REDOX 1

REDOX & Electrochemistry

REDOX 2

There are two types of chemical reactions:

1) acid-base reactions (Lewis)

2) electron-transfer reactions

Oxidation/oxidize: when you remove electrons from a material

Reduction/reduce: when you add electrons to a material

You cannot have oxidation without reduction; you cannot generally reduce a molecule without oxidizing another molecule (and the other way around).

Oxidizing agent: a chemical that causes another material to be oxidized (the oxidizing agent is reduced!)

Reducing agent: a chemical that causes another material to be reduced (the reducing agent is oxidized!)

REDOX

REDOX 3

Oxidation States

In order to understand redox reactions, we first need to be able to figure out what the oxidation state of an element is. The oxidation state is a method to indicate how many electrons are "assigned" to a particular element. For this we use a +/- system:

+n indicates that an atom has lost electrons and now has a positive charge

−n indicates that an atom has gained electrons and now has a negative charge

0 indicates that an atom has its elemental number of electrons assigned to it and, therefore, has no charge

Oxidation state is a “formalism”, that is, is may or may not reflect the actual charge on an atom.

Common reference atoms & their oxidations states:

Alkali metals = +1, alkaline earths = +2

O = −2 (exception, peroxides, H2O2, −1)

Halides = −1 (exception, oxyhalides)

REDOX 4

The key to being able to figure out the oxidation state of an element in a molecule is to note its electronegativity:

• The higher the electronegativity the more the element likes to add electrons to its valence shell.

• The lower the electronegativity the more likely an element will lose electrons.

Let's dissect an example reaction: Reaction: + 3 2PH O

PH3 2 + O

P2 5 2 + HO O

P2 5 2 + O OH

Electronegativities: O = 3.5 P = 2.1 H = 2.1

-3 +1 0 +1+5 -2 -2 Atoms that gain electrons usually gain enough to fill their valence shell (octet rule). Atoms that lose electrons only lose enough to get them down to the next lowest filled valence shell (but not always all the way!).

REDOX 5

When two elements have the same electronegativity (as with phosphorus and hydrogen) the element that is furthest to the upper right hand side of the periodic table is the one that gets the electrons.

H

Sc Mn

Li

Na

K

Rb

Cs Ba

Sr

Ca

Mg

Be

Ti V Cr Fe Co Ni Cu Zn

Y Zr Nb Mo Tc Ru Rh Pd Ag Cd

AuPtIrOsReWTaHfLa

He

Ne

Ar

Kr

Xe

Rn

F

Cl

I

At

O

S

Se

Te

Po

N

P

As

Sb

Bi

C

Si

Ge

Sn

Pb

B

Al

In

Tl

1 2

3 4 5 6 7 8 9 10

13 14 15 16 17 1811 12

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

Periodic Table of the ElementsHydrogen Helium

NeonFluorineOxygenNitrogenCarbonBoronLithium Berylium

Sodium Magnesium Aluminum Silicon Phosphorus Sulfur Chlorine Argon

Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Galium Germanium Arsenic Selenium Bromine Krypton

Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon

Cesium Barium Lanthanum Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Astatine Radon

1

2

3 4 5 6 7 8 9 10 11 12

13 14 15 16 17

18

Group 8

It is important to remember that the bond between phosphorus and hydrogen is, in reality, pretty much neutral and covalent in character (although the hydrogen atoms actually do have a small amount of positive charge on them). Only when there is a considerable difference in the electronegativities of two atoms does one see formal charge separations and polar bonds occurring.

REDOX 6

Problem: What are the oxidation state assignments for the following compounds:

a) CH4 b) CO2

c) KMnO4 d) FeS

e) H2SO4 f) HCN

g) Na2CrO4 h) N2

i) MnO2 j) H2C2O4

REDOX 7

Balancing Redox Equations

The best way of balancing redox reactions is the half cell method. Break up the overall rxn into 2 half-cell rxns: one for the reduction and one for the oxidation. We then want to multiply each half cell reaction to make the overall # of electrons the same so they cancel out.

Note that this fairly simple rxn (and many redox rxns) could have been balanced the “normal” way. But if it is a redox rxn and looks “hard”, use the half cell method.

REDOX 8

Let's do another balancing act:

Note that in the last two examples, one of the products or reactants (O2, P2O5, and Al2O3) had two redox active atoms present (O, P, and Al) and that had to be accounted for when writing out the half cell reaction. Also note that when balancing half cell rxns, we mainly worry about the redox active atoms and not the H and O atoms (assuming they are not changing oxidation states). These will be balanced at a later point (see next example).

REDOX 9

Another EXAMPLE:

This is the “core” redox balanced reaction. Now, you have to check the oxygen, and then hydrogen atom balance:

REDOX 10

To balance the rxn in basic solution you need to balance the rxn first as if it was in acidic solution (as shown above). Then you “get rid” of the H+ by adding as many OH− to each side as there are protons and reacting the H+ and OH− together to make waters:

REDOX 11

Problem: Balance the following (in water):

a) Cr2O72- + Cl- Cr3+ + Cl2(g) (acidic)

b) MnO4- + CN- MnO2(s) + CNO- (basic)

REDOX 12

c) Na2S2O3(aq) + NaOCl(s) NaCl(aq) + Na2S4O6(aq)

d) Au(s) + CN−(aq) + O2(g) Au(CN)4−(aq) + OH−(aq)

(acidic)

(basic)

REDOX 13

Electrochemical/Voltaic/Galvanic Cell

Consider the reaction of Zn(s) and Cu2+(aq):

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

This reaction involves the transfer of 2 e- between the Zn(s) and Cu2+. The Cu2+ oxidizes the Zn, and is in turn reduced to form metallic Cu, while the Zn metal dissolves into solution as Zn2+.

This reaction is spontaneous with a ΔGº = − 212 kJ/mol. Mixing Zn(s) and Cu2+ together in a beaker causes a spontaneous reaction that gives off all the energy as heat, warming the solution.

The transferring of the electrons between the Zn and Cu2+, however, can accomplish much more useful work if we can figure out a way to get the e- to flow through a wire as electricity, where the work represented by −ΔGº (free energy) can now be used as electrical work.

REDOX 14

To harness the intrinsic chemical energy stored in these oxidizing and reducing agents, we must separate them to force the electrons to flow though an external circuit and do some work.

salt bridge orsemi-permeable membrane

flow of electrons

ANODE CATHODEoxidation reduction

Cu2+

Cu2+

Cu2+

Cu2+

Cu2+Zn

Zn2+

Zn2+

Zn2+

Zn2+Zn2+

Cu

Cu

Cu

Cu Cu

Cu

CuCu

Cl -Cl -

Cl -Cl -

Cl -

Cl -

Cl -

Cl -

Cl -

Cl -Cl -

Cl -

Cl -

Cl -

Anion flow Red Cat = Reduction occurs at Cathode

REDOX 15

There are different configurations of liquid galvanic cells that one can setup:

flow of electrons

Zn

ANODE CATHODEoxidation reductionAnion flow

Zn

Cu

flow of electrons

CATHODEreduction

ANODEoxidation

semipermeablemembrane

salt bridge Daniell Cell

salt bridge orsemi-permeable membrane

flow of electrons

ANODE CATHODEoxidation reduction

Cu2+

Cu2+

Cu2+

Cu2+

Cu2+Zn

Zn2+

Zn2+

Zn2+

Zn2+Zn2+

Cu

Cu

Cu

Cu Cu

Cu

CuCu

Cl -Cl -

Cl -Cl -

Cl -

Cl -

Cl -

Cl -

Cl -

Cl -Cl -

Cl -

Cl -

Cl -

Anion flow

REDOX 16

Measuring Cell Potentials (Voltages)

salt bridge orsemi-permeable membrane

H+

H+

H+

H+

H+

H+

H+

H+ X−

X− X−

X−

X− X−

X− X−

X−

X−

X−

X−

X−

X−

X−

X−

X−

X−

X−

H2

Pt M+M+M+

M+

M+M+M+

M+ M+

M

On a platinum electrode, H2(g) and H+(aq) are in redox equilibrium with one another (1 atm H2, 1M H+). This is called a hydrogen electrode.

H2 2H+ + 2e- By setting this electrochemical potential to 0.0 V, we have a reference electrode to which we can measure the innate ability of a material in the other electrode compartment of the electrochemical cell to either accept (be reduced/cathode) or give up (be oxidized/anode) electrons from/to H2/H+.

REDOX 17

Standard Reduction Potentials at 25°C Half Cell Rxn E°red (V) F2(g) + 2e- 2F-(aq) +2.87 O3(g) + 2H+(aq) + 2e- O2(g) + H2O +2.07 Co3+(aq) + e- Co2+(aq) +1.81 Cl2(g) + 2e- 2Cl-(aq) +1.36 O2(g) + 4H+(aq) + 4e- 2H2O +1.23 Ag+(aq) + e- Ag(s) +0.80 Cu+(aq) + e- Cu(s) +0.52 Cu2+(aq) + 2e- Cu(s) +0.34 AgCl(s) + e- Ag(s) + Cl-(aq) +0.22 Cu2+(aq) + e- Cu+(aq) +0.15 2H+(aq) + 2e- H2(g) 0.00 Pb2+(aq) + 2e- Pb(s) -0.13 V3+(aq) + e- V2+(aq) -0.24 Zn2+(aq) + 2e- Zn(s) -0.76 Al3+(aq) + 3e- Al(s) -1.66 H2(g) + 2e- 2H-(aq) -2.25 Mg2+(aq) + 2e- Mg(s) -2.36 Na+(aq) + e- Na(s) -2.71 Li+(aq) + e- Li(s) -3.05

It is critically important that you learn how to read, interpret, understand, and properly use this table.

REDOX 18

Voltage (or potential) represents the driving force for “pushing” the electrons from one location to another (for example, from a reducing agent to an oxidizing agent). The higher the voltage the more strongly the electrons will be pushed through the wire (or solution).

A positive cell potential (voltage) indicates a spontaneous electrochemical reaction. A negative cell potential (voltage) indicates a non-spontaneous reaction (the opposite reaction will, therefore, be spontaneous!). Note that the sign notation here is opposite that we learned for ΔG!

Problems: Which of the following substances is the best reducing agent?

a) Cu b) Zn c) Pb d) Ag e) H2

Which of the following substances is the best oxidizing agent?

a) Cu+ b) O3 c) O2 d) Na+ e) Li+

REDOX 19

To calculate the voltage of an electrochemical cell one can simply add together the two half-cell reactions that make up the overall galvanic (electrochemical) cell.

Let’s calculate the spontaneous cell potential (voltage) for our Cu/Zn cell. First look up the two half cell reactions from the table of Standard Reduction Potentials:

Cu2+(aq) + 2e- Cu(s) +0.34 Zn2+(aq) + 2e- Zn(s) -0.76

The more positive potential (relative to the hydrogen electrode) for the Cu2+ reduction compared to the Zn2+ reduction means that Cu2+ is a stronger oxidizing agent (wants to be reduced more) than Zn2+.

REDOX 20

To add these half-cell rxns together to give us our overall spontaneous net rxn and cell potential, we must switch one of the two half-cell rxns around so that it is written as an oxidation rxn (a compound giving up electrons on the product side of the rxn). The half-cell rxn with the smaller positive, or more negative potential gets flipped around (reversed). This would be the Zn rxn:

Cu2+(aq) + 2e- Cu(s) +0.34 Zn(s) Zn2+(aq) + 2e- +0.76

When one flips one of the half-cell rxns around, one also flips (reverses) the sign of the cell potential!!

The next step is to multiply each of the half cell rxns by the appropriate # to balance the # of electrons involved so that they cancel out. For the Cu/Zn cell, there are 2e- on each side of the two half-cell rxns, so everything is already balanced:

Cu2+(aq) + 2e- Cu(s) +0.34 V + Zn(s) Zn2+(aq) + 2e- +0.76 V

Cu2+(aq) + Zn(s) Zn2+(aq) + Cu(s) +1.1 V

REDOX 21

So the net reaction has a voltage of +1.1 V, the positive potential indicates that the rxn is spontaneous.

Note that the voltage calculated for the cell is for standard conditions (1 atm, 1 M). Changing the concentrations away from 1M will change the cell potential! Similarly, changing the temperature will also affect the voltage.

The Nernst Equation allows one to calculate the voltage (potential) of a electrochemical reaction when one is not working under standard conditions!

Also note that for galvanic cells (batteries) we always want to use a spontaneous reaction that gives off energy that we can use to perform work. But on homeworks or tests I may give you an overall balanced reaction that is non-spontaneous. In this case you need to write down the half-cell rxns in the right order to give the overall balanced rxn and calculate the cell potentials from the way that these half-cell reactions are written.

REDOX 22

Example: What is the potential for the following reaction?

2Na+(aq) + 2Cl−(aq) Cl2(g) + 2Na(s) Reduction half cell: Na+(aq) + 1e- Na(s) −2.71 V

Oxidation half cell: 2Cl−(aq) Cl2(g) + 2e- −1.36 V

sum = −4.07 V

Non-spontaneous

The reverse reaction is very spontaneous with a cell potential of +4.07 V.

This half-cell was the one that was flipped around (reversed)

REDOX 23

Example: What is the potential for the electrochemical cell composed of the following 2 half cell rxns?

Cr+3(aq) + 3e- Cr(s) -0.74 MnO2 + 4H+ + 2e- Mn2+(aq) + 2H2O +1.28

The Cr+3(aq) + 3e- → Cr(s) rxn has the more negative potential, so it gets flipped around:

Cr(s) Cr+3(aq) + 3e- +0.74 MnO2 + 4H+ + 2e- Mn2+(aq) + 2H2O +1.28

Now we have to balance the # of electrons in each rxn so we can add them together. 6 e- is the common factor, so we need to multiply the Cr rxn by 2 and the MnO2 rxn by 3:

2[Cr(s) Cr+3(aq) + 3e- ] +0.74 3[MnO2 + 4H+ + 2e- Mn2+(aq) + 2H2O] +1.28

2Cr(s) + 3MnO2 + 12H+

2Cr+3(aq) + 3Mn2+(aq) + 6H2O

+2.02V

Note that you do NOT multiply the cell potentials by the numerical factors (2 or 3) used to balance the # of electrons in each half-cell rxn!!

DANGER!! Common mistake!!

When phrased like this, we are looking for a spontaneous reaction that gives a positive cell potential!

REDOX 24

Example: Zn metal reacts with HCl, but Cu metal doesn’t. Calculate the cell potentials for these rxns to see if they fit the experimental data.

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

Splitting this rxn into the two half cell rxns, we have:

Zn(s) Zn2+(aq) + 2e- +0.76 2H+(aq) + 2e- H2(g) 0.00

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

+0.76V

Note that the Zn half-cell rxn is flipped around from how it is written in the Standard Reduction Table, since it is acting as a reducing agent here. The calculated positive cell potential indicates a spontaneous rxn.

REDOX 25

For the copper rxn, we have:

Cu(s) + 2H+(aq) Cu2+(aq) + H2(g)

Cu(s) Cu2+(aq) + 2e- −0.34 2H+(aq) + 2e- H2(g) 0.00

Cu(s) + 2H+(aq) Cu2+(aq) + H2(g)

−0.34V

Once again, note that the copper half-cell rxn is flipped around from how it is written in the Standard Reduction Table, since it is acting as a reducing agent here. So here we calculate that the cell potential is negative, indicating a non-spontaneous rxn – or one that should not occur normally, which fits the experimental data.

This is because copper metal is a more inert material relative to zinc metal. One could also note the position of copper above the hydrogen half cell rxn, which indicates that the reverse rxn of Cu+2 with H2 gas would be spontaneous:

Cu2+(aq) + H2(g) Cu(s) + 2H+(aq)

+0.34V

REDOX 26

Example: What about the reaction of Cu with nitric acid? Will this be a spontaneous rxn?

Nitric acid is not like the other strong acids (HCl, HBr, HI, H2SO4) in that it is a good oxidizing acid due to the presence of the NO3

− anion, which is not just a simple inert counter-anion for H+. The combination of NO3

− and H+ makes for a rather strong oxidizing mixture.

3[Cu(s) Cu2+(aq) + 2e-] −0.34

2[NO3−(aq) + 4H+(aq) + 3e- NO(g) + 2H2O]

+0.96

3Cu(s) + 8H+(aq) + 2NO3−(aq) 3Cu2+(aq) + 2NO(g) + 4H2O

+0.62V

If you get nitric acid on your skin, you will not only feel the burning of the acid (H+), but your skin will be oxidized to a yellow-brown color! So concentrated nitric acid is doubly dangerous!

REDOX 27

Although one can add half-cell rxns to yield overall redox equations, one can not simply add two half-cell rxns to yield another half-cell rxn. For example, consider the addition of the following two half cell rxns to generate a third half cell rxn:

Cu2+(aq) + 1e- Cu+(aq) +0.16 V Cu+(aq) + 1e- Cu(s) +0.52 V

Cu2+(aq) + 2e- Cu(s) +0.68 V

The correct half-cell potential for this rxn is 0.34 V, which is exactly half of what we incorrectly attempted to calculate above. This factor of ½ comes from the fact that all the half-cell potential values are normalized to a single e- value even if multiple e- are used in the half cell rxn.

As you might expect, the electrochemical potential for a rxn is directly related to the ΔG for a reaction (only with an opposite sign relationship!):

ΔGº = −nFEº n = # of electrons being transferred, F = Faraday’s constant (96.5 kJ/Vmol), Eº = standard potential.

REDOX 28

Problems: Calculate the cell potentials for the following reactions. Are they spontaneous or not?

a) F2(g) + 2Na(s) 2Na+(aq) + 2F−(aq)

b) Cu(s) + Zn2+(aq) Cu2+(aq) + Zn(s)

c) 2Al3+(aq) + 3Pb(s) 2Al(s) + 3Pb2+(aq)

d) 2Co3+(aq) + 2Cl-(aq) 2Co2+(aq) + Cl2(g)

e) 2Mg(s) + O2(aq) + 4H+(aq) 2Mg2+(aq) + 2H2O(aq)

f) 2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g)

g) 2Cu(s) + Cl2(g) 2Cu+(aq) + 2Cl−(aq)

REDOX 29

Non-Standard Conditions: Nernst Equation

Eº values are for standard conditions (1 M or 1 atm). If the reactant/product concentrations are different the cell potential, E, will differ from Eº.

One can calculate non-standard potentials using the Nernst equation:

2.303 logRTE QnF

E= −

R = gas constant, 8.314 J/mol●K

T = absolute temperature, K

n = # of moles of electrons transferred

F = faraday, 96,485 C/mol●e-

Q = reaction quotient

mol 2.303At room temp, 25 C , 0.0592

substituting this into the Nernst eq gives :0.0592 logE Q

nE

RTF

=

= −

iV

REDOX 30

Example: Calculate the potential for the Fe3+/Fe2+ electrode when the concentration of Fe2+ is five times greater than that of Fe3+.

Look up potential in a half cell table: Fe3+ + e- Fe2+ Eº = +0.771 V

The Fe2+ concentration is five times greater than that of Fe3+, so the Q expression is:

2

3[products] [Fe ][reactants] [Fe ]

5p

rQ = = =+

+

Assume room temp (25ºC = 298K) and solve using the Nernst equation:

0.0592 log

0.05920.771 log(5)1

0.771 (0.0592)(0.7) 0.730 V

Qn

EE

E

E

= −

= −

= − =

Problem: The potential for this is lower than Eº where both concentrations are 1 M. This should make sense based on Le Chatelier’s principle. Why?

REDOX 31

Problem: A cell is constructed at 25ºC as follows. One half-cell consists of Cu2+/Cu, [Cu2+] = 0.4 M. The other half-cell involves Zn2+/Zn with [Zn2+] = 0.4 M. Apply the Nernst equation to the overall cell reaction to determine the cell potential.

Problem: Consider the following half-cell rxn: Pb(s) + SO4

2−(aq) PbSO4(s) + 2e- Eº = +0.13 V

Can the Nernst equation help us figure out how to change concentrations to increase the cell potential?

Why?

REDOX 32

Lead-Acid Storage Battery

Anode Rxn: Pb(s) + SO4

2−(aq) PbSO4(s) + 2e- Eº = +0.13 V

Cathode Rxn: PbO2(s) + 4H+(aq) + SO4

2−(aq) + 2e- PbSO4(s) + 2H2O

Eº = +1.69 V

Overall Rxn: Pb(s) + PbO2(s) + 4H2SO4(aq)

2PbSO4(s) + 2H2O

Eº = +1.82 V

REDOX 33

Dry Cell (regular battery)

GraphiteCathodeMoist Pasteof NH Cl,MnO &carbon

42

Porousseparator

Zinc anode

Anode Rxn: Zn Zn2+ + 2e- Eº = 0.76 V

Cathode Rxn: 2MnO2 + 8NH4

+ 2Mn3+ + 4H2O + 8NH3

Eº = 1.21 V Overall Rxn: Zn + 2MnO2 + 8NH4

+ 2Mn3+ + Zn2+ + 4H2O + 8NH3 Eº = 1.97 V

Note that the production of NH3(g) in this “regular” cell produces an insulating layer between the cathode and the MnO2/NH4+. This causes these batteries to fade relatively quickly after steady use. Letting them sit for a while, allows the NH3(g) to diffuse away & allow fresh reagents to come in contact with the cathode, thus the battery becomes rejuvenated.

Real batteries only produce 1.5 V because the real chemistry is far more complicated and less efficient.

REDOX 34

Alkaline Cells

GraphiteCathodeMoist Pasteof KOH,MnO &graphite

2

Porousseparator

Zinc anode

Anode Rxn:

Zn + 2OH− Zn(OH)2 + 2e-

Cathode Rxn:

2MnO2 + 2H2O 2MnO(OH) + 2OH−

Overall Rxn: Zn + 2MnO2 + H2O 2MnO(OH) + Zn(OH)2

Eº = 1.5 V

The simple replacement of NH4Cl electrolyte with a KOH gel eliminates the production of the problematic NH3 gas that occurs in the regular dry cell. This battery has much improved stability for delivering electrical current until all the reagents are used up.

REDOX 35

Corrosion of Steel/Iron

Steel Anode

Fe2+ Fe2+

O2O2Cathode

Water

Cathode: ½O2 + H2O + 2e- 2OH−

Anode: Fe Fe2+ + 2e-

Sum: ½O2 + Fe + H2O Fe(OH)2

Subsequent Redox rxn:

4Fe(OH)2 + O2 + 4OH− 2Fe2O3 • 4H2O The presence of NaCl (salt) in the water makes it more conductive to e- flow, making it easier for the O2 to oxidize the iron without having to come directly in contact with it. Instead it can react anywhere at the surface of the water drop.

RUST

REDOX 36

Standard Reduction Potentials - 25ºC

REDOX 37

REDOX 38

Tables from Chemistry, 8th Ed, Whitten, Davis, Peck & Stanley (Thomson-Brooks/Cole)