3+ + e- + ClO3- + 6e- + 6H+ Cl- + 3H2O
2MnO4¯ + 5H2O2 + 6H+ 2Mn2+ + 5O2 + 8H2O
6Fe2+ + ClO3- + 6H+ Fe3+ + Cl- + 3H2O
Complete the half-equationNa Na+ Na Na++ e-
Complete the half-equationPb4+ Pb2+ Pb4+ + 2e- Pb2+
Complete the half-equationH2 H+ H2 2H+ + 2e-
Complete the half-equationCr2O7
2- Cr3+ Cr2O72- + 6e- 2Cr3+
What’s wrong with this equation?Ce3+ + e- Ce4+ Electron is on the wrong side!
What’s wrong with this equation?Mg + 2H+ + e- Mg+ + H2 + e-
Should be Mg2+ Electrons should be cancelled out
31Balancing Equations
Balance the following in acid solution—
VO2+ + Zn ---> VO2+ + Zn2+
Step 1:Write the half-reactionsOx Zn ---> Zn2+
Red VO2+ ---> VO2+
Step 2:Balance each half-reaction for mass.Ox Zn ---> Zn2+
RedVO2
+ ---> VO2+ + H2O2 H+ +
Add H2O on O-deficient side and add H+ on other side for H-balance.
32Balancing Equations
Step 3:Balance half-reactions for charge.Ox Zn ---> Zn2+ + 2e-Red e- + 2 H+ + VO2
+ ---> VO2+ + H2OStep 4:Multiply by an appropriate factor.Ox Zn ---> Zn2+ + 2e-
Red 2e- + 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2O
Step 5:Add balanced half-reactionsZn + 4 H+ + 2 VO2
+ ---> Zn2+ + 2 VO2+ + 2 H2O
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Tips on Balancing Equations
• Never add O2, O atoms, or O2- to balance oxygen.
• Never add H2 or H atoms to balance hydrogen.
• Be sure to write the correct charges on all the ions.
• Check your work at the end to make sure mass and charge are balanced.
• PRACTICE!
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Electrochemical CellsElectrochemical Cells• An apparatus that allows a
redox reaction to occur by transferring electrons through an external connector.
• Product favoured reaction ---> voltaic or galvanic cell ----> electric current
• Reactant favoured reaction ---> electrolytic cell ---> electric current used to cause chemical change.
Batteries are voltaic cells
35
Anode Cathode
Basic Concepts of Electrochemical Cells
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
36
CHEMICAL CHANGE --->ELECTRIC CURRENT
CHEMICAL CHANGE --->ELECTRIC CURRENT
Zn metal
Cu2+ ions
Zn metal
Cu2+ ions
With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”
With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”
•Zn is oxidized and is the reducing agent Zn(s) ---> Zn2+(aq) + 2e-•Cu2+ is reduced and is the oxidizing agentCu2+(aq) + 2e- ---> Cu(s)
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•To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire.
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
CHEMICAL CHANGE --->ELECTRIC CURRENT
CHEMICAL CHANGE --->ELECTRIC CURRENT
This is accomplished in a GALVANIC or VOLTAIC cell.
A group of such cells is called a battery.
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf
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Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
•Electrons travel through external wire.•Salt bridge allows anions and cations to move between electrode compartments.
•Electrons travel through external wire.•Salt bridge allows anions and cations to move between electrode compartments.
Zn --> Zn2+ + 2e- Cu2+ + 2e- --> Cu
<--AnionsCations-->
OxidationAnodeNegative
OxidationAnodeNegative
ReductionCathodePositive
ReductionCathodePositive
RED CAT
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Terms Used for Voltaic Cells
40
CELL POTENTIAL, E
• For Zn/Cu cell, potential is +1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M.
• This is the STANDARD CELL POTENTIAL, Eo
• —a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C.
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
41
Calculating Cell Voltage
• Balanced half-reactions can be added together to get overall, balanced equation.
Zn(s) ---> Zn2+(aq) + 2e-Cu2+(aq) + 2e- ---> Cu(s)--------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Zn(s) ---> Zn2+(aq) + 2e-Cu2+(aq) + 2e- ---> Cu(s)--------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
If we know Eo for each half-reaction, we could get Eo for net reaction.
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TABLE OF STANDARD REDUCTION POTENTIALS
TABLE OF STANDARD REDUCTION POTENTIALS
2
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H 0.00
Zn 2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
To determine an oxidation from a reduction table, just take the opposite sign of the reduction!
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Zn/Cu Electrochemical Cell
Zn(s) ---> Zn2+(aq) + 2e- Eo = +0.76 VCu2+(aq) + 2e- ---> Cu(s) Eo = +0.34 V---------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Eo = +1.10 V
Cathode, positive, sink for electrons
Anode, negative, source of electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons +
44
45
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
Cd --> Cd2+ + 2e-or
Cd2+ + 2e- --> Cd
Fe --> Fe2+ + 2e-or
Fe2+ + 2e- --> Fe
Eo for a Voltaic Cell
All ingredients are present. Which way does reaction proceed?
46
From the table, you see • Fe is a better reducing
agent than Cd• Cd2+ is a better
oxidizing agent than Fe2+
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
Eo for a Voltaic Cell
47More About Calculating Cell Voltage
Assume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OH- Cathode2 I- ---> I2 + 2e- Anode-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
2 H2O + 2e- ---> H2 + 2 OH- Cathode2 I- ---> I2 + 2e- Anode-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
Assuming reaction occurs as written,
E˚ = E˚red+ E˚ox= (-0.828 V) - (- +0.535 V) = -1.363 V
Minus E˚ means rxn. occurs in opposite direction(the connection is backwards or you are recharging the battery)
48Charging a Battery
When you charge a battery, you are forcing the electrons backwards (from the + to the -). To do this, you will need a higher voltage backwards than forwards. This is why the ammeter in your car often goes slightly higher while your battery is charging, and then returns to normal.
In your car, the battery charger is called an alternator. If you have a dead battery, it could be the battery needs to be replaced OR the alternator is not charging the battery properly.
49
Dry Cell Battery
Anode (-)
Zn ---> Zn2+ + 2e-
Cathode (+)
2 NH4+ + 2e- --->
2 NH3 + H2
50
Alkaline Battery
Nearly same reactions as in common dry cell, but under basic conditions.
Anode (-): Zn + 2 OH- ---> ZnO + H2O + 2e-
Cathode (+): 2 MnO2 + H2O + 2e- --->
Mn2O3 + 2 OH-
51
Mercury Battery
Anode:Zn is reducing agent under basic conditions
Cathode:
HgO + H2O + 2e- ---> Hg + 2 OH-
52
Lead Storage Battery
Anode (-) Eo = +0.36 V
Pb + HSO4- ---> PbSO4 + H+ + 2e-
Cathode (+) Eo = +1.68 V
PbO2 + HSO4- + 3 H+ + 2e-
---> PbSO4 + 2 H2O
53
Ni-Cad Battery
Anode (-)
Cd + 2 OH- ---> Cd(OH)2 + 2e-
Cathode (+)
NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-
54
H2 as a Fuel
Cars can use electricity generated by H2/O2 fuel cells.H2 carried in tanks or generated from hydrocarbons