electrochemistry. reduction-oxidation (redox) reactions in chemical reaction bonds, both covalent...

Electrochemistry

Reduction-Oxidation (REDOX) Reactions

In chemical reaction bonds, both covalent and ionic, are made and broken by moving electrons.

So far all in reactions considered the number of electrons on each atom has been preserved

In REDOX reaction electrons are transferred between atoms as well as bonds being broken and formed.

We therefore need to keep track of the number of electrons of each atom.

The oxidation state of an atom is used for this purpose, which is related to the idea of the formal charge.

2 Na + ½ O2 → Na2O 2 Na+ + O2-

The balance between ionic/covalency of the bonds in the reactants is different from that in the product.

0 0 1+ 2-

2 Na → 2 Na+ + 2 e ½ O2 + 2 e → O2-

Oxidized Reduced

Loss e’s Oxidation = LEO

Gain e’s Reduction = GER

Rules for Assigning Oxidation States

Oxidation StateThe oxidation state of an element is its charge assuming ionic bonding

Electrons are not shared they are placed on the more electronegative element

1) Pure elements have oxidation states of 0

2) Ions have oxidation states that add up to the charge of the ion

3) Hydrogen has an oxidation state of +1 unless bonded to a less electronegative atom. When bound to metals or boron it has an oxidation state of -1.

4) Fluorine has an oxidation state of -1

5) Oxygen has an oxidation state of -2 unless bonded to fluorine or another oxygen.

6) Halogens other than fluorine have oxidation states of -1 unless bonded to oxygen or a more electronegative halogen

7) The rest are determined by the process of elimination, where the oxidation states must add up to the total charge of the molecule or ion.

Oxidation State1) Determine the oxidation state of all the element in the following molecules:

a) H2 H = 0

b) CO2O = -2 0 = C + 2(-2)

c) BF3 F= -1 0 = B + 3(-1)

d) H2SO4H = 1 O = -2 0 = S +(2(+1) + 4(-2)) = S - 6

e) SO2ClF O = -2 F = -1 Cl = -1 0 = S +(2(-2) -1 -1) = S - 6

f) IO2F2

_ O = -2 F = -1 -1 = I + (2(-2) + 2(-1)) = I - 6

C = 4

B = 3

S = 6

S = 6

g) HPO42- O = -2 H = 1

I = 5

-2 = P + (4(-2) + 1) = P - 7 P = 5

Oxidation States of Poly Atomic Ions

PO43- SO4

2-

NO3-

ClO3-

CO32-

NO2-

PO33- SO3

2- ClO2-

4 5

3

5

3

6

4

5

3

Recognizing Redox Reactions

3

3 2 2 4 2

(s) (aq)

(aq) (g) (l)

Cu + 4HNO

Cu(NO ) +N O + 2H O

Consider the following reaction:

Cu

HNO3 N = 5

Reactants Products

Cu2+

N2O4 N = 4

Cu → Cu2+ + 2 e Oxidation

4 HNO3 + 2 e → 2NO3

- + N2O4 + 2 H2O Reduction

# of e’s gained = # of e’s lost

REDOX equation seem difficult to balance

N = 0 N = 2

Recognizing Redox ReactionsThe species that is oxidized in a Redox reaction is called the reducing agent

The species that is reduced in a Redox reaction is called the oxidizing agent

In the previous example Cu is the reducing agent

In the previous example HNO3 is the oxidizing agent

Redox ReactionsThe term Oxidation originates from the fact that many of these reactions involve oxygen where

the element loses electrons to it.

For example,

Burning magnesium:

Mg Mg2+ + 2 e–

Oxygen gains electrons:

½ O2 + 2 e– O2–

and the net reaction is:

1 Mg + ½ O2 1 MgO

Oxidation Half Reaction

Reduction Half Reaction

Redox ReactionsThe term Reduction originates from the concept of reducing an ore, usually a metal oxide, to its

elemental form, usually the pure metal.

For example:

Iron is made from iron ore, the iron reaction is:

Fe2+ + 2 Fe3+ + 8 e– 3 Fe

and the carbon monoxide reaction is:

4 CO 4 CO2+ + 8 e–

Here the oxide ions are spectators, transferring from the iron to the oxidized carbon. The net reaction therefore becomes:

1 Fe3O4 + 4 CO 3 Fe + 4 CO2

Oxidation Half Reaction

Reduction Half Reaction

Balancing Redox Reactions in Solution

The method of half reactions

Step 1. Recognize the reaction as an oxidation-reduction.Step 2. Separate the overall process into half-reactions.Step 3. Balance each half-reaction by mass.

i) In acid solution, a) add H2O to the side requiring O atoms. b) add H+ to balance any remaining unbalanced H atoms. ii) In basic solution, a) add 2 OH– to the side requiring O atoms, and H2O to the other b) add H2O to the side requiring H atoms, and one OH– to the other side.

Step 4. Balance the half-reactions by charge.Step 5. Multiply the balanced half-reactions by appropriate factors to achieve common whole

number of electrons being transferred.Step 6. Add the balanced half-reactions.Step 7. Eliminate common reactants and products.Step 8. Check the final result for mass and charge balance.

Balancing Redox Reactions Sample Problems

2. Balance equations for the half-reactions. Assume these reactions occur in acidic solution, meaning that H+ or H+ and H2O may be used to balance the equation.

(a) Br2(l) Br–(aq)

Ox. St. 0 -1 reduction

Mass balance: Br2(l) 2 Br–(aq)

Charge balance: Br2(l) + 2e- 2 Br–(aq)

(b) VO2+(aq) V 3+(aq)Ox. St. +4 +3 reduction

O balance: VO2+(aq) V 3+(aq) + H2O

H balance: VO2+(aq) + 2 H+ V 3+(aq) + H2O

Charge balance: VO2+(aq) + 2 H+ + e- V 3+(aq) + H2O


3. The half-reactions here are in basic solution. You may need to use OH– or the OH– / H2O pair to balance the equation.

(a) CrO2– (aq) CrO4

2– (aq) Ox. St. 3+ 6+ oxidation

O balance: CrO2– (aq) + 4 OH- CrO4

2– (aq) + 2 H2O

Charge balance: CrO2– (aq) + 4 OH- CrO4

2– (aq) + 2 H2O + 3 e-

(b) Ni(OH)2(s) NiO2(s)Ox. St. 2+ 4+ oxidation

H balance: Ni(OH)2(s) + 2 OH- NiO2(s) + 2 H2O

Charge balance: Ni(OH)2(s) + 2 OH- NiO2(s) + 2 H2O + 2 e-

Balancing Redox Reactions Sample Problems4. The reactions here are in acid solution, meaning that H+ or H+ and H2O may be used to balance the

equation. MnO4

– (aq) + HSO3– (aq) Mn2+ (aq) + SO4

2– (aq)Ox. St. +7 +4 2+ 6+

Reduction half reaction: MnO4– / Mn2+

Mass Balance MnO4– (aq) + Mn2+ (aq)

O balance: MnO4– (aq) + 8H+ Mn2+ (aq) + 4 H2O

Charge balance: MnO4– (aq) + 8H+ + 5 e– Mn2+ (aq) + 4 H2O

Oxidation half reaction: HSO3– / SO4

2–

Mass balance: HSO3– (aq) SO4

2– (aq)O balance: HSO3

– (aq) + H2O SO42– (aq) + 3 H+

Charge balance: HSO3– (aq) + H2O SO4

2– (aq) + 3 H+ + 2 e–

2 MnO4– (aq) + 5 HSO3

– (aq) + 16H+ + 5 H2O 2 Mn2+ (aq) + 5 SO42– (aq) + 8 H2O + 15 H+

2 MnO4– (aq) + 5 HSO3

– (aq) + 1 H+ 2 Mn2+ (aq) + 5 SO42– (aq) + 3 H2O

x 2

x 5

X XX X


4. The reactions here are in acid solution, meaning that H+ or H+ and H2O may be used to balance the equation.

Cr2O72– (aq) + Fe2+ (aq) Cr3+ (aq) + Fe3+ (aq)

Ox. St. +6 +2 3+ 3+

Reduction half-reaction: Cr2O72–/ Cr3+

Mass balance: Cr2O72– (aq) 2Cr3+ (aq)

O balance: Cr2O72– (aq) + 14H+ 2 Cr3+ (aq) + 7 H2O

Charge balance: Cr2O72– (aq) + 14H+ + 6 e– 2 Cr3+ (aq) + 7 H2O

Oxidation half-reaction: Fe2+/ Fe3+

Mass balance: Fe2+ (aq) Fe3+ (aq)Charge balance: Fe2+ (aq) Fe3+ (aq) + 1 e–

1 Cr2O72– (aq) + 6 Fe2+ (aq) + 14H+ 2 Cr3+ (aq) + 6 Fe3+ (aq) + 7 H2O

x 6


5. The reactions here are in basic solution. You may need to add OH– or the OH– / H2O pair to balance the equation.

Fe(OH)2(s) + CrO42– (aq) Fe2O3(s) + Cr(OH)4

– (aq)Ox. St. +2 +6 3+ 3+ Oxidation half-reaction: Fe(OH)2/Fe2O3

Mass balance : 2 Fe(OH)2(s) + Fe2O3(s)

O balance: 2 Fe(OH)2(s) + 2 OH– Fe2O3(s) + 3 H2O

Charge balance: 2 Fe(OH)2(s) + 2 OH– Fe2O3(s) + 3 H2O + 2 e–

Reduction half-reaction: CrO42–/ Cr(OH)4

–

Mass balance: CrO42– (aq) Cr(OH)4

– (aq)

O balance: CrO42– (aq) + 4 H2O Cr(OH)4

– (aq) + 4 OH–

Charge balance: CrO42– (aq) + 4 H2O + 3 e– Cr(OH)4

– (aq) + 4 OH–

6 Fe(OH)2(s) + 2 CrO42–(aq) + 6 OH– + 8 H2O 3 Fe2O3(s) + 2 Cr(OH)4

–(aq) + 9 H2O + 8 OH-

6 Fe(OH)2(s) + 2 CrO42–(aq) 3 Fe2O3(s) + 2 Cr(OH)4

–(aq) + 1 H2O + 2 OH-

x 2

x 3

X X X X


5. The reactions here are in basic solution. You may need to add OH– or the OH– / H2O pair to balance the equation. (b) PbO2(s) + Cl– (aq) ClO– (aq) + Pb(OH)3

– (aq)Ox. St. 4+ 1- 1+ 3+

Reduction half-reaction: PbO2/ Pb(OH)3–

Mass balance: PbO2 (s) + Pb(OH)3– (aq)

O balance:PbO2(s) + 2 H2O Pb(OH)3– + OH–

Charge balance: PbO2(s) + 2 H2O + 2 e– Pb(OH)3– + OH–

Oxidation half-reaction: Cl–/ ClO– Mass balance: Cl– (aq) ClO– (aq)O balance: Cl– (aq) + 2 OH– ClO– (aq) + H2O

Charge balance: Cl– (aq) + 2 OH– ClO– (aq) + H2O + 2 e–

PbO2(s) + Cl–(aq) + 2 OH– + 2 H2O Pb(OH)3– (aq) + ClO–(aq) + OH– + H2O

1 PbO2(s) + 1 Cl–(aq) + 1 OH– + 1 H2O 1 Pb(OH)3– (aq) + 1 ClO–(aq)

X XX X

Gibbs Free Energy and Cell Potential

The Gibbs Free Energy of reaction tells us whether a certain reaction is product favored or reactant-favoured in the forward direction

The electrochemical analogue is the cell potential given by the symbol EE is expressed in the common electrical unit of volts.

The exact relationship between cell potential and Gibbs energy is given by:

in general, and for standard conditions where F = 96485 C/mol

Faradays ConstantThe cell voltage E has the opposite sign convention to that of G:

i) Product Favoured Reaction: –ve G or +ve Eii) Reactant Favoured Reaction: +ve G or –ve E

A reaction which is product favoured produces a voltage, and is therefore called a Voltaic cell, in honor of Alessandro Volta.

A reaction which is reactant favoured can be driven forward by the application of a greater opposite voltage; such reactions are called electrolytic cells, and the process is named electrolysis.

ΔG=-nFE 0 0ΔG =-nFE

Voltaic CellsConsider the reaction:

Zn(s) +Cu2+(aq) Zn2+(aq) + Cu(s)Gf (kJ mol–1) 0 65.52 –147.0 0

Grxn = –212.5 kJ mol–1 < 0 & n = 2

Anode: OxZn(s) Zn2+(aq) + 2 e–

Cathode: Red.Cu2+(aq) + 2 e– Cu(s)

> 0 under zero load conditions

00 -ΔG -212.5kJ

E = =nF 2×96.485kC

=+1.10V

ΔG=-nFE

Reversible cell reactions and cell notation

Although the reaction only requires a copper(II) salt and metallic zinc, the cell is built in such a way that both metallic zinc and copper are present, and copper and zinc sulfate.

This ensures that the reactions are reversible as accurate potentials can only be measured for reversible electrochemical cells.

A short-hand notation indicates the complete composition of a given electrochemical cell.

The notation utilizes the balanced redox reaction, including both the reactants and the products.

The notation for the previous example would be:

Zn(s) | Zn2+(1 M) || Cu2+(1 M) | Cu(s)

Left Hand Side Right Hand Side

Anode reaction Cathode reactionOxidation reaction. Reduction reaction

The vertical lines, |, indicate a phase boundary, such as between a solid and a liquid.

Double lines, ||, indicate double boundaries, such as commonly occur when a salt bridge is placed between the two half cells.

Electrochemical Cell Conventions

Standard Half-Cell Reduction Potentials

The Cu/Zn cell has a standard potential of +1.10 V

Can we relate this tothe voltage of the two“half-cells”?

Choose an arbitraryreference point:Standard HydrogenElectrode, or SHE

Thus Zn/Zn2+ hasa voltage of +0.76 Vagainst a SHE

Standard Half-Cell Reduction Potentials

The Cu/Cu2+ has half cell has a voltage of +0.34 V against a SHE

Combining the two half-cell voltages gives+0.76 + +0.34 = +1.10 V

This is in good agreementwith the measured voltage

These values have beencompiled into a generaltable

To account for the signconvention, the table iswritten only as reductionreactions

Standard Hydrogen Electrode

The glass tube is a hydrogen electrodethat can be used as real reference electrodeThe picture below is a diagram showing thefunction of this device.

+( s) ( g) ( aq)2 3Pt |H |H O ∥

Reduction Half-Reaction E (V)

F2(g) + 2 e – 2 F – (aq) +2.87

H2O2(aq) + 2 H3O + (aq) + 2 e – 4 H2O(l) +1.77

PbO2(s) + SO4 2– (aq) + 4 H3O

+ (aq) + 2 e – PbSO4(s) + 6 H2O(l) +1.685

MnO4 – (aq) + 8 H3O

+ (aq) + 5 e – Mn 2+ (aq) + 12 H2O(l) +1.52

Au 3+ (aq) + 3 e – Au(s) +1.50

Cl2(g) + 2 e – 2 Cl – (aq) +1.360

Cr2O7 2– (aq) + 14 H3O

+ (aq) + 6 e – 2 Cr 3+ (aq) + 21 H2O(l) +1.33

O2(g) + 4 H3O + (aq) + 4 e – 6 H2O(l) +1.229

Br2(l) + 2 e – 2 Br – (aq) +1.08

NO3 – (aq) + 4 H3O + (aq) + 3 e– NO(g) + 6 H2O(l) +0.96

OCl – (aq) + H2O(l) + 2 e – Cl – (aq) + 2 OH – (aq) +0.89

Hg 2+ (aq) + 2 e – Hg(l) +0.855

Ag + (aq) + e – Ag(s) +0.80

Hg2 2+ (aq) + 2 e – 2 Hg(l) +0.789

Fe 3+ (aq) + e – Fe 2+ (aq) +0.771

I2(s) + 2 e – 2 I – (aq) +0.535

O2(g) + 2 H2O(l) + 4 e – 4 OH – (aq) +0.40

Cu 2+ (aq) + 2 e – Cu(s) +0.337

Sn 4+ (aq) + 2 e – Sn 2+ (aq) +0.15

2 H3O + (aq) + 2 e – H2(g) + 2 H2O(l) 0.00

Strongest oxidizers

Reduction Half-Reaction E (V)

2 H3O + (aq) + 2 e – H2(g) + 2 H2O(l) 0.00

Sn 2+ (aq) + 2 e – Sn(s) –0.14

Ni 2+ (aq) + 2 e – Ni(s) –0.25

V 3+ (aq) + e – V 2+ (aq) –0.255

PbSO4(s) + 2 e – Pb(s) + SO4 2– (aq) –0.356

Cd 2+ (aq) + 2 e – Cd(s) –0.40

Fe 2+ (aq) + 2 e – Fe(s) –0.44

Zn 2+ (aq) + 2 e – Zn(s) –0.763

2 H2O(l) + 2 e – H2(g) + 2 OH – (aq) –0.8277

Al 3+ (aq) + 3 e – Al(s) –1.66

Mg 2+ (aq) + 2 e – Mg(s) –2.37

Na + (aq) + e – Na(s) –2.714

K + (aq) + e – K(s) –2.925

Li + (aq) + e – Li(s) –3.045

† In volts (V) versus the standard hydrogen electrode.

Such tables may have the strongest reducing agent at the top, or at the bottom, as is the case in this table. Reactions are always written as reductions, but this is correlated to the voltages. For negative potentials, the reverse reaction is predicted.

Strongestreducingagents

ExamplesCopper/Lithium cell

From the table we get the standard reduction potentials of the half-reactions:

Cu 2+ (aq) + 2 e – Cu(s) E = +0.337 V ReductionLi + (aq) + e – Li(s) E = –3.045 V

We now combine them in such a way as to get a positive overall cell potential. This means we must reverse the lithium equation, making it the anode (where the oxidation will take place.)

Li(s) Li + (aq) + e – E = +3.045 V Anode

now we have the cell potential for the overall reaction

Cu 2+ (aq) + 2 Li(s) Cu(s) + 2 Li + (aq) Ecell = +3.382 V

Note here very carefully, that the voltage was not doubled when the coefficients are doubled. However, n for this reaction = 2. Cell voltages are therefore independent of stoichiometry. The stoichiometric information is stored in the value of n.

Some examples of using the Tables

Silver/zinc cell

From the table we get the standard reduction potential for the half-reactions:

Zn2+ (aq) + 2 e– Zn(s) E = –0.763 V ReductionAg+ (aq) + e– Ag(s) E = +0.80 V

We now combine them in such a way as to get a positive overall cell potential. This means we must reverse the zinc equation, making it the anode (where the oxidation will take place.)

Zn(s) Zn 2+ (aq) + 2 e – E = +0.763 V Oxidation

now we have the cell potential for the overall reaction

Zn (s) + 2 Ag+ (aq) Zn2+ (aq) + 2 Ag (s) Ecell = +1.563 V

Note here very carefully, that the voltage was not doubled when the coefficients are doubled. However, n for this reaction = 2. Cell voltages are therefore independent of stoichiometry. The stoichiometric information is stored in the value of n.

Nernst Equation - Cells Non-Standard Conditions

Voltaic cells eventually get depleted and stop producing a voltage.

In fact, as the reactants in the cell are consumed, the voltage produces decreases.

Temperature effects cell potential – the potential is reduced upon cooling.

Walter Nernst developed an expression, the Nernst equation, defined for the general equilibrium equation:

Frequently the equation is given with all the constants applied out at 298.15K

c d

o oactual a b

C DRT RTE =E - ln =E - lnQ

nF nFA B

aA + bB cC + dD

oactual

0.0257VE =E - lnQ

n

pH meter: the glass electrodeA “pH meter” is actually a sensitive voltmeter attached to a glass

electrode

The electrode response to changes in [H3O+] by altering its voltage as described by the Nernst equation

The key to its function as a meter is the very thin and fragile glass membrane made of special materials that allow [H3O+] but not other cations or anions to be sensed on the outer surface of the membrane

The meter is calibrated in pH units by rewriting the Nernst equation in base-10 logarithms

Filling in the constants at 298.15K this becomes:

Note that calibration is required to determine the constant

Glass membrane

0.1 M HCl

Porous membrane

Silver wire

AgCl (s)

KCl (saturated)

The Glass Electrode

+

insideglass_ electrode +

outside

H0.0592E =0- log

1 H

.glass_ electrodeEpH= -const

0.0592

Electrochemical Cells and Equilibrium Constants

We have already seen that ΔG and E are related:

We also know that ΔG and K are related:

So its stands to reason that E and K are related, and you already know the relationship:An electrochemical reaction is at equilibrium when the cell voltage is zero

So using the Nernst equation:

when E= 0 V

Gives the relationship between cell potential and equilibrium constant

Many equilibria are actually determined in this way

0 0ΔG =-nFE

oactual

RTE =E - lnQ

nF

0ΔG =-RTlnK

o RT0V =E - lnK

nF

onFlnK = E

RTNote that “nF” is on top in this equation!

Electrochemistry Example Problems

1. Calculate the Ecell of a cell formed from a Zn2+/Zn half-cell in which [Zn2+] = 0.0500 M and a Cl2/Cl– half cell in which the [Cl–] = 0.0500 M and P(Cl2) = 1.25 atm. In addition, provide a cell notation. From the Standard Reduction Potentials Table:

Cl2(g) + 2 e – 2 Cl – (aq) +1.360 V

Zn 2+ (aq) + 2 e – Zn(s) –0.763 V

Rev.#2: Zn(s) Zn 2+ (aq) + 2 e – +0.763 V

Add: Cl2(g) + Zn(s) Zn 2+ (aq) + 2 Cl –(aq) +2.123 V

- 2+( s) ( g) ( s)2Pt |Cl |Cl( 0.0500M)∥Zn ( 0.0500M)|Zn

2+ - 2

oactual

2

RT [Zn ][Cl ]E =E - ln

nF p Cl

2JKmol

actual Cmol

8.314 ×298.15K ( 0.0500) ( 0.0500)E =+2.123V - ln =+2.203V

2×96485 ( 1.25atm)Voltage higherthan standard!


2. Consider the following half-reactions:Half-Reaction E (V)Ce 4+ (aq) + e – Ce 3+ (aq) + 1.61Ag + (aq) + e – Ag(s) + 0.80Hg2

2+ (aq) + 2e – 2 Hg(l) + 0.79

Sn 2+ (aq) + 2 e – Sn(s) – 0.14Ni 2+ (aq) + 2 e – Ni(s) – 0.25Al 3+ (aq) + 3 e – Al(s) – 1.66(a) Which is the weakest oxidizing agent in the list?(b) Which is the strongest oxidizing agent?(c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Does Sn(s) reduce Ag + (aq) to Ag(s)? (f) Does Hg(l) reduce Sn 2+ (aq) to Sn(s)? (g) Name the ions that can be reduced by Sn(s). (h) What metals can be oxidized by Ag + (aq)?

More oxidizing

More reducing

Al3+

Ce4+

AlCe3+

+0.80 V + 0.25 V = +0.94 V: YES

-0.79 V + -0.14 V = -0.93 V: NO

Hg22+, Ag+, Ce4+

Hg, Sn, Ni, Al


3. Calculate equilibrium constant for the following reaction:

Zn 2+ (aq) + Ni(s) Zn(s) + Ni 2+ (aq)

Locate in tables: Zn 2+ (aq) + 2 e - (aq) Zn(s)-0.763 V Ni(s) Ni 2+ (aq) + 2 e - (aq) +0.25 V (reversed,

oxidation)

So Ecell = -0.763 + 0.25 = -0.51 V (two sig. fig. from the subtraction)

We use the equation: which becomes:

onFlnK = E

RT

onFE

RTK =e

Cmol

JKmol

2×96485×-0.51V

8.314 ×298.15K -18K =e =5.7×10

Commercial Voltaic Cells and Storage Batteries

Strictly speaking, a “battery” refers to an assembly, or battery, of 2 or more individual voltaic cells.

In this sense, a typical flashlight battery is misnamed: it is really just a single cell.

If cells are arranged in series their voltages add together.

Thus for example a 12 V car battery is an assembly of six cells producing about 2 V each.

A 9 V radio battery is also an assembly of 6 cells, each delivering about 1.5 V.

Commercial voltaic cells are divided into two main categories:

i) primary cells which are essentially the throw-away single use type, and

ii) secondary cells which are capable of multiple recharges. In charging mode, these cells function as electrolysis cells. Nowadays there are many different cell technologies in the marketplace. Let us consider just a few examples of both categories of cellsWe will first look at several primary cells, then after consider a few secondary cells

Some typical Primary and Secondary Cells

Cell?

Primary Batteryand cell (throw-away)Battery

Battery

Battery

Battery

Battery

CellsCells

CellsCell

Secondary Batteryand cell (recharge-able)

Zinc-carbon cell (Leclenché cell)(Zn/MnO2 in NH4Cl)

CATEGORY: Primary (Throwaway) Zinc Family

CONSTRUCTION:The Leclanché dry cell or ordinary "flashlight battery." The anode is a zinc container covered with an insulating wrapper, the cathode is the carbon rod at the center. The electrolyte paste contains ZnCl2, NH4Cl, MnO2, starch, graphite, and water. A fresh dry cell delivers about 1.5 V.

REDOX REACTIONSPositive terminal:2MnO2(s) + 2NH4

+(aq) + 2e– Mn2O3(s) + 2NH3(aq) + H2O (l) 0.74 V

Negative terminal:Zn (s) ––––> Zn2+ (aq) + 2e– 0.76

V

COMMENTS

The original "flashlight battery" was developed in 1865 by Georges Leclanché. It is often called a dry cell because the electrolyte is incorporated into a non-flowing paste.

Nominal cell voltage = +1.5 V

Alkaline (Zinc/MnO2 in KOH)CATEGORY: Primary (Throwaway) Zinc Family

CONSTRUCTION:The alkaline dry cell. The anode is a paste of zinc,KOH, and water, which donates electrons to the cell base via a brass collector. The cathode is a paste of MnO2, graphite, and water, which takes electrons from the inner steel case. A plastic sleeve separates the inner steel case from the outer steel jacket.

REDOX REACTIONSPositive terminal:2MnO2(s) + H2O(l) + 2e– Mn2O3(s) + 2OH-(aq) 0.15 V

Negative terminal:Zn (s) + 2OH-(aq) Zn2+ (aq) + H2O(l) + 2e– 1.25 V

COMMENTS

The alkaline dry cell is more expensive than the Leclanché cell, but it is also more efficient. Again, zinc is the anode and manganese dioxide the oxidizing agent. The electrolyte is 40% KOH saturated with zinc oxide (ZnO).


Outer steel jacket

Plastic sleeve

Inner steel jacket

(-)

(+)

Cathode(+): paste containingMnO2, graphite, and water

Anode(-): Paste containingpowdered zinc, KOH, and water

Brass collector

Cell base

Alkaline Button Cell (Zinc/MnO2 in KOH)


CONSTRUCTION:The alkaline dry cell. The anode is a paste of zinc,KOH, and water, which donates electrons to the cell base via a brass collector. The cathode is a paste of MnO2, graphite, and water, which takes electrons from the inner steel case. A plastic sleeve separates the inner steel case from the outer steel jacket.

REDOX REACTIONSPositive terminal:2MnO2(s) + H2O(l) + 2e– Mn2O3(s) + 2OH-(aq) 0.15

VNegative terminal:Zn (s) + 2OH-(aq) Zn2+ (aq) + H2O(l) + 2e– 1.25

V

COMMENTS

Apart from the different type of container, the chemistry of this cell is the same as that of the standard alkali dry cell.


Lithium (Li/MnO2 in KOH)CATEGORY: Primary (Throwaway) Lithium Family

CONSTRUCTION:Very similar to alkaline cell in design, except much more MnO2 paste is used compared to Li due to the light weight of Lithium metal.

REDOX REACTIONSPositive terminal:2MnO2(s) + H2O(l) + 2e– Mn2O3(s) + 2OH-(aq) 0.15 VNegative terminal:Li (s) Li+ (aq) + – 3.04 V

COMMENTSThe electrolyte consists of between 20 and 40% by mass of KOH, which is impregnated on the absorbent material between the two half-cells. This is only one of a whole family of Li based batters. Others are:Li/SO2, Li/SOCl2, Li/CuO, Li-poly(vinyl pyridine)/I2 solid electrolyte.Thus the lithium batteries can replace the zinc family of batteries with a whole new range of high-power, low mass cells. The next 10-20 years should see substantial development in this area.

Nominal cell voltage = +3.0 VPositive terminalMnO2, graphiteKoH, and water

Absorbent fabricsat'd in KOH andmoistened

Wire mesh cont.lithium metal

(+)

(-)

Outer casing

Inner steel jacket

Mercury (Zn/HgO in KOH)CATEGORY: Primary (Throwaway) Zinc Family

CONSTRUCTION:A mercury battery. A zinc-mercury amalgam is the anode; the cathode is a paste of HgO, graphite, and water, Mercury batteries, some of which are smaller than a pencil eraser, deliver about 1.34 V.

REDOX REACTIONSPositive terminal:HgO (s) + H2O(l) + 2e– Hg(l) + 2OH-(aq) 0.09 V

Negative terminal:Zn (s) + 2OH-(aq) ZnO(s) + H2O(l) + 2e– 1.25 V

COMMENTS

The mercury cell, developed in 1942, is another zinc dry cell devised for use in small appliances such as watches, and us usually produced as a button battery. It delivers only 1.34 V, which is significantly less than the 1.5 V of a standard dry-cell. It has the advantage of maintaining a fairly constant voltage during its lifetime.


Silver oxide (Zn/Ag2O in KOH)


CONSTRUCTION:A silver oxide button battery, similar to the mercury and alkaline manganese cell. Also a reliable source of voltage. Used in medical devices, calculators, older cameras.

REDOX REACTIONSPositive terminal:Ag2O (s) + H2O(l) + 2e– 2 Ag(s) + 2OH-(aq) 0.34 V


COMMENTS

Because no solution species is involved in the net reaction, the quantity of electrolyte is very small, and the electrodes can be maintained very close together. The storage capacity of a silver-zinc cell is about six times as great as a lead-acid cell of the same size. They are expensive because of the use of silver oxide.


Zinc-Air (Zn/O2 in KOH)


CONSTRUCTION:Constructions very similar tomercury or silver oxide buttoncells. But there are holes in the base to allow air in!

REDOX REACTIONSPositive terminal:½ O2 (s) + H2O(l) + 2e– 2OH-(aq) 0.40 V


COMMENTS

Mercury is added to the zinc, forming a liquid mixture called an amalgam, which prevents the formation of an insulating layer of zinc oxide. The zinc-air cell is sold with a patch covering the air holes. This patch must be removed before the battery develops a voltage. When oxygen gas from the air percolates into the hydroxide solution in the positive half-cell, reaction commences and a voltage develops in the cell. Can you see why the voltage is less than predicted by E?


p(O2) ≈ 0.20 atm

Lead Acid (PbO2/Pb)CATEGORY: Secondary/Rechargeable

CONSTRUCTION:Shown in the picture is a 6-V (i.e. motorcycle) battery, which contains 3 individual cells in series. The anode (negative terminal) consists of a lead grid filled with spongy lead, and the cathode (positive terminal) is a lead grid filled with lead dioxide. The cellalso contains 38% (by mass) sulfuric acid.

REDOX REACTIONSPositive terminal:PbO2(s) + HSO4

-(aq) + 3H+ + 2e– PbSO4(s) + 2H2O (l) 1.69 V

Negative terminal:Pb (s) + HSO4

-(aq) PbSO4(s) + H+ + 2e– 0.34 V

COMMENTS

The cell equations shows that discharging depletes the sulfuric acid electrolyte and deposits solid lead sulfate on both electrodes. The electrolyte density can be measured by a hydrometer; it is charged when d = 1.28 (37% H2SO4) and discharged when it is below 1.15 g/mL (21% H2SO4). The lead acid battery is extremely reliable, but heavy. Electrical cars using such batteries are too inefficient.


How many cells are there ina 12 V automotive battery?How can you tell?

Nickel-Cadmium (Nicad)

CATEGORY: Secondary/Rechargeable CONSTRUCTION:

The Nicad cell consists of many layers of alternating Ni and Cd electrodes. The actual electrodes consist of steel foil with lots of holes in them, onto which the active components are smeared. Each electrode is surrounded by an absorbent pad which contains the electrolyte. The nickel hydroxide electrodes are slightly longer at the top, and are soldered to a collector along the top of the cell; the cadmium electrodes are similarly joined along the bottom.

REDOX REACTIONSPositive terminal:NiO(OH)(s) + H2O(l) + e– NiO2(s) + OH-(aq) 0.82 V

Negative terminal:Cd (s) + 2 OH-(aq Cd(OH)2 (s) 0.49 V

COMMENTS

The nickel-cadium cell, for example, is used in hand calculators, electronic flash units, and other small battery-operated devices. The nickel-cadmium cell has electrodes of cadmium and NiO(OH), and the electrolyte is a 20% aqueous KOH solution.


Concentric rings of mesh covered alternativelywith NiO(OH) and Cd

Negative terminal collector

Positive terminalcollector

Electrolysis ReactionsSo far, voltaic cells, which are product favoured reactions doing useful work

Many electrochemical reactions are truly reversible, and can be driven against the spontaneous direction by applying a greater voltage of opposite polarity

Driven electrochemical reactions are called electrolytic reactions, and the process is know as electrolysis.

Important industrial electrolytic reactions:

AluminumMagnesiumSodiumMany of the rarer metalsElectro-refining of copper depends

in part on electrolysisElemental chlorineSodium hydroxideElectroplating technologies,

e.g. chrome plating The aluminum refining industry is ofgreat significance in Canada

Electro-Refining of Copper

Electrolytic Production of Magnesium

Counting ElectronsIt is both desirable and straightforward to relate the electrical current in an electrochemical process

to the quantity of materials produced and/or consumed during an electrochemical reaction.

This is best understood as a branch of stoichiometry in which the number of moles of electrons is used as a stoichiometric factor.

The basis of this is the Faraday constant – which we have already dealt with – and the equation relating current to charge from basic physics.

This equation is: in which I is the current in Amperes, q is the charge in Coulombs and t is the time in seconds.

Note that these precise units must be used.

Rearrangment allows the calculation of the number of Coulombs of electrical charge that is consumed in any given process.

Then we remember that Faraday’s constant is 96 485 C/mol of electrons

Once we have the number of moles of electrons, and knowing the value of n in the balanced redox reaction, the whole problem boils down to simple stoichiometry.

qI=

t

Counting Electron Example Problems

1. If you electrolyze a solution of Ni 2+ (aq) to form Ni(s), and use a current of 0.15 amp for 10 min, how many grams of Ni(s) is produced?

Ni2+ (aq) + 2 e- Ni (s) n = 2

qI= ; ⇒ q=I×t

t

sminq=0.15amp×10min×60 =90C

- -4

Cmol

90C#mole = =9.327×10 mol

96485

- -4 -41#molNi= #mole =0.5×9.327×10 mol=4.66×10 mol

2

-4 gmol#gNi=4.66×10 mol×58.69 =0.027g

Counting Electron Example Problems

2. An electrolysis cell for aluminum production operates at 5.0 V and a current of 1.0 10 5 amp. Calculate the number of kilowatt-hours of energy required to produce 1 tonne (1.0 10 3 kg) of aluminum.1 tonne = 1.0 x 106 g

Al3+ (aq) + 3 e- Al (s) n = 3

- 5 10Cmol#C e =1.11×10 mol×96485 =1.07×10 C

64

gmol

1.0×10 mol#molAl= =3.706×10 mol

26.9815

104

6 JkWh

1.07×10 C×5.0V#kWh= =1.5×10 kWh

3.60×10

- 4 5#mole =3×#molAl=3×3.706×10 mol=1.11×10 mol

6J =C×V; W=J ×s; 1kWh=3.60×10 J

Assigning oxidation states, redox reactions, etc.

Using half-reactions to balance redox reactions

Voltaic cells vs. electrolytic cells

Cell potential (and how it relates to Gibbs free energy)

Components of a voltaic cell

Electrodes (cathode and anode)

Salt bridge

Cell notation

Cells under nonstandard conditions (the Nernst equation)

Relationship between E˚ and equilibrium constant

Practical considerations for batteries, pH meters, etc.

Applications of electrolytic cells

Determining efficiency of electrolytic cells

Important Concepts In Electrochemistry

electrochemistry. reduction-oxidation (redox) reactions in chemical reaction bonds, both covalent...

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