Chapter 20 - Electron Transfer
ReactionsObjectives: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 ReactionsCu(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 ReactionsDirect Redox Reaction
Oxidizing and reducing agents in direct contact.
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
Redox ReactionsIndirect 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
ElectrochemistryAlessandro Volta, 1745-1827, 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 Fe = +3Mn = +2
Balancing Redox Equations Cu + Ag+ --give--> Cu2+ + Ag
Balancing Redox Equations
Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction.OxRed
Step 2: Balance each for mass.
Step 3: Balance each half-reaction for charge by adding electrons.OxRed
Cu + Ag+ --give--> Cu2+ + Ag
Balancing Redox EquationsStep 4: Multiply each half-reaction by a factor so
that the reducing agent supplies as many electrons as the oxidizing agent requires.
Reducing agentOxidizing 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-reactionsOxRed
Step 2: Balance each half-reaction for mass.OxRed
Add H2O on O-deficient side and add H+ on other side for H-balance.
Balancing… Step 3: Balance half-reactions for charge.
OxRed
Step 4: Multiply by an appropriate factor.OxRed
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 + NO3-
Oxidation half reaction:
Balancing…MnO4
- + NO2- MnO2 + NO3
-
Reduction half reaction:
Balancing…MnO4
- + NO2- MnO2 + NO3
-
Oxidation half reaction: Reduction half reaction: Multiply by appropriate factor to cancel e- and add both half-reactions
Oxid. XRed. X
Sum:
Study Exp 11 – Procedure to balance redox reactions – practice and E calculation
Chemical Change ---> Electric Current
Oxidation: Zn(s) ---> Zn2+(aq) + 2e-Reduction: Cu2+(aq) + 2e- ---> Cu(s) -------------------------------------------------------- Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Zn metal
Cu2+ ions Electrons are transferred from Zn to Cu2+, but there is no useful electric current.
With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”
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 _____________.
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Chemical Change ---> Electric Current
• Electrons travel thru external wire.• _____________ allows anions and cations to move
between electrode compartments.
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electronsZn --> Zn2+ + 2e- Cu2+ + 2e- --> Cu
<--AnionsCations-->
OxidationAnodeNegative
ReductionCathodePositive
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.
Zn and Zn2+,anode
Cu and Cu2+,cathode
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons1.10 V
1.0 M 1.0 M
Cell Potential, EFor 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
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
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons• If we know Eo for each half-reaction, 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) + 2e-
Volts
ZnH2
Salt Bridge
Zn2+ H+
Zn Zn2+ + 2e- OXIDATION ANODE
2 H+ + 2e- H2REDUCTIONCATHODE
- +
Negative electrode
Supplier of electrons
Acceptor of electrons
Positive electrode
2 H+ + 2e- --> H2
ReductionCathode
Zn --> Zn2+ + 2e- Oxidation
Anode
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.
Volts
ZnH2
Salt Bridge
Zn2+ H+
Zn Zn2+ + 2e- OXIDATION ANODE
2 H+ + 2e- H2REDUCTIONCATHODE
- +
Cu/Cu2+ and H2/H+ Cell,E0 for the cell = + 0.34 V
Cu2+(aq) + 2e- ---> Cu(s)
Volts
CuH2
Salt Bridge
Cu2+ H+
Cu2+ + 2e- Cu REDUCTION CATHODE
H2 2 H+ + 2e-OXIDATION ANODE
-+
Eo = +0.34 V
Acceptor of electrons
Supplier of electrons
Cu2+ + 2e- --> CuReductionCathode
H2 --> 2 H+ + 2e-Oxidation
Anode
Positive Negative
e- e-
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
Volts
CuH2
Salt Bridge
Cu2+ H+
Cu2+ + 2e- Cu REDUCTION CATHODE
H2 2 H+ + 2e-OXIDATION ANODE
-+
Zn/Cu Electrochemical Cell
• Zn(s) ---> Zn2+(aq) + 2e- Eo = +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
Anode, negative, source of electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons +
Uses of Eo values• Organize half-reactions by relative ability to act as oxidizing agents. Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Cu2+(aq) + 2e- ---> Cu(s) Eo = +0.34 VZn2+(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 oxidizedCu2+ reaction (reduction) will occur at the cathodeZn reaction (oxidation) will occur at the anode.
E˚net = E˚cathode - E˚anode = (0.43) – (-0.76)
= 1.1 V
Std. Reduction Potentials• Organize half-reactions 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 Half-Reactions
Best oxidizing agents
Best reducing agents
Figure 20.14
Using Standard Potentials, Eo
Which is the best oxidizing agent: O2, H2O2, or Cl2?
Which is the best reducing agent: Hg, Al, or Sn?
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+
+ 2e- Zn -0.76
oxidizingability of agent
reducing abilityof 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+.
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
Standard Reduction Potentials
Cu2+ + 2e- --> Cu +0.34
+2 H + 2e- --> H2 0.00
Zn2+ + 2e- --> Zn -0.76
Northwest-southeast rule: product-favored reactions occur between • reducing agent at southeast corner • oxidizing agent at northwest corner
Any substance on the right will reduce any substance higher than it on the left.
Ox. agent
Red. agent
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 PotentialE˚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 CellVolts
Cd Salt Bridge
Cd2+
Fe
Fe2+
Cd --> Cd2+ + 2e-or
Cd2+ + 2e- --> Cd
Fe --> Fe2+ + 2e-or
Fe2+ + 2e- --> Fe
All ingredients are present. Which way does reaction proceed?
Eo for a CellFrom 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+Overall reaction:Fe + Cd2+ ---> Cd + Fe2+
Eo = E˚cathode - E˚anode
==
Fe/Fe2+ // Cd2+/Cd
More about Eo for a CellAssume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OH- Cathode 2 I- ---> I2 + 2e- Anode------------------------------------------------- 2 I- + 2 H2O --> I2 + 2 OH- + H2
Assuming reaction occurs as written, E˚net = E˚cathode - E˚anode
=_________ E˚ means rxn. occurs in ___________ directionIt is ____________ favored.
Eo at non-standard conditions
The NERNST EQUATIONE = potential under nonstandard conditionsR = gas constant (8.314472 J/Kmol)T = temperature (K)n = no. of electrons exchangedF = Faraday constant (9.6485338 x 104
C/mol)ln = “natural log”Q = reaction quotent (concentration of
products/concentration of reactants to the appropriate power)
E = Eo - (RT/nF) lnQ E = Eo - 0.0257/n lnQ
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˚
One ___________ is the quantity of electric charge carried by one mole of electrons.
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 = Eo
cathode – 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 VAl3+(aq) + 3 e- → Al(s) E° = -1.66 V
Eo = Eocathode – Eo
anode
CathodeAnode
E = Eo – 0.0257/n ln Q
Eo and Thermodynamics
• 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
DE = q + wThe maximum work done by an electrochemical system (ideally) is proportional to the potential difference (volts) and the quantity of charge (coulombs):Wmax = nFEE is the cell voltagenF is the quantity of electric charge transferred from anode to cathode.
Calculate DGo from Eo
• Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
Eonet for Cu/Ag+ reaction = +0.46 V
1J = 1C * 1V1000 J = 1kJ
DGo = -nFEo
Eo and the Equilibrium ConstantWhen Ecell = 0, the reactants and products are
at equilibrium, Q = KE = 0 = Eo – 0.0257/n ln K
thenln 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 > 0Eo is positive
For a reactant-favored reaction Reactants <---- Products
∆Go > 0 and so Eo < 0Eo is negative
Eo = - DG0
nF
Primary batteriesUses 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 batteriesNearly 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 VPbO2 + 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 fuelCathode (red) O2 (g) + 2 H2O (l) + 4e- 4 OH-
Anode (ox) 2H2(g) 4 H+ (aq) + 4e-
• Temperature of 70-140oC andproduce ~ 0.9 V.
• The two halves are separated bya proton exchange membrane (PEM).
• Protons combine withhydroxide ions forming water.
• The net reaction is then:2 H2 + O2 2 H2O
ElectrolysisElectric Energy ----> Chemical Change
Anode Cathode2 H2O 2 H2 + O2
________________4 OH- ---> O2(g) + 2 H2O + 4e-
________________4 H2O + 4e- ---> 2 H2 + 4 OH-
Eo for cell = -1.23 V
Electrolysis of Molten NaCl
BATTERY
+
Na+Cl-
Anode Cathode
electrons• Electrolysis of molten NaCl.
• Here a battery “pumps” electrons from Cl- to Na+.
• NOTE: Polarity of electrodes is reversed from batteries.________________2 Cl- ---> Cl2(g) + 2e-
________________Na+ + e- ---> Na
Electrolysis of Molten NaClEo 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 NaClAnode (+) 2 Cl- ---> Cl2(g) + 2e-
Cathode (-) 2 H2O + 2e- ---> H2 + 2 OH-
Eo for cell = -2.19 VNote that H2O (-0.8277) is more easily reduced than Na+ (-2.71).
BATTERY
+
Na+Cl-
Anode Cathode
H2O
electrons
Also, Cl- (1.36) is oxidized in preference to H2O (1.33) because of kinetics.
Electrolysis of Aqueous CuCl2
Anode (+)
2 Cl- ---> Cl2(g) + 2e-
Cathode (-)
Cu2+ + 2e- ---> Cu
Eo for cell = -1.02 V
Note that Cu is more easily reduced than either H2O or
Na+.
BATTERY
+
Cu2+Cl-
Anode Cathode
electrons
H2O
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
Charles Hall (1863-1914) developed electrolysis process. Founded Alcoa.
2 Al2O3 + 3 C ---> 4 Al + 3 CO2
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/sthen 1C = A *s
1 F = 9.6485338 x 104 C/mol e- 1 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 isPb(s) + HSO4
-(aq) ---> PbSO4(s) + H+(aq) + 2e-If 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 timeTime (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.
