CM 1502 1

Electrochemistry

1. Introduction 2. Oxidation Numbers 3. Balancing Redox Equations 4. Voltaic/Galvanic Cells 5. Cell Diagrams 6. Cell Potentials 7. Free Energy and Nernst Equations 8. Latimer Diagrams 9. Electrolytic Cells

CM 1502 2

Electrochemistry In electrochemical reactions, electrons are transferred from

one species to another.

Electron transfer reactions oxidation-reduction (redox) are reactions.

The electron transfer results in the generation of an electric current (electricity).

The electron transfer can be caused by imposing an electric current.

3

Definitions

Oxidation: - Loss of electrons - Increase in oxidation number - Gain of oxygen - Loss of hydrogen

Oxidising Agent: - Species is reduced

Reduction: - Gain of electrons - Decrease in oxidation number - Loss of oxygen - Gain of hydrogen

Reducing Agent: - Species is oxidised

CM 1502 4

You cannot have one… without the other! Reduction (gaining electrons) can’t happen without an oxidation to provide the electrons. You can’t have 2 oxidations or 2 reductions in the same equation. Reduction has to occur at the cost of oxidation. Hence, redox reactions!

LEO the lion says GER!

OIL RIG

GER!

CM 1502 5

Oxidation Numbers (O.N.) In order to keep track of

what loses electrons and what gains them, assign oxidation numbers.

Zn is oxidised as it loses two electrons to go from neutral Zn (O.N. = 0) metal to the Zn2+ (O.N. = 2) ion.

Each of the H+ is reduced as it gains an electron each to go from H+ (O.N. = 1) ions to combine to form H2 (O.N. = 0) gas.

CM 1502 6

Assigning Oxidation Numbers

1. Elements in their elemental form have an oxidation number of 0. e.g. Hg (O.N. = 0)

2. The oxidation number of a monoatomic ion is the same as its charge. e.g. Cu2+ (O.N. = 2)

CM 1502 7

Assigning Oxidation Numbers 3. Non-metals tend to have negative oxidation numbers, although some

are positive in certain compounds or ions. Oxygen has an oxidation number of −2, except in the peroxide ion (e.g. in H2O2) in which it has an oxidation number of

−1.

Hydrogen has an oxidation number of +1, except when bonded to a metal ( hydride ion, H-) in which it has an oxidation number of −1.

Fluorine always has an oxidation number of −1.

The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. e.g. HClO4 (Cl O.N. = 7)

CM 1502 8

Assigning Oxidation Numbers

4. The sum of the oxidation numbers in a neutral compound is 0.

5. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

*

1 + x + 4(-2) = 0 x = +7 O.N. of Mn is +7

2x + 2(-1) = 0 x = +1 O.N. of H is +1

O.N of Mn in CsMnO4 O.N of H in H2O2

O.N of l in lO65-

x + 6(-2) = -5 x = +7 O.N. of I is +7

CM 1502 9

Balancing Oxidation-Reduction Equations

Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.

This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.

1 oxidation half reaction + 1 reduction half reaction => Overall reaction

CM 1502 10

Half-Reaction Method This must be done in sequential order.

1. Assign oxidation numbers to determine what is oxidised

and what is reduced.

2. Write the oxidation and reduction half-reactions.

3. Balance each half-reaction. a. Balance elements other than H and O. b. Balance O by adding H2O. c. Balance H by adding H+. d. Balance charge by adding electrons.

CM 1502 11

Half-Reaction Method 4. Multiply the half-reactions by integers so that the

electrons lost and gained are the same for oxidation and reduction, respectively.

5. Add the half-reactions, subtracting things that appear on both sides.

6. Check that the equation is balanced on both sides according to elements present.

7. Check that the equation is balanced on both sides according to charge.

CM 1502 12

Redox Reaction

Consider the reaction between MnO4− (permanganate) and

C2O42− (oxalate):

MnO4

−(aq) + C2O42−(aq) → Mn2+(aq) + CO2(aq)

NOT BALANCED

Reaction complete

Excess permanganate

CM 1502 13

Half-Reaction Method First, assign oxidation numbers.

MnO4− + C2O4

2- → Mn2+ + CO2

+7 +3 +4 +2

Manganese is reduced.

Carbon is oxidised.

x + 4(-2) = -1 x = +7

2x + 4(-2) = -2 x = +3

x + 2(-2) = 0 x = +4

CM 1502 14

Oxidation Half-Reaction C2O4

2− → CO2 To balance the carbon, add a coefficient of 2:

C2O42− → 2CO2

The oxygen is now balanced as well. To balance the

charge, add 2 electrons to the right side.

C2O42− → 2CO2 + 2e−

CM 1502 15

Reduction Half-Reaction MnO4

− → Mn2+

The manganese is balanced; to balance the oxygen, add 4 waters to the right side.

MnO4

− → Mn2+ + 4H2O

To balance the hydrogen, add 8 H+ to the left side.

8H+ + MnO4− → Mn2+ + 4H2O

To balance the charge, add 5 e− to the left side.

5e− + 8H+ + MnO4

− → Mn2+ + 4H2O

CM 1502 16

Combining the Half-Reactions Now combine the two half-reactions together: C2O4

2− → 2CO2 + 2e− 5e− + 8H+ + MnO4

− → Mn2+ + 4H2O To attain the same number of electrons on each side, multiply the first reaction by 5 and the second by 2.

5C2O42− → 10CO2 + 10e−

10e− + 16H+ + 2MnO4− → 2Mn2+ + 8H2O

Add these together, subtracting things that appear on both sides and get:

16H+ + 2MnO4− + 5C2O4

2− → 2Mn2+ + 8H2O + 10CO2

X 5 X 2

17

1. In a redox process, H2SO4 is converted to H2SO3 and Fe is converted to Fe2+. Write the equation for the redox reaction. 2. Balance the following reaction: Cr2O7

2-(aq) + I-(aq) → Cr3+(aq) + I2(aq)

Fe + H2SO4 + 2H+ → Fe2+ + H2SO3 + H2O

14H+ + Cr2O72- + 6I- → 2Cr3+ + 3I2 + 7H2O

CM 1502 18

Balancing in Basic Solution

If a reaction occurs in basic solution, one can balance it as if it occurred in acid.

Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.

If this produces water on both sides, subtract water from each side.

CM 1502 19

+ 14OH-(aq) + 14OH-(aq)

Cr2O72- + 6I- → 2Cr3+ + 3I2 + 7H2O + 14OH- 14H2O +

Reconcile the number of water molecules.

+ 14OH- Cr2O72- + 6I- → 2Cr3+ + 3I2 7H2O +

*

7

Cr2O72-(aq) + 6I-(aq) → 2Cr3+(aq) + 3I2(s) + 7H2O(l) 14H+(aq) +

20

Energy is absorbed to drive a nonspontaneous redox reaction

Figure 21.3 General characteristics of voltaic and electrolytic cells.

VOLTAIC CELL ELECTROLYTIC CELL Energy is released from

spontaneous redox reaction

Reduction half-reaction at electrode Y++ e- Y

Oxidation half-reaction X X+ + e- from electrode

Reduction half-reaction in solution B++ e- B

Oxidation half-reaction A- A + e- in solution

Overall (cell) reaction X + Y+ X+ + Y; ∆G < 0

Overall (cell) reaction A- + B+ A + B; ∆G > 0

Inert electrodes

CM 1502 21

Electrolytic vs. Voltaic Cells

Voltaic cell: electric current produced Electrolytic cell: electric current used to cause chemical change

Cells Ecell Electrode name

Process at electrode

Sign of electrode

Electrodes

Voltaic/ Galvanic

>0 Spontaneous

Anode Cathode

Ox. Red.

- +

Redox process involves electrodes

Electrolytic <0 Non-spontaneous

Anode Cathode

Ox. Red.

+ -

Electrodes are (usually) inert

CM 1502 22

In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

The energy can be used to do electrical work when the electrons are channeled through an external device => voltaic cell.

Voltaic Cells

CM 1502 23

Voltaic Cells Separate the species (i.e. Zn and Cu) into two compartments. One for oxidation other for reduction.

Connect electrodes by a wire/voltmeter.

Connect solutions by a salt bridge.

Ox. Red.

‘-’ve

‘+’ve

24

Salt Bridge

Once even just one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop. (right half cell will be ‘-’ve, preventing further electron flow)

Therefore, a salt bridge is used, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. Left half cell: net ‘+’ ve charge in solution so anions move toward the

anode.

Right half cell: net ‘-’ ve charge in solution so cations move toward the cathode.

25

Cell Diagrams

Cell diagram: Zn(s) I Zn2+(aq) II Cu2+

(aq) I Cu(s) Ox. first then red.

If both ox. and red. species are ions, then use Pt (or graphite) as the electrode. e.g Zn(s) I Zn2+(aq) II Fe3+(aq), Fe2+ (aq) I Pt(s) Cannot dip wire into solution, need an electrode.

Phase boundary Salt bridge

26

Write the balanced equation and the cell diagram for a voltaic cell that consists of one half-cell with a Cr rod in a Cr(NO3)3 solution, another half-cell with an Ag rod in an AgNO3 solution and a KNO3 salt bridge. Measurement indicates that the Cr is the negative electrode and Ag is the positive electrode.

Oxidation half-reaction Cr(s) Cr3+(aq) + 3e-

Reduction half-reaction 3Ag+(aq) + 3e- 3Ag(s)

Overall (cell) reaction Cr(s) + 3Ag+(aq) Cr3+(aq) + 3Ag(s)

Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)

CM 1502 27

Electromotive Force (emf)

Water only spontaneously flows one way in a waterfall.

Likewise, electrons only spontaneously flow one way in a redox reaction.

Which species gives (anode) and which takes (cathode)?

28

Standard Reduction Potentials (Eºred)

Standard: g = 1 atm aq = 1 molL-1

Higher red. potential, more easily reduced, half-cell will be the cathode.

Strong oxidising agents.

Lower red. potential, more easily oxidised, half-cell will be the anode.

Strong reducing agents.

CM 1502 29

Standard Hydrogen Electrode: the other half-cell

Their values are referenced to a standard hydrogen electrode (SHE).

By definition, the standard reduction potential (Eºred) for the hydrogen ion is 0 V:

2H+ (aq, 1 M) + 2e− → H2 (g, 1 atm)

aq = 1 molL-1

i.e. pH = 0

30

V0Eored =

by convention, Eºred of SHE = 0

Standard Hydrogen Electrode

Cu2+ is more readily reduced cf. H+

Eºred (Cu2+) > 0

H+ is more readily reduced cf. Zn2+

Eºred (Zn2+) < 0

31

Standard Cell Potentials Referenced against the SHE, all species with Eºred > 0 were

reduced and all species with Eºred < 0 were oxidised.

After a half-cell is referenced against the SHE, its Eºred can be compared against other half-cells. For a given voltaic cell, the cell potential at standard conditions

can be written as this equation:

Ecell ° = Ered (cathode) − Ered (anode) ° °

Standard reduction potentials

Larger Eºred, Reduction process

Smaller Eºred, Oxidation process

32

Cell Potentials

Ecell ° = Ered ° (cathode) − Ered ° (anode)

= +0.34 V − (−0.76 V) = +1.10 V

For the oxidation (anode) in this cell,

For the reduction (cathode),

Ered = +0.34 V °

Ered = −0.76 V °

CM 1502 33

Consider the following two electrode reactions and their standard electrode potentials:

Al3+(aq) + 3e- → Al(s) E0 = -1.66 V

Cd2+(aq) + 2e- → Cd(s) E0 = -0.40 V

Write the cell reaction for a voltaic cell based on these two

electrodes, and calculate the standard cell potential, E0cell.

2Al(s) + 3Cd2+(aq) → 2Al3+

(aq) + 3Cd(s)

E0cell = -0.40 – (-1.66) = 1.26 V

34

Free Energy ∆Go for a redox reaction under standard conditions, can be

written as the equation: ∆Go = total charge x Eo

cell ∆Go = −nF x Eo

∆Go = −nFEo

where: n is the number of moles of electrons transferred per mole

of reaction F is the Faraday constant (1 F = 96,485 Cmol-1) Eo is in V = J/C

For redox reaction to be spontaneous, ∆G° < 0, so E° > 0.

CM 1502

35

Nernst Equation

By dividing both sides of −nFE = −nFE° + RT ln Q by −nF, we obtain the Nernst equation:

Ecell = E°cell − Ecell = E°cell −

RT nF ln Q 0.059

n log Q

At 298 K,

In log x 2.303

Recall that ∆G = ∆G° + RT ln Q then −nFE = −nFE° + RT ln Q

CM 1502

CM 1502 36

A voltaic cell consists of Mn/Mn2+ and Cd/Cd2+ half-cells with concentrations [Mn2+] = 0.75 M and [Cd2+] = 0.15 M. Use the Nernst equation to calculate the cell potential, Ecell, at 25 oC. (F = 96485 Cmol-1)

Data: Cd2+

(aq) + 2e- → Cd(s) Eo = -0.40 V Mn2+

(aq) + 2e- → Mn(s) Eo = -1.18 V From E°red, Cd2+ undergoes reduction and Mn2+ oxidation.

Overall: Cd2+ + Mn → Cd + Mn2+

E°cell = -0.40 – (-1.18) = 0.78 V Ecell = 0.78 – [(8.314 x 298)/(2 x 96485)] In (0.75/0.15) = 0.76 V

[ ][ ]reactantsproducts

nFRTFor a cell reaction, Ecell = E°cell − ln

CM 1502 37

Concentration Cells

Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.

For such a cell, E°cell would be 0, but Q would not.

Therefore, as long as the concentrations are different, Ecell will not be 0.

*

Ecell = E°cell − RT nF ln Q

Given the following cell: Fe(s) I Fe2+(aq, 2 x 10-2 M) II Fe2+(aq, 1 M) I Fe(s) and that the standard reduction potential of Fe2+(aq) to Fe(s) is -0.44 V, calculate Ecell.

Oxidation: Fe(s) → Fe2+(2 x 10-2 M) + 2e- Reduction: Fe2+(1 M) + 2e- → Fe(s)

Overall: Fe2+(1 M) → Fe2+(2 x 10-2 M)

Ecell = Eocell – RT/nF In Q

Ecell = 0 – RT/nF In (2 x 10-2 / 1)

Ecell = 0.05 V

CM 1502

CM 1502 39

Latimer diagrams summarise the standard potential (in V) between species of an element.

ClO4- ClO3

- HClO2 HClO Cl2 Cl- in acid +7 +5 +3 +1 0 -1

+1.20 +1.18 +1.65 +1.67 +1.36

From the Latimer Diagram, we can write the half reaction and the potential connecting any two adjacent species. For example,

Latimer Diagrams

ClO3- + 3H+ + 2e- HClO2 + H2O Eo

red = +1.18V

CM 1502 40

Suppose we are interested in the potential between two non-adjacent couples. What is the Eºred?

∆Gooverall = Σ∆Go

individual steps

-noverall FEooverall = -n1FEo

1 + -n3FEo3 + -n2FEo

2

ClO4- ClO3

- HClO2 HClO Cl2 Cl- in acid

+1.20 +1.18 +1.65 +1.67 +1.36

+7 +5 +3 +1 0 -1

What about non-adjacent species?

CM 1502 41

Eooverall =

-n1FEo1 + -n3FEo

3 + -n2FEo2

-noverallF

Eooverall =

-F (n1Eo1 + n3Eo

3) + n2Eo2

-noverallF

+ n3Eo3

n1 + n2 + n3 Eo

overall = n1Eo

1 + n2Eo2

ClO4- ClO3

- HClO2 HClO Cl2 Cl- in acid

+1.20 +1.18 +1.65 +1.67 +1.36

+7 +5 +3 +1 0 -1

CM 1502 42

HClO2 + 2H+ + 2e- HClO +H2O

2HClO + 2H+ + 2e- Cl2 + 2H2O

Cl2 + 2e- 2Cl-

For reaction: HClO2 Cl- + 2H2O + 3H+ + 4e-

2 + 1 + 1 Eo

red = 2 x 1.65 + 1 x 1.67 + 1 x 1.36

= 1.58 V

Note: per atom/ion! Or follow ΔO.N!

ClO4- ClO3

- HClO2 HClO Cl2 Cl- in acid

+1.20 +1.18 +1.65 +1.67 +1.36

+7 +5 +3 +1 0 -1

CM 1502 43

Find the Eºred of IO4- to HOI:

- V0.5352

V1.430 V1.154-3

V1.589-4 IIHOIIOIO → → → →

(+7) (+5) (+1) (0) (-1)

n1 + n2 Eo

red = n1Eo

1 + n2Eo2

6

2 x 1.589 + 4 x 1.154 =

= 1.299 V

CM 1502 44

Electrolysis Electrolysis is the splitting (lysis) of a

compound using electrical energy, typically to form elements from ions.

Say +1 V is applied at the anode (active) to promote oxidation, -1 V will occur at the cathode (passive).

MgBr2 and NaCl molten salt mixture electrolysis. What is the order in which the species will undergo oxidation or reduction? Species that can be reduced: Mg2+ (Eo

red = -2.37 V) and Na+ (Eored = -2.71 V) (cations)

Species that can be oxidised: Br- (Eo

red = 1.06 V) and Cl- (Eored = 1.36 V) (anions)

1. Br-

2. Cl- 3. Mg2+

4. Na+

CM 1502 45

A technician is plating a faucet with 0.86 g of Cr from an electrolytic bath containing aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is needed?

1 A = 1 Cs-1 Current = Charge per sec

divide by M

96485 Cmol-1 e-

3 mol e-/mol Cr

divide by time

mass of Cr needed

mol of Cr needed

mol of e- transferred

current (A)

charge (C)

Cr3+(aq) + 3e- Cr(s)

0.86g x 3 mol e-

52.00 gmol-1 = 0.050 mol e-

0.050 mol e- (96485 Cmol-1 e-) = 4787 C

4787 C

12.5 min x 60 s = 6.4 A

Electroplating

i = Q/t