Electrochemistry

Applications of Redox

Review

Oxidation reduction reactions involve a transfer of electrons.

OIL- RIG Oxidation Involves Loss Reduction Involves Gain LEO-GER Lose Electrons Oxidation Gain Electrons Reduction

Applications

Moving electrons is electric current. 8H++MnO4

-+ 5Fe+2 +5e- Mn+2 + 5Fe+3 +4H2O

Helps to break the reactions into half reactions. 8H++MnO4

-+5e- Mn+2 +4H2O

5(Fe+2 Fe+3 + e- ) In the same mixture it happens without doing

useful work, but if separate

H+

MnO4-

Fe+2

Connected this way the reaction starts Stops immediately because charge builds up.

H+

MnO4-

Fe+2

Galvanic Cell

Salt Bridge allows current to flow

H+

MnO4-

Fe+2

e-

Electricity travels in a complete circuit Instead of a salt bridge

H+

MnO4-

Fe+2

Porous Disk

Reducing Agent

Oxidizing Agent

e-

e-

e- e-

e-

e-

Anode Cathode

Cell Potential

Oxidizing agent pushes the electron. Reducing agent pulls 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

Zn+2 SO4-

2

1 M HCl

Anode

0.76

1 M ZnSO4

H+

Cl-

H2 in

Cathode

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.

Cell Potential

Zn(s) + Cu+2 (aq) Zn+2(aq) + Cu(s) The total cell potential is the sum of the

potential at each electrode. Eº cell = EºZn Zn+2 + Eº Cu+2 Cu

We can look up reduction potentials in a table.

One of the reactions must be reversed, so change it sign.

Cell Potential

Determine the cell potential for a galvanic cell based on the redox reaction.

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

Fe+3(aq) + e- Fe+2(aq) Eº = 0.77 V Cu+2(aq)+2e- Cu(s) Eº = 0.34 V Cu(s) Cu+2(aq)+2e- Eº = -0.34 V 2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V 0.77 V + - 0.34V = 0.43V

Homework

17.27 and 17.28

Line Notation

solidAqueousAqueoussolid Anode on the leftCathode on the right Single line different phases. Double line porous disk or salt bridge. If all the substances on one side are

aqueous, a platinum electrode is indicated.

For the last reaction Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)

Galvanic Cell

The reaction always runs spontaneously in the direction that produced a positive cell potential.

Four things for a complete description.1) Cell Potential2) Direction of flow3) Designation of anode and cathode4) Nature of all the components-

electrodes and ions

Practice

Completely describe the galvanic cell based on the following half-reactions under standard conditions.

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

Eº=1.51 Fe+2 +2e- Fe(s) Eº=0.44V Reverse the smaller valued reaction Fe(s) Fe+2 +2e- Eº= - 0.44V

Solution

Balance the cell reaction: Balance the electrons to balance the

equation: The First reaction requires 5 e-, the second

one supplies 2 e-, so 10 must be exchanged.

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

5(Fe(s) Fe+2 +2e-) Add the reactions together

Solution Continued

2MnO4-(aq) + 16 H+ (aq)+5Fe (s)

5 Fe +2 (aq) + 2 Mn+2 (aq) + 8H2O

(l) Determine the Cell Potential:

2(1.51) + 5(-0.44) = 3.02 – 2.2 = 0.8

Fe(s) l Fe+2(aq) ll MnO4-(aq), Mn+2(aq) l Pt(s)

More Homework

17.29 and 17.30

Potential, Work and G

emf = potential (V) = work (J) / Charge(C) E = work done by system / charge E = -w/q Charge is measured in coulombs. -w = qE

Faraday = 96,485 C/mol e-

q = nF = moles of e- x charge/mole e-

w = -qE = -nFE = G

Potential, Work and G

Gº = -nFE º if E º < 0, then Gº > 0 nonspontaneous if E º > 0, then Gº < 0 spontaneous In fact, reverse is spontaneous. 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

Practice

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

Fe(s) Fe+2(aq) + 2e- Eº = 0.44 V

Cu+2(aq)+2e- Cu(s) Eº = 0.34 V 0.44 V + 0.34 V = 0.78V Gº = -nFE º Gº = -(2 mol e-)96,485 C/mol e-(0.78J/C)

Gº = -1.5 x105 J

Practice Predicting Spontaneity

Will 1M HNO3 dissolve gold metal to form a 1M Au3+ solution?

HNO3 half reaction: NO3

- + 4H+ + 3e- NO + 2H2O E=0.96V Au3+ Half Reaction: Au Au3+ + 3 e- - E = -1.50V Au(s) + NO3

-(aq) + 4H+(aq) Au3+(s) + NO(g) + 2H2O(l)

Ecell = 0.96V – 1.50V = -0.54V Since Ecell is negative, it will NOT occur under

standard conditions.

Homework

17.37, 17.39, 17.49

Cell Potential and Concentration

Qualitatively - Can predict direction of change in

E from LeChâtelier.

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 LESS if [Al+3] = 1.0 M and [Mn+2] = 1.5M GREATER if [Al+3] = 1.5 M and [Mn+2] = 1.5 M SLIGHTLY

GREATER

Homework

17.55

The Nernst Equation

G = Gº +RTln(Q) -nFE = -nFEº + RTln(Q)

E = Eº - RTln(Q) nF

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

Eº = 0.48 V Always have to figure out n by balancing. If concentration can gives voltage, then from

voltage we can tell concentration.

The Nernst Equation

As reactions proceed concentrations of products increase and reactants decrease.

Reach equilibrium where Q = K and Ecell = 0

0 = Eº - RTln(K) nF

Eº = RTln(K) nF

nFEº = ln(K) RT

Example Using Nernst

Describe the cell based on the following: VO2

+ + 2H+ + e- VO22+ + H2O Eo = 1.00V

Zn2+ + 2e- Zn Eo = -0.76V

T = 25 C [VO2

+] = 2.0 M [H+] = 0.50 M

[VO22+] = 0.01 M [Zn2+] = 0.1 M

Starting Solution

In Order to balance the electrons we must double the first equation, and reverse the second equation, so we end up with:

2VO2+ + 4H+ + 2e- 2VO2

2+ + 2H2O Zn Zn2+ + 2e-

2VO2+ (aq) + 4H+ (aq) + Zn (s)

2VO22+ (aq) + 2H2O (l) + Zn2+ (aq)

Ecell = 1.76 V Note: Because we are using the Nernst

Equation we use base values for E

Final Solution

At 25 C R = 0.0521 E = Ecell –R/n log (Q)

E = 1.76 –(0.0521/2) log ([Zn2+][VO22+]2)

[VO2+]2[H+]4

E = 1.76 –(0.0521/2) log ([0.10][0.010]2)

[2.0]2[0.50]4

E = 1.76 –(0.0521/2) log 4.0 x 10-5

E = 1.76 + 0.13 = 1.89V

Homework

17.61 and 17.63

Batteries are Galvanic Cells Car batteries are lead storage batteries. Pb +PbO2 +H2SO4 PbSO4(s) +H2O

Dry Cell Zn + NH4

+ +MnO2 Zn+2 + NH3 + H2O

Alkaline Zn +MnO2 ZnO+ Mn2O3 (in base)

NiCad NiO2 + 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 Fe+2 +2e- 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- Fe Fe+2

e-

Salt speeds up process by increasing conductivity

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 galvanic 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

Calculating 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+

Other uses

Electroysis of water. Seperating mixtures of ions. More positive reduction potential means

the reaction proceeds forward. We want the reverse. Most negative reduction potential is

easiest to plate out of solution.