Electrochemistry

Chapter 20

• Electrochemistry = the study of relationships between electricity and chemical reactions

• Topics Include:– Batteries– Spontaneity of reacation– Corrosion of metals– Electroplating

Redox Reactions (Chapter 4.4)

• REDOX reactions always involve a transfer of electrons

• Oxidation = loss of electrons• Reduction = gain of electrons2 ways to remember:• LEO says GER• Loss Electrons Oxidation; Gain Electrons

Reduction

• OIL RIG• Oxidation Involves Loss; Reduction Involves

Gain

Redox Terminology• Oxidation Number a positive or

negative value assigned to each atom in a compound or ion for the purpose of keeping track of electrons

• Oxidizing Agent the substance that makes it possible for another substance to be oxidized

• Oxidizing agents remove electrons. They pick them up and become reduced (become more negative)

• Reducing Agent the substance that gives up electrons

• Reducing agents lose electrons, become oxidized, and become less negative/more positive

Rules for Assigning Ox. #’s Summary

1. Ox. # of the atom of a free element is Zero

Element = 0

2. Ox # of a monatomic ion equals its charge

Calcium = +2

Sulfide = -2

3. In compounds, Oxygen is always (-2), except in peroxides (-1)

H2O; O=-2

H2O2; O = -1

4. In compounds, Hydrogen is always +1

Hydrogen = +1

5. In compounds, Fluorine is ALWAYS -1

Fluorine = -1

6. Sum of the oxidation states for an electrically neutral compound must be zero

Helpful Number Line

+4+3+2+1

oxidation 0 reduction1234

Reducing Agent

OxidizingAgent

Example• The nickel-cadmium battery, a rechargeable

“dry cell” used in battery-operated devices, uses the following redox reaction to generate electricity:

Cd(s) + NiO2(s) + H2O(l) Cd(OH)2(s) + Ni(OH)2(s)

• Assign all the oxidation numbers• Identify the substances that are oxidized and

reduced.• Identify which are oxidizing agents and which

are reducing agents.

Redox• Every redox reaction has both oxidation

and reduction occurring.• What is being oxidized? reduced ?

Cd(s) + NiO2(s) + 2H2O(l) --> Cd(OH)2(s) + Ni(OH)2(s)

Identify the oxidation numbers

0 +4 2 +1 2 +2 2 +1 +22+1

Identify the substances which change in oxidation no.

0 +2 +4 +2Cd ---> Cd + 2 e Ni + 2 e --> Ni Cd is oxidized Nickel is reducedNi is the oxidizing agent Cd is the reducing agent

Practice

• Label the oxidation numbers for all elements on both sides of the reaction.

• Determine the Oxidizing Agent and the Reducing Agent in each reaction.

• Zn(s) + 2H+(aq) Zn+2(aq) + H2(g)

• 2H2 (g) + O2 (g) 2H2O(l)

Balancing Redox• Redox reactions are sometimes difficult

to balance.• Must balance both atoms of each

element AND charge.• We only show the substances involved

in the reaction, and add H, OH and/or H2O

• Requires following a series of steps and practice!

Half Reactions• Although REDOX takes place

simultaneously, it is easier to look at it as two separate processes.

For Example:Sn+2 + 2Fe+3 Sn+4 + 2Fe+2

• Oxidation: Sn+2 Sn+4 + 2 e-• Reduction: 2Fe+3 + 2 e- 2Fe+2

• Notice: In oxidation, electrons are products, while in reduction electrons are reactants

Balancing using Half Reactions

• MnO4- + C2O4

-2 Mn+2 + CO2

• Lets look at the half reactions:– MnO4

- Mn+2

– C2O4-2 CO2

• First add coefficients to balance out atoms (this may need to be done in an acidic or basic solution)

• Note: if the process is done in acidic solution, H+ and H2O can be added; in basic solution, OH - and H2O can be added.

MnO4- + C2O4

-2 Mn+2 + CO2

• MnO4- Mn+2 + H2O (added water)

• H+ + MnO4- Mn+2 + H2O (added H)

• Now we can balance:

• 8H+ + MnO4- Mn+2 + 4H2O

• Now equal types of atoms on both sides, but charges aren’t equal, so need to add electrons:

• 5e- + 8H+ + MnO4- Mn+2 + 4H2O

• Now to balance the oxalate ½ reaction…

• C2O4-2 CO2 (add coefficients to balance)

• C2O4-2 2CO2(add electrons to balance

charge)

• C2O4-2 2CO2 + 2e-

• Now we have to put both balanced half reactions together to get the overall balanced equation:

5e- + 8H+ + MnO4- Mn+2 + 4H2O

C2O4-2 2CO2 + 2e-

• We need the same number of electrons to appear on both sides of the reaction so they will cancel out.

10e- + 16H+ + 2MnO4- 2Mn+2 + 8H2O

5C2O4-2 10CO2 + 10e-

16H+ + 2MnO4- + 5C2O4

-2

2Mn+2 + 8H2O + 10CO2

• The best way to do this is through practice and we will do various examples in class and for HW

AP Type question• For the following: 1) write a balanced

equation and 2) answer the question about the reaction.

• 1)Calcium metal is heated strongly in nitrogen gas. 2) What is the change in oxidation number of the nitrogen in this RXN?

• 1)Acidified potassium dichromate solution is mixed with potassium iodide solution. 2) What is the reducing agent in the above reaction?

*Hint: Water is one of the products for the above reaction.

Redox Reactions - the reality: electricity

• Moving electrons is electric current.• A voltaic cell (or galvanic cell) is a

device which allows the transfer of e

to take place.• Voltaic cells produce electric current.• A battery is a voltaic cell. • Batteries contain chemicals which

undergo redox reactions.

Features of a Voltaic Cell• Two different substances

connected by a conductor in 2 separate compartments containing an electrolytic solution which is allowed to flow between the 2 compartments

• When assembled: a complete circuit forms which allows electrons to flow in a one-way direction. Direct Current (DC)

• The one-way direction is the direction of the spontaneous reaction.

• Each substance is called an electrode.• Anode - electrode where oxidation

occurs• Cathode - electrode where reduction

occurs• Electrons always flow from anode to

the cathode.

Both metals in same mixture?• Metal ions of one metal is deposited on another • Reaction happens without doing useful work

but if separate, the electricity is available to do work

Zn Cu

Cu+2 H+ Zn+2 SO4

-2

Zn+2

NO3-

Cu+2

NO3-

• Connected this way the reaction starts, but• Stops immediately because charge builds

up. For a cell to work, the solutions in each half cell must remain electrically neutral.

CuZn

A half cell

Zn (s) --> Zn+2 + 2e Cu+2 + 2e --> Cu(s)

2e

2e

2e

Zn+2

NO3-

Cu+2

NO3-

A bridge is needed

Salt Bridge allows current to flow

Zn

Cu

2e

2e 2e

2e

2e

salt bridgeSalt bridge

Zn+2

NO3-

Cu+2

NO3-

• Now, Electricity travels in a complete circuit

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

Zn Cu

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

2e 2e

2e

2e

2e

2e

e e

Zn+2

NO3-

Cu+2

NO3-

Porous Disk or Barrier works too

Instead of a salt bridge

Reducing Agent Zn

Oxidizing Agent Cu+2

e-

e-

e- e-

e-

e- +

e are available to do work

Cathode (Cu)Is reduced

Anode (Zn)Is oxidized

Zn (s) --> Zn+2 + 2e Cu+2 + 2e --> Cu(s)

Reduction occurs at the cathode, Oxidation at the anode

Porous barrier

The Complete Picture

Zn (s) + Cu+2 --> Cu(s) + Zn+2 • Red. at cathode: Cu+2 + 2e --> Cu(s) • Oxid. at anode: Zn (s) --> Zn+2 + 2e

CuZn

NO3 Cu+2

SO4-2 Na+

CuSO4

Solution(electrolyte)

ZnSO4

Solution

NO3 Zn+2

SO4-2 Na+

Cathode

Anode

salt bridgeNaNO3

All + ions flow this way All ions flow this way

2e

2e

2e

While Operating . . .• The Zn electrode is oxidized and loses

mass, while [Zn+2] increases in solution.• The Cu electrode is reduced, gains mass,

while the [Cu+2] decreases in solution.• Salt bridge allows ions to flow in both

directions. ions move toward the anode +ions move toward the

cathode• Salt bridge contains an electrolyte that

doesn’t react with anything in the cell.• Salt bridge connects the 2 cells keeping +

and ions neutral.• The actual charge on an electrode is zero.

Cr2O7-2 + 14H+ + 6I- ---> 2Cr+3 +3I2 +

7H2O• For the above reaction, a simple voltaic cell is assembled.

• Which is the anode? Which is the cathode? • I is the anode ; Cr2O7

-2 is the cathode• What reaction occurs at the anode?

6 I ---> 3I2 + 6e• What reaction occurs at the cathode?

Cr2O7-2 + 14 H+ + 6e ---> 2Cr+3 + 7 H2O

• What is the direction of electron flow?Electrons flow from the anode, I , to the

cathode, Cr2O7-2

• What is the sign at each electrode?I Anode is () and Cr2O7

-2, cathode is ()

Cell Potential Ecell • Electrons flow from anode to cathode

because there is a potential energy difference between them.

• Oxidizing agent pulls the electron.• Reducing agent pushes the electron. • The push or pull (“driving force”) is called

the cell potential, Ecell• Also called the electromotive force (EMF) • Unit is the volt(V) = 1 joule of work/coulomb of

charge• Measured with a voltmeter

The magnitude of Ecell depends upon:

• Specific reactions at each electrode• Concentrations of reactants / products• TemperatureThe value of Ecell . . .• Is what the voltmeter reads.• Is the difference between the 2 half cell

potentials (which we look up in tables)

• Eºcell = Eºcathode Eºanode

1 M HCl

H+

Cl-

H2 in

Standard Hydrogen Electrode

• This is the reference all other oxidations are compared to

• Eº = 0• º indicates standard

states of 25ºC, 1 atm, 1 M solutions.

Platinum wire

Platinum electrode

Zn+2 SO4-

2

1 M HCl

Anode

0.76

1 M ZnSO4

H+

Cl-

H2 in

Cathode

For exampleZn ---> Zn+2 + 2e-

Cell Potential•The total cell potential is the sum of the

potential at each electrode. Eºcell = Eºcathode + Eºanode

• We can look up reduction potentials in the table. Tables are for reduction potential. The oxidation reaction must be reversed, so we change its sign.

• Since we always change the sign of the oxidation reaction, the formula is expressed as:

Eºcell = Eºreduction Eºoxidation

OR

Eºcell = Eºcathode Eºanode

Calculating Ecell Co(s) --> Co+2 (aq) + 2e Ered = 0.277 V

AgCl(s) + e --> Ag(s) + Cl Ered = +0.222 V

What reaction occurs at the anode?

What reaction occurs at the cathode?

What is the standard cell potential, Ecell?

Oxidation Co(s) --> Co+2 (aq) + 2e

Reduction AgCl(s) + e --> Ag(s) + Cl

Eºcell = Eºcathode Eºanode

Eºcell = +.222 (0.277) = + 0.499 V

Note: multiplying a reaction by a coefficient doesn’t effect the Value of Eo.

Cell Potential• Determine the cell potential for a voltaic cell

based on the redox reaction:

Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)

• Fe+3(aq) + e- Fe+2(aq) Eºred = +0.77 V

• Cu+2(aq)+2e- Cu(s) Eºred = +0.34 V

Eºcell = Eºcathode Eºanode

Eºcell = 0.77 (+0.34) = +0.43 V

Remember, to operate a cell must have a (+) cell potential.

In Summary• A voltaic cell operates on a driving force

called the cell potential, Ecell or EMF.• An operating voltaic cell must have a +

Ecell value.• In any voltaic cell, the reaction at the

cathode (reduction occurs) has a more + E°red value than does the reaction at the anode (oxidation occurs).

• The strongest oxidizing agents (they gain electrons) have the most + E°red values.

• Standard Reduction Potential tables can be used to mix and match the strongest voltaic cells possible.

Spontaneity of Redox Reactions

• A complete circuit always flows in a one way direction.

• A Direct Current (DC) results.• The one-way direction is the direction of

the spontaneous reaction.• To be spontaneous, Ecell must be (+)• If Ecell is (-), it will not work!!• The stronger oxidizing agent has the

greater “driving force” and forces the reducing agent to run in reverse as an oxidation reaction

Activity Series and Eºred • “The ions of the metal below will oxidize the

solid metal above.” The activity series is experimentally determined with the strongest reducing agents listed at the top.

• The activity series is really a list of oxidation potentials - the flipped up-side-down list of our standard reduction potentials. The Ered table has the strongest oxidizing agents listed at the top.

• We now know why certain metal ions react with other metals: potential energy exists as a driving force, + Ecell

• And, why certain metals don’t react, Ecell

Working Voltaic Cells-Why ?

• Recall that ∆G, Gibbs Free Energy determines spontaneity.

∆G = spontaneous reaction + Ecell = spontaneous redox reaction

• Michael Faraday determined the relationship between Free Energy and cell potential:

∆G = nFE where n = no. of electrons transferred (always (+) )

F = Faraday’s constant: 96,500 J/V*mol e-

E = Ecell or emf

G makes the cell operational

•if E º = +, then Gº spontaneous• if E º = , then Gº nonspontaneous, the reverse is

spontaneous.

Gº = -nFEº

Practice Problem• Calculate Gº for the following reaction:• Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq)

• Fe+2(aq) + 2e-Fe(s) Eº = 0.44 V• Cu+2(aq)+2e- Cu(s) Eº = 0.34 V

Eºcell = Eºcathode Eºanode

Eºcell = +0.34 (0.44) Eºcell = +0.78 VGº = -nFEºGº = -(2)(96,500J/V•mol)(0.78V)

Gº = 150,540J/mol = 150.4 kJ/mol

Practice Problem3Ni+2 + 2Cr(OH)3 (s) + 10 OH ---->

3 Ni (s) + 2CrO42 + 8 H2O

Calculate ∆G for this reaction.3Ni+2 + 6e ----> 3 Ni (s) E = 0.25V

CrO42 + 4H2O + 3e ---> Cr(OH)3 + 5 OH

E = 0.13V

Gº = -nFEº

Eºcell = 0.25 (0.13) = -0.12V)

Gº = -(6)(96,500J/V•mol)(-0.12V)Gº = 69,480 J/mol 69 kJ/mol

Why don’t batteries last forever?

• As a voltaic operates, reactants are consumed and products are generated.

• As the conc. of reactants decreases, so does the Ecell.

• At some point the battery goes “dead” and the Ecell = 0

• Walther Nernst determined the relationship between the conc. and EMF

The Nernst Equation• G = Gº + R T lnQ (standard free energy change)

• -nFE = -nFEº + RTlnQ (substituting Gº = -nFEº)

E = Eº RT lnQ nF

• At standard conditions, RT/F = 0.0592 V• So, E = Eº 0.0592 V log Q n• Q is the reaction quotient, [prod]/[react] Remember, we leave solids out of the equilibrium

expression

The Nernst Equation

• Remember, always determine n by balancing the redox reaction.

• What it means:

Equation relates concentrations to the emf (voltage).

If concentration can give us voltage,

then from voltage we can tell

concentration.

Equilibrium and Cell emf• When E =0, the cell reaction has reached

equilibrium and no net reaction is occurring in the voltaic cell.

• Recall that at equilibrium, ∆G = 0 and Q = KeqSubstituting, simplifying and rearranging the

Nernst Equation gives us: nE

log Keq = (at 25C) 0.0592• Thus, the equilibrium constant for a redox

reaction can be determined from the E value.

Equilibrium and Cell emf• As a voltaic operates, reactants are

consumed and products are generated. When “dead”, the cell at E =0 has reached equilibrium, ∆G =0.

• Where’s the equilibrium?• Qualitatively - we can predict direction

of change in E from LeChâtelier.

E = Eº 0.0592 V log Q

n

Practice Problem

E = Eº 0.0592 V log Q n

2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s)

Predict if Ecell will be greater or less than Eºcell

• if [Al+3] = 1.5 M and [Mn+2] = 1.0 M

E = Eº 0.0592 V log [Al+3]2 n [Mn+2]3

• if [Al+3] = 1.0 M and [Mn+2] = 1.5M• if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

Greater

Less than Ecell

greater

Practice Problem• Calculate the emf at 298K generated by a

cell for the equation below when [Cr2O7-

2]=2.OM, [H+]=1.0M, and [Cr+3]=1.0x10-5M.Cr2O7

-2(aq) + 14H+(aq)+ 6I-(aq) 2Cr+3 (aq) + 3I2(s) + 7H2O(l)

• Solve for Q• Q = 5.0 x 10-11

• Figure out n• n = 6• Plug into Nernst Equation and get 0.89 V

Batteries are Voltaic Cells• Car batteries are lead storage

batteries.Pb +PbO2 +H2SO4 PbSO4(s) +H2O

• Dry CellZn + NH4

+ +MnO2 Zn+2 + NH3 + H2O+ MnO(OH) (s)

• AlkalineZn +MnO2 ZnO+ Mn2O3 (in base)

• NiCadNiO(OH) s + Cd + 2H2O Cd(OH)2 +Ni(OH)2

Corrosion• Rusting - spontaneous oxidation.• Most structural metals have reduction

potentials that are less positive than O2 .

• Fe+2 +2e- Fe Eº= 0.44 V

• O2 + 2H2O + 4e- 4OH- Eº= 0.40 V

• Fe+2 + O2 + H2O Fe2 O3 + H+

• Reaction happens in two places.

Water

Rust

Iron Dissolves in Water: Fe Fe+2

e-

Salt speeds up process by increasing conductivity

Fe+2

Fe+2Fe+2

O2

O2Fe2O3

Preventing Corrosion

• Coating to keep out air and water.• Galvanizing - Putting on a zinc coat• Has a lower reduction potential, so it is

more easily oxidized.• Alloying with metals that form oxide

coats.• Cathodic Protection - Attaching large

pieces of an active metal like magnesium that get oxidized instead.

• Running a voltaic cell backwards.• Put a voltage bigger than the potential

and reverse the direction of the redox reaction.

• Used for electroplating.

Electrolysis

1.0 M

Zn+2

e- e-

Anode Cathode

1.10

Zn Cu1.0 M

Cu+2

1.0 M

Zn+2

e- e-

AnodeCathode

A battery >1.10V

Zn Cu1.0 M

Cu+2

Calculations for plating• Have to count charge.• Measure current I (in amperes)• 1 amp = 1 coulomb of charge per second• q = I x t• q/nF = moles of metal• Mass of plated metal

• How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+?

How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+?

q = I x t and qt = I = 5.00 amp

I

13,862 Ct = = 2772 sec = 46.2

min. 5.00 amp

q = moles x Fn

q = 0.144 mol(96,500) (1)q = 13,862 C