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Electrochemistry Electrochemistry and Its Applicationsand Its Applications

Chapter 19Chapter 19

Humphry Davy1778-1829.Prepared metallic K, Na, Sr, Ca, B, Ba, Mg, Li by electrolysis.

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Oxidation-Reduction ReactionsOxidation-Reduction ReactionsOXIDATION = loss of electrons

Examples:

Na Na+ + e-

Al Al3+ + 3e-

S2- S + 2e-

OXIDATION = increasing the oxidation number (more positive)

Example:

NO NO2 -2+2 -2+4

change = +2

+2 to +4 : N is oxidized

LEO the lion goes GER

Oxid. Nos:

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Oxidation-Reduction ReactionsOxidation-Reduction Reactions

REDUCTION = gain of electrons

Examples: N + 3e- N3-

Fe3+ +e- Fe2+

REDUCTION = decreasing the oxidation number (more negative)

Example:

MnO4- Mn2+

Oxid. Nos: +7 +2

change = -5

+7 to +2 : Mn is reduced

LEO the lion goes GER

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Review of Oxidation NumbersReview of Oxidation NumbersO.N. = 0 for atom in element form (C, Ag, O2, H2, P4, etc.) = charge for any monoatomic ion (e.g., +1 for Na+, -2 for S2- , +3 for Al3+) = -2 for oxygen in compound or ion (except peroxides and when bonded to fluorine) = +1 for hydrogen when bonded to non-metals (e.g., H2O, CH4, H3N, HCl) = -1 for hydrogen when bonded to metals (e.g., NaH) = -1 for oxygen in peroxides (e.g., HOOH)

Sum of O.N. = 0 for neutral compound (e.g., for Na2CO3, Na = +1, O = -2, C = +4)Sum of O.N. = ion charge for polyatomic ion (e.g., for CO3

-2, O = -2, C = +4)

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Oxidation-Reduction Reactions = “REDOX”Oxidation-Reduction Reactions = “REDOX”

What we have looked at are called half-reactions.

Half-reactions are fully balanced with respect to both mass (atoms) and electrons (net charge) and are either reduction or oxidation (but not both), e.g.,Sn2+ + 2e- Sn (Sn is reduced)FeCl3 + e- FeCl2 + Cl- (Fe is reduced)Mn2+(aq) + 4 H2O (l) 8H+(aq) + MnO4

-(aq) + 5e-

(Mn is oxidized) But half-reactions do not occur by themselves in thereal world; reduction cannot occur without oxidation (and vice versa).

Hence, we have REDOX occurring (both oxidation and reduction) in a chemical reaction.

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Oxidation-Reduction ReactionsOxidation-Reduction ReactionsHalf reactions are combined to get REDOX reactions.

When combining, • charges must balance (this is conservation of electrons) and• atoms must balance (this is conservation of mass).

Half-reaction method of balancing REDOX equations Consider the (unbalanced) reaction:

MnO4-(aq) + C2O4

2-(aq) Mn2+(aq) + CO2(g)(1) Write out separate reduction and oxidation half-reactions;(2) Add H2O and/or H+ as needed for mass balance, and add electrons (e-) for charge balance;(3) Multiply each half reaction by a common denominator so that electrons (e-) will cancel;(4) Add the half reactions and (mathematically) simplify.

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1. The two incomplete half reactions are:

MnO4-(aq) Mn2+(aq)

C2O42-(aq) 2CO2(g)

2a. Balance 1st half-reaction:

MnO4-(aq) Mn2+(aq)

5e- + 8H+ + MnO4- Mn2+ + 4H2O

2b. Balance 2nd half-reaction:

C2O42-(aq) CO2(g)

C2O42- 2CO2 + 2e-

MnO4-(aq) + C2O4

2-(aq) Mn2+(aq) + CO2(g)

Balancing Equations by the Method of Half-ReactionsBalancing Equations by the Method of Half-Reactions

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Balancing Equations by the Method of Half-ReactionsBalancing Equations by the Method of Half-Reactions

5e- + 8H+ + MnO4- Mn2+ + 4H2O

C2O42- 2CO2+ 2e-

The common denominator is 10.

x 2

x 5

3a. Find common denominator so that electrons willcancel.

3b. Multiply equations so that electrons will cancel.

10e- +16H+ + 2MnO4- 2Mn2+ + 8H2O

5C2O42- 10CO2 + 10e-

The 10 e- on each side cancel.

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Balancing Equations by the Method of Half-ReactionsBalancing Equations by the Method of Half-Reactions

4. Add equations and simplify*:

16H+ +2MnO4- +5C2O4

2- 2Mn2++8H2O + 10CO2

5. Check to make sure both mass and charges balance.

Left side: 16 H, 2 Mn, 28 O, 10 C, (+16-2-10) = +4

Right side: 16 H, 2 Mn, 28 H, 10 C, +4

*In some instances H+ and/or H2O appear on both sides of theEquation, so that simplification can be performed by subtracting H+ and/or H2O from the equation.

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Balancing Equations by the Method of Half-ReactionsBalancing Equations by the Method of Half-Reactions

Reactions Occurring in Basic SolutionReactions Occurring in Basic Solution

•We use OH- and H2O rather than H+ and H2O.

•First, balance as usual, then add OH- to both sides so that all H+ is consumed.

MnO4- (aq) + Br-(aq) MnO2 (s) + BrO3

-(aq) (basic soln)

2x [ 3e- + 4H+ + MnO4- MnO2 + 2H2O]

1x [ Br- + 3H2O BrO3- + 6H+ + 6e- ]

Add half reactions: 2H+ + 2MnO4

- + Br- 2MnO2 + BrO3- + H2O

Add 2OH- to both sides to remove 2H+ on left 2H+ + 2OH- + 2MnO4

- + Br- 2MnO2 + BrO3- + H2O + 2OH-

2H2O + 2MnO4- + Br- 2MnO2 + BrO3

- + H2O + 2OH-

H2O + 2MnO4- + Br- 2MnO2 + BrO3

- + 2OH-

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Consider, for example, the following REDOX reaction:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

undergoing oxidation

undergoing reduction

Reducing agent is oxidized. Oxidizing agent is reduced.

Oxidizing and Reducing AgentsOxidizing and Reducing Agents

Note: Reactions that are not redox include acid-base rxs. (e.g., HCl + NaOH H2O + NaCl); and metathesis (replacement, or ppt.) rxs. (e.g., NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl (s)).

That which is oxidized is the reducing agent;That which is reduced is the oxidizing agent.

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Electrolytic CellsElectrolytic CellsThere are essentially two types of cells:

(a)Voltaic (or galvanic): spontaneous REDOX rxn → electricity(b) Electrolytic electricity → nonspontaneous REDOX rxn

We’ll start with spontaneous processes.Consider copper (Cu) metal in silver ion (Ag+) soln

Cu

What happens?

2. White whiskers grow on Cu surface (production of metallic silver, Ag)

1. soln becomes blue (production of Cu2+)

Cu

Ag+

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Cu

Ag+

Blue solution due toproduction of Cu2+ ion

Silver whiskers (Ag) growing on Cu metal

What are the half reactions?

What is the overall REDOX reaction?

Ag+(aq) + e- Ag(s) Cu(s) Cu+2(aq) + 2e-

2Ag+(aq) + Cu(s) 2Ag(s) + Cu+2(aq)

Silver ions are reduced to metal silver.Copper metal is oxidized to cupric ions.Obviously, the process is spontaneous(because we observe it to happen).

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Another setup: Voltaic Cells Another setup: Voltaic Cells (external flow of electrons)(external flow of electrons)

CuAg

Cu2+ Ag+Electrons flow fromCu on left to Ag on right

Cu will be oxidized to produce Cu2+ {Cu(s) Cu+2(aq) + 2e-}.Ag+ will be reduced to produce Ag {Ag+(aq) + e- Ag(s)}. This needs to be balanced by negative (-) charges. But how?Add a salt bridge (inverted U tube with solution of ions, e.g. Na+ NO3

-)

Saltbridge

e-

Through salt bridge, anions (NO3-) can now move into left

chamber to balance Cu2+ being produced.Similarly, Na+ will move into right compartment to replace the Ag+ being reduced.

Ag

Same half reactions

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(Same) Voltaic Cell(Same) Voltaic Cell

Cu Ag

Cu2+ Ag+

Saltbridge

e-

Cu → Cu2+(aq) + 2e- Ag+ +(aq) + e- → Ag

Anode Cathode

(oxidation) (reduction)

The individual metals are the anode and cathode.Here, Cu is the anode.

anode

Ag is the cathode

cathode

Overall redox reaction:

Cu (s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag (s)

Oxidation always takes place at the anode.Reduction always takes place at the cathode.Electrons always flow in the external circuit from the anode (marked “-” to the cathode (marked “+”).

Ag

- +This is a completecell (made up oftwo half cells)

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Voltaic CellsVoltaic Cells(a porous barrier can be used instead of salt bridge)

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Voltage of CellsVoltage of Cells(Same Cu-Ag Cell)(Same Cu-Ag Cell)

Cu Ag

Cu2+ Ag+

Saltbridge

e-

Cu(s) → Cu2+(aq) + 2e- Ag+ +(aq) + e- → Ag (s)

Anode Cathode

(oxidation) (reduction)

anode

cathode

Cu (s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag (s) If the solutions of Cu2+ and Ag+ were “standard” (i.e., 1.0 M), and we placed a voltmeter in the electron flow line, we would get a reading of +0.46 volt (at 25oC).This is the standard cell potential or Eo

cell

(This is also called electromotive force (emf)

[Cu2+]=1.0 M [Ag+]=1.0 M

V

Why is this reaction spontaneous?Why is the voltage 0.46 volt?

Ag- +

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Choose any two entries; the top one will act as the

cathode and will be reduced; the bottom one will act as the anode and

will be oxidized.

This electromotive series presents data showing thetendency of substances to

gain or lose electrons.

The standard voltage will be the algebraic difference between the two respective

potentials.

0.46 V

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The Ag|Ag+ half-cell has a higher reduction potential than the Cu|Cu2+ half-cell. This means that the Ag/Ag+ half-cell will more readily undergo reduction when compared to the Cu/Cu2+ half-cell, and the Cu|Cu2+ half-cell will undergo oxidation.

The Ag|Ag+ standard reduction potential is Eored = +0.80 v.

The Cu|Cu2+ standard reduction potential is Eored =+0.34 v.

Standard Reduction Potentials (SRPs) are all based on and compared to the hydrogen (H2|H+) half-cell which is assigned Eo

red =0 volts by convention.

The overall cell potential is given by: Eo

cell= Eored (higher) – Eo

red (lower)

The overall cell potential is always positive if it is spontaneous (which it will be if the higher reaction in the table is the cathode and the lower reaction in the table is the anode).

Caution: spontaneous processes have positive voltages (E°, or V°), but negative free energies (ΔG).

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Cell EMF – different electrodesCell EMF – different electrodesStandard Reduction Potentials (SRPs)Standard Reduction Potentials (SRPs)

In this voltaic cell, hydrogen is the cathode and zinc is the anode.The measured voltage is 0.76 V.

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Hydrogen is the cathode and the zinc is the anode. The algebraic difference

between the standard potentials is 0.76 V.

Note: EMF (electromotive force), voltage, and cell

potential are all synonyms0.76 V

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F2 is the strongest oxidizing agent; F- is the weakest reducing agent.

Li is the strongest reducing agent; Li+ is the weakest oxidizing agent.

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Using the Standard Reduction Potentials Using the Standard Reduction Potentials (RSP) to Predict Spontaneity(RSP) to Predict Spontaneity

Some metals are easily oxidized whereas others are not.e.g., Fe is oxidized by Ni2+ but Ni is not oxidized by Fe2+.

Remember, the SRP table is an Activity series (see Ch 5), a list of metals arranged in order of ease of oxidation.* The lower a metal is on the SRP table, the more activethat metal is, i.e., the more easily it is oxidized. Any metalcan be oxidized by the ions of elements above it.

*Caution! The Activity series of Ch 5 is in reverse orderof the SRP table of this chapter.

The reaction: Fe + Ni2+ Fe2+ + Ni occurs, but Ni + Fe2+ Ni2+ + Fe does not.

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The voltage of the reaction Ni + Fe2+ Ni2+ + Feis calculated by adding the reactions:

Let’s repeat the exercise using specific values of V:Let’s repeat the exercise using specific values of V:

The voltage of the reaction Fe + Ni2+ Fe2+ + Ni is calculated by adding the reactions:

Ni Ni2+ (V = +0.28) Fe2+ Fe (V = -0.44)

The summed voltage is V = –0.16 and the reaction is nonspontaneous.

Ni2+ Ni (V = -0.28) Fe Fe2+ (V = +0.44)

The summed voltage is V = +0.16 and the reaction is spontaneous.

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Nernst Equation – for Nernst Equation – for nonnonstandard solutions –standard solutions –explains why batteries “run down”explains why batteries “run down”

0.0592 n

Cu Ag

Cu2+ Ag+

Saltbridge

e-

Cu → Cu2+(aq) + 2e- Ag+ +(aq) + e- → Ag

Anode Cathode

(oxidation) (reduction)

Ag

- +

As Cu2+ builds up,the Cu(s) Cu2+(aq)equilibrium shifts to the left (le Châtlier’s principle)

As Ag+ is depleted,the Ag+(aq) Ag(s)equilibrium shifts to the left (le Châtlier’s principle)

Also, if the Cu(s) is completely consumed, the reaction cannot proceed.

E = Eo - log Q

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BatteriesBatteriesBatteries are the most practical applications of voltaic cell.All batteries have self contained anode/cathode compartments.All operate using the same principles already discussed.

The Classic “dry” (LeClanché) cell.

carboncathode

zincanode

Mushy pasteOf MnO2 andNH4Cl

Overall reaction:

Zn + 2 MnO2 + 2NH4+→

Zn2++2MnO(OH)+ 2NH3

E~1.5 v.

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BatteriesBatteries

Alkaline Battery (similar to dry cell but more efficient)Alkaline Battery (similar to dry cell but more efficient)Anode: (Zn cap) Zn(s) Zn2+ (aq) + 2e-

Cathode: MnO2, NH4Cl and C paste:

2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + 2H2O(l)

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Lead-Acid BatteryLead-Acid Battery

Pb

PbO2

Sulfuric acid, H2SO4

(provides H+ ions)

anodecathode

At cathode: PbO2(s) + HSO4

-(aq) +3H+ +2e- → PbSO4(s) + 2H2O (l)

At anode: Pb(s) + HSO4

-(aq) → PbSO4(s) + H+(aq) + 2e-

Overall:

PbO2(s) + Pb(s) + 2HSO4-(aq) + 2H+(aq) → 2 PbSO4(s) + 2H2O(l)

Eocell = 2.04 V

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BatteriesBatteriesLead-Acid BatteryLead-Acid Battery

Six cells in seriesgive a total voltageof ~12 volts in anautomobile battery.

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BatteriesBatteriesFuel CellsFuel Cells• Direct production of electricity from fuels occurs in a

fuel cell.

• On Apollo moon flights, the H2 - O2 fuel cell was the primary source of electricity.

• Cathode: reduction of oxygen:

2H2O(l) + O2(g) + 4e- 4OH-(aq)

• Anode:

2H2(g) + 4OH-(aq) 4H2O(l) + 4e-

• Total reaction:

2H2(g) + O2(g) 2H2O(l)

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BatteriesBatteriesFuel CellsFuel Cells

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CorrosionCorrosion is the entropy monster’s greatest weapon.It is the evil side of REDOX.It costs 100s of billions of dollars yearly to prevent and correct.Basically, it is the spontaneous process (oxidation) of iron:

Fe → Fe3+ + 3e-

nice shiny metal (steel) ugly brownish-red

powder ……….RUST!

Rusting cannot occur by itself.

Can’t have only the OX in REDOX; So, what gets reduced?

Usually H2O or O2

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CorrosionCommon type of “rusting” redox: (Eo ~ 0.8 V)

O2(g) + 4H+(aq) + 2Fe(s) → 2Fe2+(aq) + 2H2O(l)Easy, but even more favorable in acid conditions. There are similar equations also involving water.

Stopping Corrosion1. Galvanize it (coat with Zn). Fe has higher SRP than

Zn. Coupled with Zn, Fe is the cathode (cathodic protection) (look for a “matte” appearance of Zn).

2. Use “sacrificial metal” such as Mg – (this is also cathodic protection).

3. Cover it (paint).4. Create rust-resistant alloys, e.g., stainless steel

(Fe/Ni/Cr), or nickel steels (Fe/Ni).

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CorrosionCorrosionPreventing the Corrosion of IronPreventing the Corrosion of Iron

Also used onships to prevent

corrosion

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ElectrolysisElectrolysis occurs in electrolytic cells, i.e., in cells where: electricity → chemical reaction takes place.This is the opposite of a voltaic cell.

In electrolysis, current is forcedinto the cathode by external power (like a battery)

cathode anode

• Widely used for electroplating (silver, gold, copper)• Also for production of certain metals from ores and salts (e.g., Na, Al)

external power

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ElectrolysisElectrolysis of water

H2O2

cathode anode

At cathode (V = 0.00):

[2H+(aq) + 2e- → H2(g)] x 2

At anode (V = -1.23):

2 H2O →O2(g) + 4H+(aq) + 4e-

Overall:

2 H2O → 2H2(g) + O2(g)

Eo = -1.23 v.--not spontaneous

external power needed

Why is the volume of H2 twice that of O2?

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Note: EMF (electromotive force), voltage, and cell

potential are all synonyms

1.23 V

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Electrolysis Electrolysis – Electrolysis of Molten Salts– Electrolysis of Molten Salts

Cathode: 2Na+(l) + 2e- 2Na(l)

Anode: 2Cl-(l) Cl2(g) + 2e-

Industrially, electrolysis is used to produce metals like aluminum (Hall-Héroult process, where Al2O3 is electrolyzed in molten cryolite, Na3AlF6, with a carbon electrode to give an overall reaction of 2Al2O3 + 3C 4Al + 3CO2)

Decomposition of molten NaCl

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cathode:Au+(aq) +e-→Au

anode:Au→Au+(aq) +e-

external power source

Au

Au+(aq) Au+(aq)

Au

external power source

Electrolysis with Active Electrodes –Electrolysis with Active Electrodes –

Gold plating – protects against corrosionGold plating – protects against corrosion

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