Download Chapter 20 - Electron Transfer Reactions

Chapter 20 Electron Transfer
Reactions
Objectives:
1. Carry out balancing of redox reactions in acidic or basic
solutions;
2. Recall the parts of a basic and commercial voltaic cells;
3. Perform cell potential calculations from standard
reduction potentials;
4. Classify oxidizing and reducing agents;
5. Apply the Nerst equation to redox problems;
6. Determine K from Ecell;
7. Perform electrolysis calculations.
Introduction
• NADH + (1/2)O2 + H+ -----> NAD+ + H2O
• Medicinal Biochemistry:
• http://web.indstate.edu/thcme/mwking/home.html
Introduction
• Pyruvate + CoA + NAD+ ------> CO2 + acetyl-CoA + NADH + H+
Redox Reactions
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
Electron transfer reactions are
________________ or redox reactions.
Redox reactions can result in the generation
of an electric current or be caused by
imposing an electric current.
Therefore, this field of chemistry is often
called _____________________.
Review
• OXIDATION
– _______________________________
• REDUCTION
– _______________________________
• OXIDIZING AGENT
– _______________________________
• REDUCING AGENT
– _______________________________
Redox Reactions
Direct Redox Reaction
Oxidizing and reducing agents in direct contact.
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
Redox Reactions
Indirect Redox Reaction
A battery functions by transferring
electrons through an external wire from
the reducing agent to the oxidizing agent.
Electrochemical Cells
• An apparatus that allows a
redox reaction to occur by
transferring electrons through
an external connector.
• Product favored reaction --->
________________cell ---->
electric current.
• Reactant favored reaction --->
________________ cell --->
electric current used to cause
chemical change.
Batteries are voltaic cells
Electrochemistry
Alessandro Volta, 17451827, Italian scientist and
inventor.
Luigi Galvani, 1737-1798, Italian
scientist and inventor.
Why study electrochemistry?
• Batteries
• Corrosion
• Industrial production
of chemicals such as
Cl2, NaOH, F2 and Al
• Biological redox
reactions
The heme group
Balancing Redox Equations
• 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 = +2
Mn = +2 Fe = +3
Balancing Redox Equations
Cu + Ag+ --give--> Cu2+ + Ag
Balancing Redox Equations
Cu + Ag+
--give--> Cu2+ + Ag
Step 1: Divide the reaction into half-reactions,
one for oxidation and the other for reduction.
Ox
Red
Step 2: Balance each for mass.
Step 3: Balance each half-reaction for charge
by adding electrons.
Ox
Red
Balancing Redox 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
Oxidizing agent
Step 5: Add half-reactions to give the overall equation.
The equation is now balanced for both
charge and mass.
Reduction of VO2+ with Zn
Balance the following in ACID solution:
VO2+ + Zn ---> VO2+ + Zn2+
Step 1: Write the half-reactions
Ox
Red
Step 2: Balance each half-reaction for mass.
Ox
Red
Add H2O on O-deficient side and add H+ on
other side for H-balance.
Balancing…
Step 3: Balance half-reactions for charge.
Ox
Red
Step 4: Multiply by an appropriate factor.
Ox
Red
Step 5: Add balanced half-reactions
Tips on Balancing
• 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!
Balance the following in basic solution:
MnO4- + NO2-  MnO2 + NO3Oxidation half reaction:
Balancing…
MnO4- + NO2-  MnO2 + NO3Reduction half reaction:
Balancing…
MnO4- + NO2-  MnO2 + NO3Oxidation half reaction:
Reduction half reaction:
Multiply by appropriate factor to cancel e- and add both half-reactions
Oxid. X
Red. X
Sum:
Study Exp 11 – Procedure to balance redox reactions – practice and E calculation
Chemical Change ---> Electric Current
With time, Cu plates out onto Zn metal
strip, and Zn strip “disappears.”
Electrons are transferred from Zn to Cu2+,
but there is no useful electric current.
Oxidation:
Zn(s) ---> Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e- ---> Cu(s)
-------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Chemical Change ---> Electric Current
• To obtain a useful
current, we separate
the oxidizing and
reducing agents so that
electron transfer
occurs thru an external
wire.
• This is accomplished in a GALVANIC or VOLTAIC
cell.
• A group of such cells is called a _____________.
Chemical Change ---> Electric Current
• Electrons travel thru external wire.
• _____________ allows anions and cations to move
between electrode compartments.
Zn --> Zn2+ + 2eOxidation
Anode
Negative
Cu2+ + 2e- --> Cu
<--Anions
Cations-->
Reduction
Cathode
Positive
The Cu|Cu2+ and Ag|Ag+ Cell
Electrochemical Cell
• _________ move from anode to cathode in the wire.
• _______ & _________move thru the salt bridge.
Terminology
Figure 20.6
What Voltage does a Cell Generate?
• Electrons are “driven” from anode to cathode by
an _______________________or emf.
• For Zn/Cu cell, this is indicated by a voltage of
1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M.
1.10 V
Zn and Zn2+,
anode
Cu and Cu2+,
cathode
1.0 M
1.0 M
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.
Calculating Cell Voltage
• Balanced half-reactions can be added together to
get overall, balanced equation.
Zn(s) ---> Zn2+(aq) + 2eCu2+(aq) + 2e- ---> Cu(s)
--------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
• If we know Eo for each halfreaction, we could get Eo for net
reaction.
• But we need a reference!
Cell Potential: SHE
• Can’t measure 1/2 reaction Eo directly. Therefore,
measure it relative to a ______________________,
SHE.
2 H+(aq, 1 M) + 2e- <----> H2(g, 1 atm)
Eo = 0.0 V
Zn/Zn2+ half-cell hooked to a SHE.
Eo for the cell = +0.76 V
Zn(s) ---> Zn2+(aq) + 2eNegative
electrode
Positive
electrode
Supplier of
electrons
Acceptor of
electrons
Zn --> Zn2+ + 2eOxidation
Anode
2 H+ + 2e- --> H2
Reduction
Cathode
Reduction of Protons (H+) by Zn
Overall reaction is reduction of H+ by
Zn metal.
Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g)
Eo = +0.76 V
Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is +0.76 V
Zn is a (better) (poorer) reducing agent than H2.
Cu/Cu2+ and H2/H+ Cell,
E0 for the cell = + 0.34 V
Cu2+(aq)
Eo = +0.34 V
+ 2e- ---> Cu(s)
e-
e-
Positive
Negative
Acceptor of
electrons
Supplier of
electrons
Cu2+ + 2e- --> Cu
Reduction
Cathode
H2 --> 2 H+ + 2eOxidation
Anode
Overall reaction is reduction of Cu2+
by H2 gas.
• Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H+(aq)
• Measured Eo = +0.34 V
• Therefore, Eo for Cu2+ + 2e- ---> Cu is
+ 0.34 V
Zn/Cu Electrochemical Cell
+
Anode,
negative,
source of
electrons
• Zn(s) ---> Zn2+(aq) + 2eEo = +0.76 V
• Cu2+(aq) + 2e- ---> Cu(s)
Eo = +0.34 V
--------------------------------------------------------------• Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Cathode,
positive, sink
for electrons
Uses of Eo values
• Organize half-reactions by
relative ability to act as oxidizing
agents.
Cu2+(aq) + 2e- ---> Cu(s) Eo = +0.34 V
Zn2+(aq) + 2e- ---> Zn(s) Eo = –0.76 V
Note that when a reaction is reversed the sign of E˚ is reversed!
Cu2+ is better oxidazing agent than Zn2+;
Cu2+ will be reduced and Zn will be oxidized
Cu2+ reaction (reduction) will occur at the cathode
Zn reaction (oxidation) will occur at the anode.
E˚net = E˚cathode - E˚anode
= (0.43) – (-0.76)
= 1.1 V
Std. Reduction
Potentials
• Organize halfreactions by relative
ability to act as oxidizing
agents. (All references
are written as reduction
processes). Table 20.1
• Use this to predict
direction of redox
reactions and cell
potentials.
Potential Ladder for Reduction HalfReactions
Best
oxidizing
agents
Figure 20.14
Best
reducing
agents
Using Standard Potentials, Eo
Which is the best oxidizing agent: O2, H2O2, or Cl2?
Which is the best reducing agent: Hg, Al, or Sn?
oxidizing
ability of agent
Cu2+ + 2e-
Eo (V)
Cu
+0.34
+
2 H + 2e-
H2
0.00
2+
Zn + 2e-
Zn
-0.76
reducing ability
of agent
Standard Reduction Potentials
Any substance on the right will
reduce any substance higher
than it on the left.
Zn can reduce H+ and Cu+.
H2 can reduce Cu2+ but not Zn2+
Cu cannot reduce H+ or Zn2+.
Standard Reduction Potentials
Ox. agent
Cu2+ + 2e- --> Cu
+
2H
+ 2e- --> H2
Zn2+ + 2e- --> Zn
+0.34
0.00
-0.76
Red. agent
Any substance on the right will reduce any substance
higher than it on the left.
Northwest-southeast rule: product-favored reactions
occur between
• reducing agent at southeast corner
• oxidizing agent at northwest corner
Using Standard Reduction Potentials
In which direction do the following reactions
go?
• Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
• 2 Fe2+(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s)
• What is Eonet for the overall reaction?
Calculating Cell Potential
E˚net = “distance” from “top” half-reaction (cathode)
to “bottom” half-reaction (anode)
E˚net = E˚cathode - E˚anode
Eonet for Cu/Ag+ reaction = +0.46 V
Eo for a Cell
Cd --> Cd2+ + 2eor
Cd2+ + 2e- --> Cd
Fe --> Fe2+ + 2eor
Fe2+ + 2e- --> Fe
All ingredients are present. Which
way does reaction proceed?
Eo for a Cell
From the table, you see
• Fe is a better reducing agent
than Cd
• Cd2+ is a better oxidizing agent
than Fe2+
Overall reaction:
Fe + Cd2+ ---> Cd + Fe2+
Eo = E˚cathode - E˚anode
=
=
Fe/Fe2+ // Cd2+/Cd
More about Eo for a Cell
Assume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OHCathode
2 I- ---> I2 + 2eAnode
------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
Assuming reaction occurs as written,
E˚net = E˚cathode - E˚anode
=
_________ E˚ means rxn. occurs in ___________
direction
It is ____________ favored.
Eo at non-standard conditions
E = Eo - (RT/nF) lnQ
The NERNST
E = Eo - 0.0257/n lnQ
EQUATION
E = potential under nonstandard conditions
R = gas constant (8.314472 J/Kmol)
T = temperature (K)
n = no. of electrons exchanged
F = Faraday constant (9.6485338 x 104 C/mol)
ln = “natural log”
Q = reaction quotent (concentration of
products/concentration of reactants to
the appropriate power)
One ___________ is the
quantity of electric charge
carried by one mole of
electrons.
If [P] and [R] = 1 mol/L, then E = E˚
If [R] > [P], then E is ______________ than E˚
If [R] < [P], then E is ______________ than E˚
A voltaic cell is set up at 25oC with the following half-cells:
Al3+(0.0010 M)/Al and Ni2+(0.50 M)/Ni. Write an equation for the
reaction that occurs when the cell generates an electric current.
a)
Determine which substance is oxidized (decide which is the
better reducing agent).
Al is best reducing agent.
Then Al is oxidized and Ni2+ is reduced.
Ox (Anode):
Red (Cathode):
b)
Add the half-reactions to determine the net ionic equation.
Net eq:
c)
Calculate Eo and use Nernst eq. to calculate E.
Eo = Eocathode – Eoanode
E = Eo – 0.0257/n ln Q
Calculate the cell potential, at 25 °C, based upon the overall
reaction: 3 Cu2+(aq) + 2 Al(s)  3 Cu(s) + 2 Al3+(aq)
if [Cu2+] = 0.75 M and [Al3+] = 0.0010 M.
The standard reduction potentials are as follows:
Cu2+(aq) + 2 e- → Cu(s)
E° = +0.34 V Cathode
Al3+(aq) + 3 e- → Al(s)
E° = -1.66 V Anode
Eo = Eocathode – Eoanode
E = Eo – 0.0257/n ln Q
Eo and Thermodynamics
DE = q + w
The maximum work done by an electrochemical system (ideally) is
proportional to the potential difference (volts) and the quantity
of charge (coulombs):
Wmax = nFE
E is the cell voltage
nF is the quantity of electric charge transferred from anode to
cathode.
• Eo is related to ∆Go, the free energy change for the reaction
(energy released by the cell); under standard conditions:
∆Go = -nFEo
where F = Faraday constant
= 9.6485 x 104 J/V•mol of e(or 9.6485 x 104 coulombs/mol)
and n is the number of moles of electrons transferred
Calculate DGo from Eo
• Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
Eonet for Cu/Ag+ reaction = +0.46 V
DGo = -nFEo
1J = 1C * 1V
1000 J = 1kJ
Eo and the Equilibrium Constant
When Ecell = 0, the reactants and products are
at equilibrium, Q = K
E = 0 = Eo – 0.0257/n ln K
then
ln K = n Eo / 0.0257
(at 25oC)
For: Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
Eonet for Cu/Ag+ reaction = +0.46 V
Eo and the Equilibrium Constant
∆Go
= - n F
Eo
For a product-favored reaction
Reactants ----> Products
∆Go < 0 and so Eo > 0
Eo is positive
For a reactant-favored reaction
Reactants <---- Products
∆Go > 0 and so Eo < 0
Eo is negative
Eo = - DG0
nF
Primary batteries
Uses redox reactions that cannot be restored
by recharge.
* Indicate which reaction goes in
the anode which in the cathode.
Dry cell battery:
_____________
Zn ---> Zn2+ + 2e_____________
2 NH4+ + 2e- ---> 2 NH3 + H2
Alkaline batteries
Nearly same reactions as in common dry cell,
but under basic conditions.
_______________
Zn + 2 OH- ---> ZnO + H2O + 2e_______________ 2 MnO2 + H2O + 2e- ---> Mn2O3 + 2 OH-
Secondary batteries
• Uses redox
reactions that can
be reversed.
• Can be restored by
recharging.
Lead storage batteries
___________ Eo = +0.36 V
Pb + HSO4- ---> PbSO4 + H+ + 2e___________ Eo = +1.68 V
PbO2 + HSO4- + 3 H+ + 2e- ---> PbSO4 + 2 H2O
Ni-Cd battery
______________
Cd + 2 OH- ---> Cd(OH)2 + 2e______________
NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-
Fuel Cell: H2 as fuel
•
•
•
•
Reactants are supplied continuously from an external source.
Cars can use electricity generated by H2/O2 fuel cells.
H2 carried in tanks or generated from hydrocarbons.
Used in space rockets.
Fuel Cell: H2 as fuel
Cathode (red)
Anode (ox)
O2 (g) + 2 H2O (l) + 4e-  4 OH2H2(g)  4 H+ (aq) + 4e-
• Temperature of 70-140oC and
produce ~ 0.9 V.
• The two halves are separated by
a proton exchange membrane (PEM).
• Protons combine with
hydroxide ions forming water.
• The net reaction is then:
2 H2 + O2  2 H2O
Electrolysis
Electric Energy ----> Chemical Change
2 H2O  2 H2 + O2
________________
4 OH- ---> O2(g) + 2 H2O + 4e________________
4 H2O + 4e- ---> 2 H2 + 4 OHEo for cell = -1.23 V
Anode
Cathode
Electrolysis of Molten NaCl
• Electrolysis of molten
NaCl.
• Here a battery “pumps”
electrons from Cl- to Na+.
+
• NOTE: Polarity of
Anode
electrodes is reversed
from batteries.
________________
2 Cl- ---> Cl2(g) + 2e________________
Na+ + e- ---> Na
electrons
BATTERY
Cathode
Cl- Na+
Electrolysis of Molten NaCl
Eo for cell (in water)
=E˚c - E˚a
= - 2.71 V – (+1.36 V)
= - 4.07 V (in water)
External energy needed
because Eo is (-).
Electrolysis of Aqueous NaCl
Anode (+)
2 Cl- ---> Cl2(g) + 2eCathode (-)
2 H2O + 2e- ---> H2 + 2 OHEo for cell = -2.19 V
Note that H2O (-0.8277)
is more easily reduced
than Na+ (-2.71).
Also, Cl- (1.36) is oxidized in
preference to H2O (1.33) because
of kinetics.
Electrolysis of Aqueous CuCl2
Anode (+)
2 Cl- ---> Cl2(g) + 2eCathode (-)
Cu2+ + 2e- ---> Cu
Eo for cell = -1.02 V
Note that Cu is more easily
reduced than either H2O or
Na+.
Electrolytic Refining of Copper
Impure copper is oxidized to Cu2+ at the _________.
The aqueous Cu2+ ions are reduced to Cu metal at the
_______________.
Electrolysis of Aqueous SnCl2
Sn2+(aq) + 2 Cl-(aq) ---> Sn(s) + Cl2(g)
Eocell = Eocathode-Eoanode
=
Al production
2 Al2O3 + 3 C ---> 4 Al + 3 CO2
Charles Hall (1863-1914) developed
electrolysis process. Founded Alcoa.
Counting electrons
• The number of e- consumed or produced in an
electron transfer reaction is obtained by
measuring the current flowing in the circuit in
a given time.
• The current flowing is the amount of charge
(coulombs, C) per unit time, the unit is the
ampere (A).
1 A = 1 C/s
then 1C = A *s
1 F = 9.6485338 x 104 C/mol e1 mol e- = 96,500 C
1.50 amps flow thru a Ag+(aq) solution for 15.0
min. What mass of Ag metal is deposited?
a) Calculate the charge
Charge (C) = current (A) x time (t)
b) Calculate moles of e- used
c) Calculate the mass
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have 454 g of Pb,
how long will the battery last?
a) Calculate moles of Pb
b) Calculate moles of e-
c) Calculate charge (C):
d) Calculate time
Time (sec) = Charge (C)
I (amps)
End of Chapter
• Go over all the contents of your
textbook.
• Practice with examples and with
problems at the end of the chapter.
• Practice with OWL tutor.
• Practice with the quiz on CD of
Chemistry Now.
• Work on your OWL assignment for
Chapter 20.