Electrochemistry

Brown, LeMay Ch 20AP Chemistry

Monta Vista High School

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20.1: Oxidation-reduction reactions

“Redox”: rxns with a change in oxidation number• At least one element’s oxidation number will increase

and one will decrease• Consists of two half-reactions:

oxidation: loss of e-; oxidation number increasesreduction: gain of e-; oxidation number decreases (“reduces”)

Oxidizing agent or oxidant: causes another substance to be oxidized; gains e-

Reducing agent or reductant: causes another substance to be reduced; loses e-

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Ex: Determine the half-reactions.Cr2O7

2- (aq) + Cl1- (aq) → Cr3+ (aq) + Cl2 (g)

+6 -2 -1 +3 0

Reduction half-reaction:

Cr2O72- (aq) → Cr3+ (aq)

Oxidizing agent

Oxidation half-reaction:

Cl1- (aq) → Cl2 (g)

Reducing agent

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20.2: Balancing Redox Reactions“Method of Half-Reactions” steps:1. Assign oxidation numbers to all species; based on this,

split the half-reactions of reduction and oxidation.2. For each, balance all elements except H and O.

3. Balance oxygen by adding H2O to the opposite side.4. Balance hydrogen by adding H+ to the opposite side.5. Balance charges by adding e- to side with overall

positive charge.6. Multiply each half-reaction by an integer so there are

an equal number of e- in each.7. Add the half reactions; cancel any species; check final

balance.

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Ex: Balance the following reaction, which takes place in acidic solution:

Cr2O72- (aq) + Cl1- (aq) → Cr3+ (aq) + Cl2 (g)

Reduction:

Cr2O72- (aq) → Cr3+ (aq)2 + 7 H2O (l)14 H+ (aq) +6 e- +

Oxidation:

Cl- (aq) → Cl2 (g)

6 Cl- (aq) → 3 Cl2 (g) + 6 e-

2 + 2 e-3 [ ]

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6 e- + 14 H+ (aq) + Cr2O72- (aq) → 2 Cr3+ (aq) + 7 H2O (l)

6 Cl- (aq) → 3 Cl2 (g) + 6 e-

14 H+ (aq) + Cr2O72- (aq) + 6 Cl- (aq) →

2 Cr3+ (aq) + 7 H2O (l) + 3 Cl2 (g)

• Since occurs in acidic solution, H+ appears as a reactant

• If reaction occurs in basic solution, balance using same method, but neutralize H+ with OH- as a last step.

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Ex: In basic solution,

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

- (aq)

+7 -2 -1 +4 +5-2 -2

Reduction: MnO4- (aq) → MnO2 (s)

MnO4- (aq) → MnO2 (s) + 2 H2O (l)

4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O (l)

3 e- + 4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O (l)

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Oxidation: Br- (aq) → BrO3- (aq)

3 H2O (l) + Br- (aq) → BrO3- (aq)

3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq)

3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq) + 6 e-

2[3 e- + 4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O (l)]

6 e- + 8 H+(aq) + 2 MnO4-(aq) → 2 MnO2(s) + 4 H2O(l)

3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq) + 6 e-

2 H+(aq) + 2 MnO4- (aq) + Br- (aq) →

2 MnO2 (s) + BrO3- (aq) + H2O (l)

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Neutralize with OH-:

2 H+(aq) + 2 MnO4-(aq) + Br- (aq) + 2 OH- (aq) →

2 MnO2 (s) + BrO3- (aq) + H2O (l) + 2 OH- (aq)

2 H2O (l) + 2 MnO4-(aq) + Br- (aq) →

2 MnO2 (s) + BrO3- (aq) + H2O (l) + 2 OH- (aq)

Simplify:

H2O (l) + 2 MnO4-(aq) + Br- (aq) →

2 MnO2 (s) + BrO3- (aq) + 2 OH- (aq)

Introduction to Electrochemistry

• An electric cell converts chemical energy into electrical energy– Alessandro Volta invented the first electric cell but got his inspiration from Luigi

Galvani. Galvani’s crucial observation was that two different metals could make the muscles of a frog’s legs twitch. Unfortunately, Galvani thought this was due to some mysterious “animal electricity”. It was Volta who recognized this experiment’s potential.

– An electric cell produces very little electricity, so Volta came up with a better design:

• A battery is defined as two or more electric cells connected in series to produce a steady flow of current– Volta’s first battery consisted of several bowls of brine (NaCl(aq))

connected by metals that dipped from one bowl to another

– His revised design, consisted of a sandwich of two metals separated by paper soaked in salt water.

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20.3: Voltaic (or Galvanic) Cells

• Device that utilizes energy released in a spontaneous electrochemical reaction by directing e- transfer along an external pathway, rather than directly between reactants

Electrodes: metals connected by an external circuit AnOx RedCatAnode: half-cell where oxidation occursCathode: half-cell where reduction occurs

Luigi Galvani(1737–1798)

Alessandro Volta(1745–1827)

Voltaic Cells (aka Galvanic Cell)• A device that spontaneously produces electricity by spontaneous

redox reactions.

– The reduction half-reaction (SOA) will be above the oxidation half-reaction (SRA) in the activity series to ensure a spontaneous reaction.

• Composed of two half-cells; which each consist of a metal rod or strip immersed in a solution of its own ions or an inert electrolyte.

– Electrodes: solid conductors connecting the cell to an external circuit

– Anode: electrode where oxidation occurs (-)

– Cathode: electrode where reduction occurs (+)

-The electrons flow from the anode to the cathode (“a before c”) through an electrical circuit rather than passing directly from one substance to another-A porous boundary separates the two electrolytes while still allowing ions to flow to maintain cell neutrality. Often the porous boundary is a salt bridge, containing an inert aqueous electrolyte (such as Na2SO4(aq) or KNO3(aq)), Or you can use a porous cup containing one electrolyte which sits in a container of a second electrolyte.

Voltaic Cells (aka Galvanic Cell)

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Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)Oxidation half-cell: Zn (s) → Zn2+ (aq) + 2 e-Reduction half-cell: Cu2+ (aq) + 2 e- → Cu (s)

anode

Zn Cu

(-) (+)

salt bridge, saturated with NaNO3 (aq)

cathode

e- flow

Na1+NO31-

Zn2+ (aq) Cu2+ (aq)

Notation:

Anode| Anode product|Salt |Cathode|Cathode productZn(s)|Zn2+(aq, 1 M)|NaNO3 (saturated)|Cu2+(aq, 1 M)|Cu(s)

Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Cell Notation for Voltaic CellsVoltaic cells can be represented using cell notation:

The single line represents a phase boundary (electrode to electrolyte) and the double line represents a physical boundary

(porous boundary)

The single line represents a phase boundary (electrode to electrolyte) and the double line represents a physical boundary

(porous boundary)

Write cell notation for Zn-Cu cell

For a redox reaction to be spontaneous, the metal with higher reduction potential should be made cathode.

For a redox reaction to be spontaneous, the metal with higher reduction potential should be made cathode.

Voltaic Cell Summary• A voltaic cell consists of two-half cells

separated by a porous boundary with solid electrodes connected by an external circuit

• Oxidizing agent undergoes reduction at the cathode (+ electrode) – cathode increases in mass

• Reducing agent undergoes oxidation at the anode (- electrode) – anode decreases in mass

• Electrons always travel in the external circuit from anode to cathode

• Internally, cations move toward the cathode, anions move toward the anode, keeping the solution neutral

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20.4: Electromotive force (emf or E)

• Potential difference between cathode and anode that causes a flow of e-.– The result of the difference in electric fields produced at

each electrode

• Also called cell potential or cell voltage, Ecell

• Units = Volt (1 V = 1 J / 1 C)

• Standard conditions (25°C, 1 atm, 1 M): Eºcell

• Eºcell is unique to each set of half-cells involvedCharles-Augustin de Coulomb(1736–1806)

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Standard Reduction Potentials• Assigned to each reduction reaction

Eºcell = Eºreduction(cathode) - Eº reduction(anode)

• All Eºred are referenced to standard hydrogen electrode (S.H.E.):

2 H+ (aq, 1 M) + 2 e- → H2 (g, 1 atm)

Eºreduction = 0 V

Standard Cells and Cell Potentials

• Standard Cell Potential, E0 cell = the electric potential difference of the cell (voltage)

E0 cell = E0r cathode – E0

r anode

Where E0r is the standard reduction potential, and is a measure of a

standard ½ cell’s ability to attract electrons.• The higher the E0

r , the stronger the OA

• All standard reduction potentials are based on the standard hydrogen ½ cell being 0.00V. This means that all standard reduction potentials that are positive are stronger OA’s than hydrogen ions and all standard reduction potentials that are negative are weaker.

• If the E0 cell is positive, the reaction occurring is spontaneous.

• If the E0 cell is negative, the reaction occurring is non-spontaneous

Rules for Calculating Cell Potential

1. Determine which electrode is the cathode. The cathode is the electrode where reduction occurs- marked by the higher reduction potential of the metal

i.e. The OA that is closet to the top on the left side of the redox table = SOA

If required, copy the reduction half-reaction for the strongest oxidizing agent and its reduction potential

2. Determine which electrode is the anode. The anode is the electrode where the strongest reducing agent present in the cell reacts.

i.e. The RA that is closet to the bottom on the right side of the redox table = SRA

If required, copy the oxidation half-reaction (reverse the half-reaction)

3. Determine the overall cell reaction. Balance the electrons for the two half reactions (but DO NOT change the E0

r) and add the half-reaction equations.

4. Determine the standard cell potential, E0cell using the equation:

E0 cell = E0r cathode – E0

r anode

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Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

• Anode: Zn (s) → Zn2+ (aq) + 2 e-– This is oxidation, but Eºreduction is recorded as the reduction

reaction:Zn2+ (aq) + 2 e- → Zn (s), where Eºreduction = - 0.76 V

• Cathode:Cu2+ (aq) + 2 e- → Cu (s) Eºreduction = + 0.34 V

Eºcell = Eºreduction (cathode) - Eºreduction (anode)= 0.34 V – (- 0.76 V)

Eºcell = +1.10 V

Practice Problem # 1

Practice Problem # 2

• Example: What is the standard potential of the cell represented below:

1. Determine the cathode and anode

2. Determine the overall cell reaction

3. Determine the standard cell potential

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• Positive voltages = reaction is spontaneous (cell will produce voltage, G is -)

• Changing the stoichiometric coefficients of a half-cell does not affect the value of Eº, since potential is a measure of the energy per electrical charge.

• However, having more moles means the reaction will continue for a greater time.

Some Important Facts to Remember

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Using the Standard Reduction Potential table

• When Eºred is very negative, reduction is difficult (but oxidation is easy):

Li1+ (aq) + e- → Li (s) Eºred = - 3.05 V– Li (s) oxidizes easily, so it is the strongest reducing agent;

lower on the Standard Reduction Potential chart signifies the anode.

• When Eºred is very positive, reduction is easy:

F2 (g) + 2 e- → 2 F1- Eºred = + 2.87 V– F2 reduces easily, so it is the strongest oxidizing agent;

higher on the chart signifies the cathode.

• Predicting metal “activity series”: most reactive metals are near bottom

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20.5: Spontaneity of Redox Reactions

• Gibbs free energy and emf:DGo = - n F Eo

n = # of moles of e- transferredFaraday’s constant, F

= 96,500 C/mol of e-

• 1F = 1 mole of e-, which has a charge of 96,500 C• Units = (mol e-)(C/mol e-)(V) = (mol e-)(C/mol e-)(J/C) = J

• As Eºcell increases, DGº decreases (becomes more spontaneous)

Michael Faraday(1791-1867)

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20.6: Effect of Concentration on Ecell

The Nernst Equation: Used to figure out cell potential under experimental conditions (E)

G = Go + RT ln Q and Go = - n F Eo

(R = 8.314 J/(mol•K))

- n F E = - n F Eº + RT ln Q

Qn

RTEE ln0

F

Walther Nernst(1864-1941) Q

nEE log

0591.00

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• At equilibrium, G = 0

Go K Eocell Reaction

Negative > 1 Positive Spontaneous

0 1 0 Equilibrium

Positive < 1 Negative Non-spontaneous

Kn

RTE ln0

F

Kn

E log0591.00 since T º = 298 K

Equilibrium composition measurements

Calorimetry and entropy: S0 and S0

Electrical measurements: E0 and E0

cell

Relationship Between Thermodynamic Quantities

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E0cell of cells

• Model of a portion of a cell membrane shows a K+

channel (blue) and a Na+ channel (red). The area outside the cell (top) is rich in Na+ and low in K+. Inside the cell, the fluids are relatively rich in K+ and low in Na+.

Electrolytic Cells• The term “electrochemical cell” is often used to refer to a:

– Voltaic Cell – one with a spontaneous reaction

The metal that is higher on activity series, should be made cathode for a spontaneous reaction

Eocell greater than zero = spontaneous

– Electrolytic cell – one with a nonspontaneous reaction

Eocell less than zero= nonspontaneous

• Why would anyone be interested in a cell that is not spontaneous?

– This would certainly not a good battery choice, but by supplying electrical energy to a non spontaneous cell, we can force this reaction to occur.

– This is especially useful for producing substances, particularly elements. I.e. the zinc sulfate cell discussed above is similar to the cell used in the industrial production of zinc metal.

Electrolytic Cells

• Electrolytic Cell – a cell in which a nonspontaneous redox reaction is forced to occur; a combination of two electrodes, an electrolyte and an external power source.

– Electrolysis – the process of supplying electrical energy to force a nonspontaneous redox reaction to occur

– The external power source acts as an “electron pump”; the electric energy is used to do work on the electrons to cause an electron transfer

Electrons are pulled from the anode and pushed to the

cathode by the battery or power supply

Electrons are pulled from the anode and pushed to the

cathode by the battery or power supply

Comparing Electrochemical Cells: Voltaic and Electrolytic

It is best to think of “positive” and “negative” for electrodes as labels, not charges.

It is best to think of “positive” and “negative” for electrodes as labels, not charges.

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20.8-9: Electrolytic cells• Electrical energy is used to cause a non-

spontaneous reaction to occur.• Current (I): a measure of the charge (Q) per unit

time (t)

• Units = amps or amperes (A)1 A = 1 C / 1 s

t

QI

André-Marie Ampère(1775 – 1836)

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Ex: Na+ (l) + Cl- (l) → Na (l) + Cl2 (g)Oxidation half-cell: 2 Cl- (l) → Cl2

(g) + 2 e-

Reduction half-cell: Na+ (l) + e- → Na (l)

anode

Pt Pt(-)(+)

cathode

e- flow

NaCl (l)

Cl2 (g) Na (l)

• In a voltaic cell: anode is (-), cathode is (+)

• In an electrolytic cell: anode is (+), cathode is (-)

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20.9: Quantitative ElectrolysisEx: A current of 0.452 A is passed through an

electrolytic cell containing molten NaCl for 1.5 hours.

• Write the electrode reactions.

• Calculate the mass products formed at each electrodes

Oxidation: 2 Cl- (l) → Cl2 (g) + 2 e-

Reduction: Na+ (l) + e- → Na (l)

Strategy: calculate the total charge of e- transferred, and use F to convert to moles.

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C 2441sec) A)(5400 (0.452Q

Na mol 1

g 22.99

1

Na mol 1

C 96,500

1C) (2441Na Mass

FF

g 0.58

2

22 Cl mol 1

g 70.90

2

Cl mol 1

C 96,500

1C) (2441Cl Mass

FF

g 0.90

tIQor t

QI

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Electrolysis of Aqueous Mixtures

• If the voltage on an aqueous mixture of CuCl2 and ZnCl2 in an electrolytic cell is slowly increased, what products will form at each electrode?

o What exists in solution that could be reduced?Cu2+, Zn2+, and H2O (not Cl-)

Cu2+ + 2 e- → Cu Eºred = 0.34 V

Zn2+ + 2 e- → Zn Eºred = - 0.76 V

2 H2O + 2 e- → H2 + 2 OH- Eºred = - 0.83 V

o Since Eºred (Cu) is the most positive, it is the most probable reaction, and therefore will occur most easily (at the cathode):Cu (s) > Zn (s) > H2 (g)

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• If the voltage on an aqueous mixture of CuCl2 and ZnCl2 in an electrolytic cell is slowly increased, what products will form at each electrode?

o What exists in solution that could be oxidized?Cl- and H2O (not Cu2+ and Zn2+)

2 H2O → O2 + 4 H+ + 4 e- Eºox = - 1.23 V

2 Cl- → Cl2 + 2 e- Eºox = - 1.36 V

o Since Eºox (H2O) is the most positive, it is the most probable reaction, and therefore will occur most easily (at the anode):O2 (g) > Cl2 (g)

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Electroplating• The anode is a

silver bar, and the cathode is an iron spoon.

Anode:Ag (s) → Ag+ (aq)

Cathode:Ag+ (aq) → Ag (s)