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ELECTROCHEMISTRY

Review: Balancing REDOX reactions

• Balance the following redox reactions:

1. MnO4-(aq) + Cl-(aq) ⇆ Mn2+

(aq) + Cl2(aq) acidic solution

2. H2O2(aq) + ClO2(aq) ⇆ ClO2-(aq) + O2(g) basic solution

3. NO2-(aq) + Cr2O7

2-(aq) ⇆ Cr3+

(aq) + NO3-(aq) acidic

solution

Electrochemistry

• Oxidation-Reduction reactions.

– LEORA (loose electron oxidation reducing agent)

– GEROA (gain electron reduction oxidizing agent)

Cu2+ + Zn � � � � Zn2+ + Cu

– Reduction: Cu2+ + 2e- � Cu

– Oxidation: Zn � Zn2+ + 2e-

(Half-reactions of the redox process)

• The two parts of the reaction can be physically

separated.

– The oxidation reaction occurs in one cell .

– The reduction reaction occurs in the other cell.

– A “cell” is a compartment for the half-reaction

• The cell must contain all physical forms of the species

involved

– Reduction half-reaction cell contains aqueous Cu2+ and solid Cu

– Oxidation half-reaction cell contains aqueous Zn2+ and solid Zn

• The combination of two cells (reduction and

oxidation cell) is called an electrochemical cell

Electrochemistry

There are two kinds electrochemical cells.

1. Electrochemical cells containing in

nonspontaneous chemical reactions are

called electrolytic cells.

2. Electrochemical cells containing

spontaneous chemical reactions are called

voltaic or galvanic cells.

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Electrical Conduction

• Metals conduct electric currents well in a process called metallic conduction.

• In metallic conduction there is electron flow with no atomic motion.

• In ionic or electrolytic conduction ionic motion transports the electrons.

– Positively charged ions, cations, move toward the negative electrode (cathode)

– Negatively charged ions, anions, move toward the positive electrode (anode)

Electrodes

• The surface in which oxidation or reduction half-reaction occurs is called the ELECTRODE.– Electrode part of reaction: active electrode

– Electrode not a part of reaction: INERT electrode

• Convention for electrodes (correct for either electrolytic or voltaic cells):– Cathode : reduction

• Negative in electrolytic cells and positive in voltaic cells.

– Anode : oxidation• Positive in electrolytic cells and negative in voltaic cells.

Voltaic or Galvanic Cells

• Electrochemical cells in which a spontaneous chemical reaction produces electrical energy.

• Cell halves are physically separated so that electrons (from redox reaction) are forced to travel through wires and creating a potential difference.

• Examples of voltaic cells include:

Construction of Simple Voltaic Cells

• Voltaic cells consist of two half-cells which contain the oxidized and reduced forms of a substances in contact with each other.

• A simple half-cell consists of:

– A piece of metal immersed in a solution of its ions.

– A wire to connect the two half-cells.

– And a salt bridge to complete the circuit, maintain neutrality, and prevent solution mixing.

The Zinc-Copper Cell

• Cell components for the Zn-Cu cell are:

1. A metallic Cu strip immersed in 1.0 M copper (II) sulfate.

2. A metallic Zn strip immersed in 1.0 M zinc (II) sulfate.

3. A wire and a salt bridge to complete circuit

• The cell’s initial voltage is 1.10 volts

• In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode).

The Zinc-Copper Cell

• Short hand notation for voltaic cells.

– The Zn-Cu cell provides a good example.

Zn | (1.0 M) Zn2+ || Cu2+ (1.0 M) | Cu

Reduction half-reactionOxidation half-reaction

Double

vertical

bar : Salt

bridge

Single vertical

bar: electrode

Single vertical

bar: electrode

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The Copper - Silver Cell• Cell components:

1. A Cu strip immersed in 1.0 M copper (II) sulfate.

2. A Ag strip immersed in 1.0 M silver (I) nitrate.

3. A wire and a salt bridge to complete the circuit.

• The initial cell voltage is 0.46 volts.

The Copper - Silver Cell

• Compare the Zn-Cu cell to the Cu-Ag cell– The Cu electrode is the cathode in the Zn-Cu cell.

– The Cu electrode is the anode in the Cu-Ag cell.

• Whether a particular electrode behaves as an anode or as a cathode depends on what the other electrode of the cell is.

The Copper - Silver Cell

• These experimental facts demonstrate that Cu2+ is a stronger oxidizing agent than Zn2+.

– In other words Cu2+ oxidizes metallic Zn to Zn2+.• Similarly, Ag+ is is a stronger oxidizing agent than Cu2+.

– Because Ag+ oxidizes metallic Cu to Cu 2+.• If we arrange these species in order of increasing strengths, we see

that:

Standard Electrode Potential

• Potential means the tendency to transfer

electrons

• To measure relative electrode potentials, we

must establish an arbitrary standard.

• That standard is the Standard Hydrogen

Electrode (SHE).

– The SHE is assigned an arbitrary voltage

of 0.00000000… V

– To determine the ability of a reaction to

transfer or accept electrons relative to

the SHE

The Zinc-SHE Cell

• For this cell the components are:

1. A Zn strip immersed in 1.0 M zinc (II) sulfate.

2. The other electrode is the Standard Hydrogen Electrode.

3. A wire and a salt bridge to complete the circuit.

• The initial cell voltage is 0.763 volts.

• The cathode is the Standard Hydrogen Electrode.

– In other words Zn reduces H+ to H2.

• The anode is Zn metal.

– Zn metal is oxidized to Zn2+ ions.18

The Zinc-SHE Cell

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The Copper-SHE Cell• The cell components are:

1. A Cu strip immersed in 1.0 M copper (II) sulfate.

2. The other electrode is a Standard Hydrogen Electrode.

3. A wire and a salt bridge to complete the circuit.

• The initial cell voltage is 0.337 volts.

• In this cell the SHE is the anode

– The Cu2+ ions oxidize H2 to H+.

• The Cu is the cathode.

– The Cu2+ ions are reduced to Cu metal.20

The Copper-SHE Cell Uses of Standard Electrode Potentials

• Electrodes that force the SHE to act as an anode are assigned

positive standard reduction potentials.

• Electrodes that force the SHE to act as the cathode are assigned

negative standard reduction potentials.

• Standard electrode (reduction) potentials tell us the tendencies of

half-reactions to occur as written.

Potassium (K0 ) has greater

tendency to give electrons (to

be OXidized relative to SHE)

Fluorine (F20 ) has greater

tendency to accept electrons

(to be REDuced relative to SHE)

Uses of Standard Electrode Potentials

• Standard electrode potentials are used to predict

whether an electrochemical reaction (at standard

state conditions) will occur spontaneously.

• Standard electrode potentials are used to predict

which redox couple (if there are many) will occur at

the specified standard conditions

Strongest

Oxidizing

Agent

Strongest

Reducing

Agent

Uses of Standard Electrode Potentials

1. Choose the appropriate half-reactions from a table of standard reduction

potentials.

2. Write the half-reaction equation with the more positive E0 value first, along

with its E0 value.

3. Write the other half-reaction equation as an oxidation (reverse the tabulated

reduction half-reaction) and change the sign of the tabulated E0.

4. Balance the electron transfer. Do not multiply the E0 values by the coefficient!

(E0 values are INTENSIVE properties)

5. Add the reduction and oxidation half-reactions and their potentials. This

produces the equation for the reaction for which E0cell is positive, which

indicates that the forward reaction is spontaneous (galvanic). If the system has

a negative E0cell , the system must be nonspontaneous (electrolytic).

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• Will permanganate ions, MnO4-, oxidize iron (II) ions to iron (III) ions, or

will iron (III) ions oxidize manganese(II) ions to permanganate ions in acidic solution? (Which is more spontaneous?)

• Will silver ions (Ag+) oxidize metallic zinc (to Zn2+) or will zinc ions

(Zn2+) oxidize metallic Ag (to Ag+)? (Which is more spontaneous?)Gibbs Free Energy and Electrical

Potential at Standard Conditions

• For a redox reaction at standard states

ΔG0 = -nFE0cell

n � # of electrons involved in the balanced reaction

F � 96 485 J/V•mol e- (Faraday’s Constant)

E0cell � cell potential at standard conditions

ΔG0 � standard Gibbs Free Energy

• Relationship of E0cell and Equilibrium constant Keq

ΔG0 = -nFE0cell = -RTlnKeq

∴ E0cell = RTlnKeq / nF or lnKeq = nFE0

cell / RT

kJ/mol 302.23-or rxn J/mol 302230- G

V 1.5662e mole V

J 485 96e mole 2- G

0

-

-0

=∆

+⋅⋅=∆

kJ/mol 285.60-or rxn J/mol 285595- G

V 0.74e mole V

J 485 96e mole 4- G

0

-

-0

=∆

+⋅⋅=∆

• Consider the following reduction half-reactions involving water

V 0.83- OH2He2O2H

V 0.40 4OHe4O2HO

V 1.23 O2H 4e 4H O

-

(aq)2(g)

-

(l)2

-

(aq)

-

(l)22(g)

(l)2

-

(aq)2(g)

+→+

→++

→+++

V 2.06 O2H 2H O

V 0.83 )e2OH2 OH22(H

V 1.23 O2H 4e 4H O

(i)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2

-

(aq)2(g)

→+

+→+

→+++

V 1.23 O2H 2H O

V 0.83 )e2OH2 OH22(H

V 0.40 4OH 4e O2H O

(i)2(g)22(g)

-

(l)2

-

(aq)2(g)

-

(aq)

-

(l)22(g)

→+

+→+

→++

More

spontaneous, but

cathode requires

H+ (will come from

H2 gas)

Less spontaneous,

but cathode

requires H2O

V 2.06 O2H 2H O

V 0.83 )e2OH2 OH22(H

V 1.23 O2H 4e 4H O

(l)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2

-

(aq)2(g)

→+

+→+

→+++

Electrolytic Cells• Nonspontaneous electrochemical cells

• E0cell is negative

• ΔG0 is positive [ΔG0 = -nF(-E0cell value)= + value]

Eletrolysis means forcing a nonspontaneous redox reaction to

occur (at the electrodes) by applying potential

Spontaneous, releases 212.3

kJ/mole of energy

Nonspontaneous, requires

212.3 kJ/mole of energy

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Electrolytic Cells

• Electrolytic cells requires external potential source

• In electrolytic cells, reduction still occurs at cathode, oxidation

at anode

• Polarity is reversed

o cathode: negative

o anode: positive

• Electrolysis can be done in 1-compartment or 2-compartment

cells

• Two examples of commercial electrolytic reactions are:

– Electroplating of jewelry and auto parts.

– Electrolysis of chemical compounds.

• Voltaic/Galvanic cellso Cathode (+): Most Positive

E0red

o Anode (-): Most Negative E0

red

o The MOST positive E0cell will

occur first

• Electrolytic cellso Cathode (-): Least

Negative E0red

o Anode (+): Least Positive E0

red

o The LEAST negative E0cell

will occur first

In both cases, REDUCTION occurs at CATHODE, OXIDATION occurs at ANODE

(REDCAT, OXAN)

V 0.83- OH2He2O2H

V 0.40 4OHe4O2HO

V 1.23 O2H 4e 4H O

-

(aq)2(g)

-

(l)2

-

(aq)

-

(l)22(g)

(l)2

-

(aq)2(g)

+→+

→++

→+++

V 2.06 O2H 2H O

V 0.83 )e2OH2 OH22(H

V 1.23 O2H 4e 4H O

(i)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2

-

(aq)2(g)

→+

+→+

→+++

V 1.23- 2H OO2H

V 0.83- )OH2He2OH22(

V 0.40 - 4e O2H O4OH

(g)22(g)(i)2

-

(aq)2(g)

-

(l)2

-

(l)22(g)

-

(aq)

+→

+→+

++→

Electrolytic conditionsVoltaic/Galvanic conditions

Electrolytic Cell: 1-compartment

A container for the

reaction mixture.

Two electrodes

immersed in the

reaction mixture.

A voltage source of

direct current.

The motion of the ions through the solution = electric current.

Electrolytic conduction

o Positive ions migrate toward the negative electrode (cathode).

o Negative ions migrate toward positive electrode (anode).

Electrolytic Cell: Electrolysis of NaCl

• Requires molten (melted) NaCl since ions conduct electricity– Consists of Na+ and Cl- ions.

• If direct current (greater than 3.852 V) is applied (by way of two inert electrodes) through the cell containing the molten NaCl, we observe the following:– Pale green gas (Cl2) is formed in one electrode.

– Molten, silvery Na forms at the other electrode. The Na particles floats on top of the molten NaCl.

• Liquid Na is produced at the cathode (-).

2{Na+ + e� Na(l)*}

• Gaseous Cl2 is produced at the anode (+).

2Cl- � Cl2 + 2e-

---------------------------------

2Na++2Cl-�2Na(l)+Cl2(g)

*Na(l) � Na(s)

• This is non-spontaneous except at very high T (>801°C)

• Direct current (dc) source supplies energy to force the reaction forward.

AnodeCathode

• Electrons are used in the cathode half-reaction (reduction) and produced in the anode half-reaction (oxidation).

• Travel of e- : from ANODE(+) to CATHODE(-).

• The dc source forces the e- to flow non-spontaneously from the positive electrode to the negative electrode.

• Na and Cl2 must NOT be allowed to come into contact with each other because they react spontaneously, rapidly and explosively to form NaCl.

• In Downs’ Cell, the liquid Na is drained off, cooled and cast into blocks, then stored in inert mineral oil to prevent reaction with atmosphere (O2) or H2O.

Downs’ Cell – Industrial

production of Na(s)

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Electrolytic Cells: Electrolysis of Aqueous KI

K+ (aq) + e- ���� K(s) -2.924 V

2H2O + 2e- ���� 2H2 (g) + OH- -0.828 V

I2(s) + 2e- ���� 2I- (aq) +0.535 V

O2 (g) + 4H+ +4e- ���� 2H2O +1.229 V

Calculations of different redox combinations and the minimum applied voltages required for electrolysis:

E1 = E(K+/K) – E(I2/I) = (-2.924 V) – (+0.535 V)

= -3.459V

E2 = E(H2O/H2,OH) – E(I-/I2) = (-0.828V) – (0.535 V)

= -1.363V

E3 = E(H2O/H2,OH) – E(H2O/O2, H+)=(-0.828V) – (1.229)

= -2.057

2nd case has smallest negative V and hence, this

combination will occur first.

cathode anode

Electrolytic Cells: Electrolysis of Aqueous NaCl

Na+ (aq) + e- � Na(s) -2.714 V

2H2O + 2e- � 2H2 (g) + OH- -0.828 V

Cl2(s) + 2e- � Cl- (aq) +1.360 V

Calculations of different redox combinations and the minimum applied voltages required for electrolysis:

E1 = E(Na+/Na) – E(Cl-/Cl2)

= (-2.714 V) – (+1.360 V) = -4.074 V

E2 = E(H2O/H2,OH) – E(Cl2/Cl-)

= (-0.828V) – (1.360 V) = -2.188 V

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Counting Electrons: Coulometry and Faraday’s

Law of Electrolysis

• Faraday’s Law –

mole substance oxidized or reduced ∝ mole e-

• A faraday is the amount of electricity that reduces one equivalent of a species at the cathode and oxidizes one equivalent of a species at the anode.

• A coulomb is the amount of charge that passes a given point when a current of one ampere (A) flows for one second.

• 1 amp = 1 coulomb/second

Counting Electrons: Coulometry and

Faraday’s Law of Electrolysis

• Faraday’s Law states that during electrolysis, one faraday of electricity (96,487 coulombs) reduces and oxidizes, respectively, one equivalent of the oxidizing agent and the reducing agent.

– This corresponds to the passage of one mole of electrons

through the electrolytic cell.

Counting Electrons: Coulometry and

Faraday’s Law of Electrolysis

• Calculate the mass of palladium produced by the reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.

Counting Electrons: Coulometry and

Faraday’s Law of Electrolysis

• Calculate the volume of oxygen (measured at STP)

produced by the oxidation of water.

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Commercial Applications of Electrolytic Cells

Electrolytic Refining and Electroplating of Metals

• Impure metallic copper can be purified

electrolytically to ≈ 100% pure Cu.

– The impurities commonly include some active metals plus

less active metals such as: Ag, Au, and Pt.

• The electrolytic solution is CuSO4 and H2SO4

• The impure Cu dissolves to form Cu2+.

• The Cu2+ ions are reduced to Cu at the cathode.

Commercial Applications of Electrolytic Cells

• Any active metal impurities are oxidized to cations that are more difficult to reduce than Cu2+.

– This effectively removes them from the Cu metal.

Effect of Concentrations (or Partial

Pressures) on Electrode Potentials

The Nernst Equation

• The Nernst equation describes the electrode potentials at nonstandard conditions.

The Nernst EquationThe Nernst Equation

• Substitution of the values of the constants into the

Nernst equation at 25o C gives:

The Nernst Equation

• For this half-reaction:

• The corresponding Nernst equation is:

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The Nernst Equation

• Substituting E0 into the above expression gives:

• If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard conditions, then E = E0 because the concentration term equals zero.

The Nernst Equation

• Calculate the potential for the Cu2+/Cu+ electrode at

250C when the Cu+ ion concentration is 1/3 of the Cu2+

ion concentration.

The Nernst Equation

• Calculate the initial potential of a cell that consists of

an Fe3+/Fe2+ electrode in which [Fe3+]=1.0 x 10-2 M

and [Fe2+]=0.1 M connected to a Sn4+/Sn2+ electrode

in which [Sn4+]=1.0 M and [Sn2+]=0.10 M . A wire

and salt bridge complete the circuit.

• Calculate the E0 cell by the usual procedure.

• Substitute the ion concentrations into Q to calculate

Ecell.