Chapter 17 Electrochemistry

Study Of Interchange Of Chemical And Electrical EnergyUsing RedOx chemistry to generate an electrical

current – moving electrons

ReviewReDox reactions = transfer electrons

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ReviewRules for Assigning Oxidation States p. 156• Atom in an element = 0• Monoatomic ion = same as charge• Fluorine = -1 in compound• Oxygen = -2 in compound• Hydrogen = +1 in covalent compounds• of oxidation numbers = overall charge

Apply what you just relearned:What are the oxidation states for each element? O2, Cl-1, Mg+2, NH3, NO2, NO3

-, CrO4-2, MnO4

-1

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Oxidation – Reduction ReactionsIdentify the atoms that are oxidized and reduced,

and specify the oxidizing and reducing agents in the following reactions.

2Al(s) + 3I2(s) → 2AlI3(s)

2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g)

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Method of Balancing Redox Reactions in Acidic Solutions

1. Write separate equations for the oxidation and the reduction half-reactions

2. For each half-reactiona. Balance the elements except hydrogen and oxygenb. Balance the oxygen using H2Oc. Balance the hydrogen using H+

d. Balance the charge using electrons

3. If necessary, multiply one of both balanced half-reaction by an integer to equalize the number of electrons transferred in the half-reactions

4. Add the half-reactions and cancel identical species

5. Check that the elements and charges are balanced

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Method of Balancing Redox Reactions in Basic Solutions

1. Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ were present.

2. To both sides of the equation, add a number of (OH)- that is equal to the number of H+.

3. Form H2O on the side containing both H+ and (OH)-, eliminate the number of H2O that appear on both sides of the equation.

4. Balance this equation:

Al(s) + MnO4-(aq) → MnO2(s) +Al(OH)4

-(aq)

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POP Quiz

• Complete and balance this equation for a redox reaction that takes place in basic solution.

7

( ) 4( ) ( ) 2( )aq aq aq sCN MnO CNO MnO

ApplicationsIs the following a redox reaction?

(i.e. do oxidation states change from reactant to product)• Moving electrons is electric current.

• 8H++MnO4-+ 5Fe+2 +5e- Mn+2 + 5Fe+3 +4H2O

• Break the reactions into half reactions.

• Reduction: MnO4- Mn+2

• Oxidation: Fe+2 Fe+3

• In the same mixture electrons transfer without doing useful work, but if you separate oxidation rxn. from reduction rxn. we can force the electrons to flow through a wire.

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

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Galvanic Cell

Galvanic Cell (Voltaic Cell) can contain a salt bridge or a porous disk which allows the electrons to flow in a complete circuit.

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Galvanic Cell

An electrochemical process involves electron transfer. The species that acting as a reducing agent supplies electrons to the anode. The species acting as the oxidizing agent receives electrons from the cathode. Galvanic Cell Animation 11

Constructing a Galvanic Cell

H2O + PO3-3 + MnO4

- → 3PO4-3 + MnO2(s) + 2OH-

Draw a Galvanic Cell and Identify the components• What is being reduced?• What is being oxidized? • Which direction does the electrons flow?• Which electrode is the cathode?• Which electrode is the anode?

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Anode -

e-

e- e- e-e-

e-

Cathode+

Oxidation Occurs at Anode (Anode Oxidation is neg.)

Reduction Occurs at Cathode (Red Cat is pos.)Electrons flow An. Ox. to Red. Cat.

PO3-3 MnO4

-

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POP Quiz

3Zn2+ + 2Al → 3Zn + 2Al3+

Draw a Galvanic Cell and Identify the components• What is being reduced?• What is being oxidized? • Which direction does the electrons flow?• Which electrode is the cathode?• Which electrode is the anode?

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Cell Potential

Def. Pull (“driving force”) causes electron flow = Ecell

Called electromotive force (emf) or cell potential• Oxidizing agent pulls the electron.• Reducing agent pushes the electron. • Unit is the volt(V) • V = 1 joule of work/coulomb of charge• 1V = 1J/C

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Digital Voltmeters Draw only a Negligible Current and are Convenient to Use

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Reaction in a Galvanic Cellwith Standard Hydrogen Electrode

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Standard Hydrogen Electrode

• This is the reference all other oxidations are compared to

E o = 0• This is the set up that is

used to measure Reduction Cell Potential and calculate the cell potential for all other metals.

e-

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A Galvanic Cell involving the Half-Reactions

Mnemonic: e- flow

An. Ox. to

Red. Cat.

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Standard Reduction Potential

Potential of a given species to become reduced when all species are in their standard state.

The Standard Hydrogen Potential is the reference potential against which all half-reactions potentials are assigned.

Standard States: [H+] = 1.0 MPH2 = 1 atmT = 298K

IUPAC has universally accepted the half-reaction potentials based on the assignment of zero volts to this rxn. 2H+ + 2e- → H2

Hydrogen Reduction Cell Potential is E o = 020

Cell Potential = E ocell

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

each electrode.• • We can look up reduction potentials in a table

(A25) & combine using this formula:1. One of the reactions must be reversed, so some signs

must be reversed.2. Electrons lost = Electrons gain, so don’t need to

multiply value of Eo by integers when balancing equations

2 2

o ocell Zn Zn Cu Cu

E E E

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Cell Potential Eo

Determine the cell potential for a galvanic cell

Cu(s) + Fe+3(aq)

Cu+2(aq)

+ Fe+2(aq)

Step 1: Write the Half Reactions1. Fe+3(aq) + e- Fe+2(aq) E o = 0.77 V

2. Cu+2(aq)+2e- Cu(s) E o = 0.34 V

Step 2: Reverse equation #2 & change the signCu(s) Cu+2(aq)+2e- E o = -0.34 V

Step 3: Multiply equation #1 by integer 22Fe+3(aq) + 2e- 2Fe+2(aq) E o = 0.77 V

Step 4: Eocell

= Eoreduction + Eo

oxidation 22

Important Facts1. The higher the reduction potential the greater the

tendency to be reduced.2. Between 2 species, the species with the largest Eo

will be written as reduction and the other will be oxidization reaction, therefore it must be flipped.

3. The standard potential for an Galvanic Cell (Eocell)

comes from adding Eo for each half-reaction.4. When multiplying a half-reaction by an interger to

equalize e- number, DO NOT MULTIPLY the Eo by that number. Eo is an intensive property which does not depend on quantity. 23

Time To Get Ready For Carnegie Hall

• For the following cell & data, identify the cathode & anode, write the balanced reaction and calculate Eo

cell

Al3+ + 3e- → Al Eo = -1.66 VMg2+ + 2e- → Mg Eo = -2.37 V

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Line Notation

• solid (anode)AqueousAqueoussolid (cathode)

• Anode on the leftCathode on the right• Single line = phases difference (solid | liquid).• Double line || porous disk or salt bridge.• If all the substances on one side are aqueous, a

platinum electrode is indicated.• For the reaction on slide 20:

Cu(s)Cu+2(aq)

aqFe+2(aq),Fe+3

(aq)Pt(s)

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Complete Galvanic Cell Description Given Just Half Reactions

• The cell will always runs spontaneously in the direction that produces + Eo

• Four things for a complete description.1) Cell Potential Eo & Balanced Equation2) Direction of flow gives positive Eo

3) Designation of anode and cathode e- flow An. Ox. to Red. Cat.

4) Nature of all the components: electrodes and ions

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Schematic of Galvanic Cell Involving Half-Reactions

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Practice• Completely describe the galvanic cell based on the

following half-reactions under standard conditions. Include line notation, Eo

cell and a labeled sketch of the cell.

Cu+2 + 2e- Cu Eo = 0.34VFe+2 + 2e- Fe Eo = -0.44V

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Speaking of Electricity

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Potential, Work and G

• q is charge • Work is viewed as being done by system• Work flow out is minus sign• When cell produces a current, Eo

cell is Positive therefore Eo & w have opposite signs

• Charge is measured in Coulombs(C)

• Solve the equation for w

( force) difference (V)=charge

work Jemf driving potential

C

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o wE

q

ow qE

• Maximum work in a cell is equal to the maximum cell potential, theoretically

• The charge on 1 mole of electrons is a constant called a Faraday (F)

1F = 96,500 C/mol e-

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max maxow qE

96,500(# e )( )

e

Cq nF moles

moles

• if Eo < 0, then Go > 0 not spontaneous, no current• if Eo > 0, then Go < 0 spontaneousUnder standard conditions:

• 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 VCan also use the equation to predict spontaneity, if Go > 0

the process will not occur, Study Ex. 17.4 p. 80332

max max maxw qE nFE G

o oG nFE

Cell Potential and Concentration

• Qualitatively - Can predict direction of change in Eo from LeChâtelier.

2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s) Eocell= 0.48V

Predict if Eooriginal cell will be > or < Eo

cell for the following conditions:

1. if [Al+3] = 2.0 M and [Mn+2] = 1.0 M

2. if [Al+3] = 1.0 M and [Mn+2] = 3.0M

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The Nernst EquationG = Go +RTln(Q)

-nFE = -nFEo + RTln(Q) (RT/F = 0.02567 @ 25oC)

• Calculate E for the following reaction:2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s) Eo = 0.48 V

[Mn2+] = 0.50M & [Al3+] = 1.50M• Always have to figure out n by balancing.• If concentration can gives voltage, then from voltage

we can tell concentration.

0.0591logo Q

nE=E

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The Nernst Equation

• As reactions proceed concentrations of products increase and reactants decrease.

• Reach equilibrium where Q = K and Eocell = 0

which is a “dead battery”

• Therefore lno RTK

nFE

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Batteries are Galvanic Cells

• Car batteries are lead storage batteries.– Pb + PbO2 + H2SO4 PbSO4(s) + H2O

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Common Dry Cell Zn + NH4

+ + MnO2 Zn+2 + Mn2O3 + NH3 + H2O

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Alkaline version uses KOH and NaOH in paste instead of NH4Cl

NiCad NiO2 + Cd + 2H2O Cd(OH)2 +Ni(OH)2

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NiCad batteries can be recharged multiple times before the battery is no good.

Corrosion

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

that are less positive than O2

• Reduction Potentials:– Fe+2 +2e- Fe Eo= -0.44 V

– O2 + 2H2O + 4e- 4OH- Eo= 0.40 V

– Al+3 + 3e- Al Eo= 0-1.66 V

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Corrosion of Iron• Oxid. Fe Fe+2 +2e- Eo= 0.44 V

• Red. O2 + 2H2O + 4e- 4OH- Eo= 0.40 V

Fe+2 + O2 + H2O Fe2 O3 + H+

•Reaction happens in two places.

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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.– Stainless Steel – Cr + Ni

• Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

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Electrolysis• Running a galvanic cell backwards.• Put a voltage bigger than the potential and reverse the

direction of the redox reaction.• Used for electroplating, charging a battery, producing

aluminum, electrolysis of water

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Electrolysis Calculating Plating

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Calculate mass of Cu plated out when apply 10.0 amps for 30.0 minutes. 1.Since 1 amp = 1 C / s Solve for C

• C = 10.0 amps x 1800 s = 18,000 C2.1F = 96,500C/mole of e- Solve for # of mole e-

• 18,000 C/96,500 = 0.187 moles e-

3.Each Cu+2 needs 2 e- to become an atom• 0.187 moles e- x 1mol Cu/2mol e- =

0.0935 mol Cu4.Multiple mol of Cu by Cu molar mass

• 0.0935 mol Cu x 63.55gCu/1 mol Cu = 5.94g Cu

It is all StoichiometryLast Carnegie Time p.834 #77