Chapter 20 electron transfer reactions
This presentation is the property of its rightful owner.
Sponsored Links
1 / 74

Chapter 20 - Electron Transfer Reactions PowerPoint PPT Presentation


  • 285 Views
  • Uploaded on
  • Presentation posted in: General

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;

Download Presentation

Chapter 20 - Electron Transfer Reactions

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


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

Introduction

  • NADH + (1/2)O2 + H+ -----> NAD+ + H2O

  • Medicinal Biochemistry:

  • http://web.indstate.edu/thcme/mwking/home.html


Introduction1

Introduction

  • Pyruvate + CoA + NAD+ ------> CO2 + acetyl-CoA + NADH + H+


Redox reactions

Redox Reactions

Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)

Electron transfer reactions are ________________or redoxreactions.

Redox reactions can result in the generation of an _________________ or be caused by imposing ___________________.

Therefore, this field of chemistry is often called _____________________.


Why study electrochemistry

Why study electrochemistry?

  • Batteries

  • Corrosion

  • Industrial production of chemicalssuch as Cl2, NaOH, F2 and Al

  • Biological redox reactions

The heme group


Review

Review

  • OXIDATION

    • _______________________________

  • REDUCTION

    • _______________________________

  • OXIDIZING AGENT

    • _______________________________

  • REDUCING AGENT

    • _______________________________


Redox reactions1

Redox Reactions

Direct Redox Reaction

Oxidizing and reducing agents in direct contact.

Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)


Redox reactions2

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

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

Electrochemistry

Alessandro Volta, 1745-1827, Italian scientist and inventor.

Luigi Galvani, 1737-1798, Italian scientist and inventor.


Balancing redox equations

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


Balancing redox equations1

Balancing Redox Equations

Cu + Ag+ --give--> Cu2+ + Ag


Balancing redox equations2

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 equations3

Balancing Redox Equations

Step 4: Multiply each half-reaction by a factor so that the reducing agent supplies as many electronsas 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 vo 2 with zn

Reduction of VO2+ with Zn


Balance the following in acid solution

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

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

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

Balance the following in basic solution:

MnO4- + NO2- MnO2 + NO3-

Oxidation half reaction:


Balancing1

Balancing…

MnO4- + NO2- MnO2 + NO3-

Reduction half reaction:


Balancing2

Balancing…

MnO4- + NO2- MnO2 + NO3-

Oxidation 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

Electrons are transferred from Zn to Cu2+, but there is no useful electric current.

Chemical Change --->Electric Current

With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”

Oxidation: Zn(s) ---> Zn2+(aq) + 2e-

Reduction: Cu2+(aq) + 2e- ---> Cu(s)

--------------------------------------------------------

Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)


Chemical change electric current1

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 current2

Chemical Change --->Electric Current

  • • Electrons travel thru external wire.

  • _____________ allows anions and cations to move between electrode compartments.

Zn --> Zn2+ + 2e-

Cu2+ + 2e- --> Cu

Oxidation

Anode

Negative

Reduction

Cathode

Positive

<--Anions

Cations-->


The cu cu 2 and ag ag cell

The Cu|Cu2+ and Ag|Ag+ Cell


Electrochemical cell

Electrochemical Cell

  • _________ move from anode to cathode in the wire.

  • _______ & _________move thru the salt bridge.


Terminology

Terminology

Figure 20.6


What voltage does a cell generate

1.10 V

1.0 M

1.0 M

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


Cell potential e

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

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)

  • If we know Eo for each half-reaction, we could get Eo for net reaction.

  • But we need a reference!


Cell potential she

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 zn 2 half cell hooked to a she e o for the cell 0 76 v

Negative electrode

Positive electrode

Zn/Zn2+ half-cell hooked to a SHE.Eo for the cell = +0.76 V

Zn(s) ---> Zn2+(aq) + 2e-

Supplier of electrons

Acceptor of electrons

Zn --> Zn2+ + 2e-

Oxidation

Anode

2 H+ + 2e- --> H2

Reduction

Cathode


Reduction of protons h by zn

Reduction of Protons (H+) by Zn


Overall reaction is reduction of h by zn metal

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 cu 2 and h 2 h cell e 0 for the cell 0 34 v

Cu/Cu2+ and H2/H+ Cell,E0 for the cell = + 0.34 V

Eo = +0.34 V

Cu2+(aq) + 2e- ---> Cu(s)

e-

e-

Positive

Negative

Acceptor of electrons

Supplier of electrons

Cu2+ + 2e- --> Cu

Reduction

Cathode

H2 --> 2 H+ + 2e-

Oxidation

Anode


Overall reaction is reduction of cu 2 by h 2 gas

Overall reaction is reduction of Cu2+ by H2 gas.

  • Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H+(aq)

  • Measured Eo = +0.34 V

  • Therefore, Eofor Cu2+ + 2e- ---> Cu is + 0.34 V


Zn cu electrochemical cell

+

Zn/Cu Electrochemical Cell

Anode, negative, source of electrons

Cathode, positive, sink for electrons

  • 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)


Uses of e o values

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


Std reduction potentials

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

Potential Ladder for Reduction Half-Reactions

Best oxidizing agents

Figure 20.14

Best reducing agents


Using standard potentials e o

2+

Cu

+ 2e- Cu

+0.34

+

2 H

+ 2e- H2

0.00

2+

Zn

+ 2e- Zn

-0.76

reducing ability

of agent

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

Eo (V)


Standard reduction potentials

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 potentials1

2+

Cu

+ 2e- --> Cu

+0.34

+

2 H

+ 2e- --> H2

0.00

2+

Zn

+ 2e- --> Zn

-0.76

Standard Reduction Potentials

Ox. agent

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

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

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


E o for a cell

Eo for a Cell

Cd --> Cd2+ + 2e-

or

Cd2+ + 2e- --> Cd

Fe --> Fe2+ + 2e-

or

Fe2+ + 2e- --> Fe

All ingredients are present. Which way does reaction proceed?


E o for a cell1

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 e o for a cell

More about Eo for a Cell

Assume 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 ___________ direction

It is ____________ favored.


E o at non standard conditions

Eo at non-standard conditions

E = Eo

- (RT/nF) lnQ

E = Eo

- 0.0257/n lnQ

The NERNST 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˚


Chapter 20 electron transfer reactions

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.

  • 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


Chapter 20 electron transfer reactions

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

Al3+(aq) + 3 e- → Al(s)E° = -1.66 V

Eo = Eocathode – Eoanode

Cathode

Anode

E = Eo – 0.0257/n ln Q


E o and thermodynamics

Eoand 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 d g o from e o

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


E o and the equilibrium constant

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


E o and the equilibrium constant1

Eo and the Equilibrium Constant

Eo = - DG0

nF

  • ∆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


Primary batteries

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

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

Secondary batteries

  • Uses redox reactions that can be reversed.

  • Can be restored by recharging.


Lead storage batteries

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

Ni-Cd battery

______________

Cd + 2 OH- ---> Cd(OH)2 + 2e-

______________

NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-


Fuel cell h 2 as fuel

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 h 2 as fuel1

Fuel Cell: H2 as fuel

  • Cathode (red) O2 (g) + 2 H2O (l) + 4e-  4 OH-

  • Anode (ox) 2H2(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

Electrolysis

Electric Energy ----> Chemical Change

2 H2O  2 H2 + O2

________________

4 OH- ---> O2(g) + 2 H2O + 4e-

________________

4 H2O + 4e- ---> 2 H2 + 4 OH-

Eo for cell = -1.23 V

Anode

Cathode


Electrolysis of molten nacl

Electrolysis of Molten NaCl

•  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 nacl1

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

Electrolysis of Aqueous NaCl

Anode (+)

2 Cl- ---> Cl2(g) + 2e-

Cathode (-)

2 H2O + 2e- ---> H2 + 2 OH-

Eo 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 cucl 2

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+.


Electrolytic refining of copper

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 sncl 2

Electrolysis of Aqueous SnCl2

  • Sn2+(aq) + 2 Cl-(aq) ---> Sn(s) + Cl2(g)

Eocell = Eocathode-Eoanode

=


Al production

Al production

2 Al2O3 + 3 C ---> 4 Al + 3 CO2

Charles Hall (1863-1914) developed electrolysis process. Founded Alcoa.


Counting electrons

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 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

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


Chapter 20 electron transfer reactions

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?

  • Calculate moles of Pb

  • Calculate moles of e-

  • c)Calculate charge (C):

  • d) Calculate time

Time (sec) = Charge (C)

I (amps)


End of chapter

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.

  • Work on your OWL assignment for Chapter 20.


  • Login