Electrochemistry

1. Balancing Redox Reactions

2. Using Half Cell Potentials and Latimer Diagrams 100 measured half cell potentials generate 10,000 full reactions

3. E as a Thermodynamic state function ∆G = -n E F

4. Applying the Nernst Equation - non standard conditions

5. Electrochemical Cells Voltaic and Electrolytic

6.Batteries

7. Applications to Analytical and Industrial Chemistry

Oxidation State Assignments rules on p-82

Assign O as -2 except for peroxides and O2

Assign H as +1 except for hydrides and H2

Assign halides -1 except in X2 and oxyhalide species

HClO, HClO2 ClO4- have Cl in +1, +3 ,and +7 ox states

Sum of ox numbers adds up to net charge

In Lewis structures there are 3 ways to divide up the bonding

electrons.

For octet - count them twice- once for each atom

For formal charge - divide them equally

For OX number - award both bonding electrons to more electronegative element.

LATIMER DIAGRAMS SHOW REDOX CHEMISTRY OF EACH ELEMENT

+6 +4 +2 0 -2

+5 +4 +3 +2 +1 0 -1 -2 -3

+7 +5 + 3 +1 0 -1

0 0 -1/2 -1 -2

HO2 H-O-O. [:S-SO3]2- [:N=N=N:] -

+1 -1 0 -1 +5 or +2ea -1 +1 -1 -1/3ea

Balancing Redox Reactions (review) *Assign oxidation states and write skeletal OX and RED half

reactions Ox leo RED ger

Multiply to eliminate electrons on adding OX and RED

Balance Charge with H+ in acid or OH- in base

Balance H or O with H2O , check with O or H.

* If only full rxn is needed it is easier to combine ox and red first

Work examples from Chapter 5. `47-58

* Make sure skeletal atoms are balanced

red 6 e- + Cr2O72- → 2 Cr3+ X 1 not fully balanced

ox Fe2+ → Fe3+ + 1 e- X 6

6 Fe2+ + Cr2O72- → 2 Cr3+ + 6 Fe3+

14 H+ + 6 Fe2+ + Cr2O72- → 2 Cr3+ + 6 Fe3+ + 7 H2O

Zn + NO3-→ NH3 + Zn(OH)4

2-

ox 4 OH- + Zn → Zn(OH)42- + 2e- x 4

red 6 H2O + NO3- + 8 e- → NH3 + 9 OH-

7 OH- + 6 H2O + NO3- + 4 Zn → NH3 + 4 Zn(OH)4

2-

Disproportionation. Cl2 is both oxidized and reduced

Cl2 → OCl- + Cl-

ox 2 OH- + ½ Cl2 → OCl- + 1e- + H2O

red ½ Cl2 + 1 e- → Cl-

2 OH- + Cl2 → OCl- + Cl- + H2O

Activity series Na,K Mg Al Zn Fe Sn Pb H2 Cu Ag Au

Zn + Cu2+ → Cu + Zn2+ Cu + 2 Ag+ → Cu2+ + 2 Ag

Eo cell = 0.34 + 0.76= 1.10 V E0cell = -0.34 + 0.80 = + 0.46 V

anode is Zn, reverse sign of -0.76V anode is Cu, reverse sign of +0.34

SHE H+ (1 M) / H2 (1 atm) Pt E1/2 = 0.00 V

H2 + Cu2+ → 2 H+ + Cu Zn + 2H+ → H2 + Zn2+

Eo½ redn of Cu2+ = +0.34 V Eo

½ ox of Zn = + 0.763 V

Eo1/2 red of Zn2+ = - 0.763 V

A CAUTION ON SIGNS and COMBINING HALF RXNS.

TEXT Method Ecell = Ecathode - Eanode

or Erxn = E1/2 redn - E1/2 redn potential for the oxidn half rxn

how confusing is that?

Recommended Method

Write both half reactions, reverse one of them and change

the sign of E1/2 . Then multiply as needed and add making sure the electrons drop out.

red 6 e- + Cr2O72- → 2 Cr3+ X 1 Eo

1/2 = +1.33 V

ox Fe2+ → Fe3+ + 1 e- X 6 Eo1/2 = -0.77 V

6 Fe2+ + Cr2O72- → 2 Cr3+ + 6 Fe3+ +0.56 V

14 H+ + 6 Fe2+ + Cr2O72- → 2 Cr3+ + 6 Fe3+ + 7 H2O

*

Eo ‘s are the potential at standard conditions

GASES 1 atm

solutes 1 M

solids or liquids unit activity

In ACID - the species which exists at 1 M acid is shown

HClO4, HClO3 are strong acids and do not exist in 1 M H+

HClO2 and HClO are present as the acids

IN BASE all are present as the conjugate bases.

pH dependence of Eo : species differing by H or O will require H+ or OH- to balance the rxn and are pH dependent

ACID

BASE

Redn of ClO4-, ClO3

- ClO2, and Cl2 are pH independent

in base H2O + ClO3- + 2e- → ClO2

- + 2 OH- is balanced

∆G = - n E F

G is an extensive quantity kJ/mol E is an intensive quantity Volts

( Note T vs. heat Q : 1 ml of 50o water vs. 1000 ml of 50o water)

forward E>0, G< 0 ; at equil E = 0, G = 0; reverse E< 0, G>0

Nernst Equation

∆G = ∆Go + RT ln Q

-n E F = -n Eo F + RT ln Q

E = Eo - RT/nF ln Q

at T = 298 and using log10 2.303 RT/F = 0.0592

E = Eo - 0.0592/n log Q

The naught has exactly the same meaning as it did for ∆G

E is the potential at the conditions of concentration given by the reaction quotient Q. Eo is the potential for all species in their standard states

3 criteria for thermodynamic equilibrium.

K = Q E = 0 ∆G = 0

If K > Q E >0 or G <0 reaction goes forward

If K < Q E < 0 or G > 0 reaction goes in reverse

Which you use is your choice.

When a system is at equilibrium E = 0 = E0 - 0.0592/n log Keq

Eo = 0.0592/n log Keq in volts and ∆Go = -RT ln K in KJ/mol

at T =298 2.303 x 8.314 J K-1mol-1 x 298 = 5.7 kJ/mol

Keq = 10+n∆Eo/0.0592 = 10-∆Go/5.7

SKIP STEP POTENTIAL

Skip step Eo1/2 . Note that when combining half cell potentials to

generate another half cell, they are not additive. The correct net half cell potential is the weighted average of the components.

Thus for Cr 3+ + 3e- � Cro (s)

Eo1/2 = {Eo

1/2 ( 1 e- + Cr3+ � Cr2+ ) + 2 Eo1/2 ( 2e- + Cr2+� Cro) }/3

Eo1/2 = { -0.41 V + 2 (-0.91V)}/3 = - 0.74 V

Think of E as the driving force per electron.

We can prove this is the case by combing ∆G’s which are additive.

Eo1/2 ∆G0

1/2 = -n EoF

1e- + Cr3+ � Cr2+ -0.41 V + 0.41 F

2e- + Cr2+ � Cro -0.91 V + 2(0.91)F

3e- + Cr 3+ � Cro (s) -0.74 V + 2.23 F

Eo1/2 = ∆G0/nF = -2.23 F /3F = -0.74 V

E1/2 is the weighted average of the E1/2‘s weighted by the number of electrons

HF or F- ?? PICK YOUR REFERENCE STATE

1) ½ F2 (g) + 1 e- → F- Eoacid = Eo

base = +2.87 V

2) ½ F2 (g) + 1 e- + H+ → HF Eoacid = +3.06 V

3) HF � H+ + F- pKa = 3.14

the third can be computed from the other two,

at 1 atm F2 gas E1 = +2.87 -0.0592 log [F-]

E2 = +3.06 -0.0592 log [HF]/[H+]

but E1 = E2 so 0.19 = -0.0592 log [F-][H+]/[HF]

0.19 / 0.0592 = pKa = 3.2

eq 1) is pH independent, eq 2) depends on pH

If we want the E at pH = pka with 1 M total HF + F- we can use either. E1 = +2.87 -0.0592 log (1/2) = + 2.89

E2 = +3.06 -0.0592 log ( 1/2/ 7.2 e-4) = + 2.89

Adding Half RXNS to give Full REACTIONS

A bonus of E being an intensive quantity is that you don’t need to consider the number of electrons in each half IF you produce a proper FULL RXNS ( which will not have e’s in it !!!)

In the oxidation of iron in air ∆Go = -nEoF

O2 + 4 H+ + 4 e- → 2 H2O Eo1/2 = + 1.23 V - 4.92 F

4Fe2+ → 4 Fe3+ + 4 e- Eo1/2 = - 0.77 V +3.08 F

4 Fe2+ + O2 + 4 H+ → 4 Fe3+ + 2H2O + 0.46 V - 1.84 F

Eo = - (-1.84)F / 4F = 0.46 V you get this by simply combining the Eo

1/2’s without regard for balancing or number of e’s.

E is the driving force per electron

-∆G is the driving force per mole of rxn

∆Gof for ions. Taking SHE as a reference, we assign ∆Go

f for H+ (aq) = 0

Then for Zn (s) + 2H+ → Zn2+ + H2 1 Coul Volt = Joule

∆Go = - n EoF = -2 (+ 0.76 V) 94,485 C/mol = 146 kJ/mol

= Gof H2 + Go

f Zn2+ - 2 Gof H+ - Go

f Zn (s)

0 + Gof Zn2+ - 0 - 0 = -146 kJ/mol ( see App D)

From ∆Gof in Appendix D obtain Eo

1/2 for Fe2+ + 2 e- → Fe0

∆Go = ∆Gof (Feo) - ∆Go

f (Fe2+) = 0 - (-78.90) = +78.90 kJ/mol

Eo = -∆Go/nF = - 78.90 x 103 J/mol / (96,485 C/mol x 2) = -0.41 V

a Joule = Coulomb Volt

or explicitly include reference electrode to give same result.

Fe2+ + H2 → Fe0 + 2 H+ ∆Go = 0 + 0 - 0 - ∆Gof (Fe2+)

1) species oxidizes water to O2 Eredn > +1.23 V

x4 Co3+ + 1 e- → Co2+ Eo1/2 = 1.82 V

2 H2O → 4 H+ + 4 e- + O2 Eo1/2 = -1.23 V

4 Co3+ + 2 H2O → 4 H+ + 4 Co2+ + O2 + 0.59 V

2) species reduces water to H2 Eox > 0

2 H+ + 2 e- → H2 Eo1/2 = 0.00 V

Mn → Mn2+ + 2 e- Eo1/2 = + 1.20 V

+ 1.20 V

3) species is oxidized by O2 Eox > - 1.23 V

O2 + 4 H+ + 4 e- → 2 H2O Eo1/2 = + 1.23 V

Fe2+ → Fe3+ + 1 e- Eo1/2 = - 0.77 V

+ 0.46 V

reactions involving gases may have overpotential ( kinetics slow)

4. disproportionation

MnO42- + 2 e- → MnO2 (s) E

o1/2 = - 0.56 V

x 2 MnO42- → MnO4

- + 1 e- Eo1/2 = + 2.26 V

4 H+ + 3 MnO42- → 2 MnO4

- + MnO2 (s) + 2 H2O + 1.7 V

Cu+ + 1 e- → Cuo Eo1/2 = + 0.521 V

Cu+ → Cu2+ + 1 e- Eo1/2 = - 0.153 V

2 Cu+ → Cu2+ + Cuo + 0.378 V

EX 1. How much Cu+ remains at equilibrium if 0.1 mol of CuCl is dissolved in a litre of water ?

2 Cu+ → Cu2+ + Cuo

since E >>0 assume reaction goes “all” the way to right.

this would give 0.05 M Cu2+ and 0.05 mol of Cuo metal plated out

E= 0 = 0.378 -0.0592 log ([Cu2+]/[Cu+]2

6.385 = log (0.05) - 2 log [Cu+] [Cu+] = 1.4 x 10-4 M

EX 2. Cl2 water. When Cl2 gas is bubbled thru water at 1 atm it

absorbs 0.09 moles of gas per litre. What is present in chlorine water ? What is the solubility of Cl2 in water?

disproportionation

Cl2 + H2O → HOCl + Cl- + H+ ∆Eo = -1.63 +1.36 = - 0.27 V

1 atm x x x

we are not at standard conditions, we are at equilibrium, let x = …

E = 0 = -0.27 - 0.0592 log [HOCl][H+][Cl-] / PCl2

0.27 = - 0.0592 log x3 x = 0.030 M

soln contains 0.03 M H+, 0.03 M Cl- and 0.03 M HOCl

it also contains 0.09 - 0.03 = 0.06 M Cl2 (aq) = solubility of Cl2the pH = 1.52 and pKa for HOCl = 7.54,

we can neglect [OCl-] = Ka [HOCl]/[H+] = 2.8 x 10-8 M

note especially the importance of standard states and proper equilibria

The cell shown provides the Ksp of AgIE = 0.80 -0.059 log (0.1/ [Ag+] = 0.417 [Ag+] = 9.1 x 10-9 MKsp = [Ag+][I-] = x2 = 8.5 x 10-17

Eo1/2 for AgI (s) + 1 e- → Ago + I- is neither of these but requires

saturated AgI in 1 M I- where [Ag+] = 10-16 M

What if we omit

the AgI ?

concentration

cells

The standard AgCl/Ag potential is listed as +0.22 V

This refers to the half reaction AgCl (s) + 1e- → Ago + Cl-

Using the standard Ag+/Ag potential of 0.80 V and applying

the Nernst eqn for non standard conditions

E ½ = 0.80 - 0.0592 log (1 / [Ag+])

The standard AgCl/Ag potential refers to 1 M [Cl-]

from Ksp = 1.8 x 10-10 = [Ag+][Cl-] = [Ag+][1]

in saturated AgCl in 1 M Cl- [Ag+] = Ksp

E ½ = 0.80 - 0.0592 log (1/[Ksp]) = 0.22 V

pKsp = 0.58/0.0592 = 9.8 Ksp = 10-9.8 = 1.6 x 10-10

The AgCl/Ag electrode measures [Cl-]

E1/2 = 0.22 - 0.0592 log [Cl-]

Titration of Halides with Ag+/Ag electrode

Titrate 100 ml of 0.03 M Cl-, 0.02 M Br- and 0.01 M I- using 0.1 M

AgNO3. Given Ksp AgCl 10-10 , AgBr 10-13 , AgI 10-16

for simplicity we will neglect volume change, and use approx Ksp’s etc.

one drop ppts AgI : [Ag+] = Ksp/ [I-] = 10-14 E = 0.8 + 0.06 log[Ag+] = -0.04V

9.90 ml ppts 99% of I- [Ag+] = Ksp/ [10-4] = 10-12 E = + 0.08 V

at this point we have yet to exceed Ksp of AgBr or AgCl Q= [Ag+][Br-] = 0.02 x 10-12

first ppt of AgBr; [Ag+] = 10-13/0.02 = 5 x 10-12; E = 0.8 + 0.06log (5x10-12) = + 0.12 V

99% of Br- is ppt after 10 ml + 19.8 ml added Ag+; E = 0.8 + 0.06 log(5x10-10) = + 0.24 V

we have yet to exceed Ksp of AgCl Q = (5 x 10-10)(0.03) < 10-10

first ppt of AgCl appears when [Ag+] = 3.3 x 10-9 or E = 0.80 + 0.06 log (3.3 x 10-9)

E = 0.29 V and when 99% has ppt (10 + 20 + 29.7 ml) ; E = 0.8 + 0.06log(3.3 x 10-7)

E = 0.41 V and at eq pt (30 ml) Ag+ = 10-5 and E = 0.5 V

Note that a single drop of AgNO3 gives ∆E = 0.04 V when AgBr starts to ppt and 0.05 V

when AgCl begins to ppt and 0.09 V when complete.

Non-standard conditions vs. Eo predictions

Ag+ + e- → Ago(s) + 0.80 V

I- → ½ I2 + e- -0.536 V

Ag+ + I- → Ago(s) + ½ I2 (s) + 0.264 V spontaneous

BUT we know Ag+ + I- → AgI (s) 1/Ksp = 10+16

in saturated AgI solution [Ag+] = [I-] = 10-8 M

E = Eo -0.0592 log 1 / [Ag+][I-] not spontaneous

E = 0.264 - 0.0592 log 10+16 = 0.264 - 0.95 = - 0.686 V

In an electrochemical cell with Ag+ and I- in separate beakers the redox

reaction is spontaneous with e’s flowing from anode to cathode.

< I- (1M) / I2 (s) // Ag+ (1M) /Ag> Eo = + 0.264 V

Using ∆Gof / standard states

A) AgI (s) → Ago(s) + ½ I2 (s)

-66.19 0.0 0.0 ∆Go = + 66.19 kJ/mol no go !

B) Ag+ + I- → Ago(s) + ½ I2 (s)

+77.11 -51.57 0.0 0.0 ∆Go = - 25.54 kJ/mol GOES

or ∆Go = - n Eo F = -0.264 F = 25.54

C) Ag+ + I- → AgI (s)

+ 77.11 -51.57 -66.19 ∆Go = - 91.73 kJ/mol GOES

or ∆Go = - 5.7 log 1/Ksp = - 91.73

for Ksp ∆Go = -5.7 log Ksp = + 91 kJ/mol , the reverse of the above rxn

NONSTANDARD CONDITIONS AT 10-8 M each ion

for A) ∆G = +66.19 + 5.7 log 1 NO

for B ∆G = -25.54 + 5.7 log 10+16 = + 66 kJ/mol NO for C ∆G = -91.73 + 5.7 log 10+16 = 0.0 EQ

Stoichiometry in electrochemical cells

Faraday constant F = 96,485 coulombs mol-1

a Faraday is a mole of electrons , 1 amp = C s-1

1 coulomb-volt = joule

1. How much Cu will be plated out if 10 amps flow for 10 minutes in the cell <Zn\Zn2+ (1M) \\ Cu2+ (1M)\Cu > ?

mol e’s = 10 C s-1 x 600 s / 96,485 C mol-1 = 0.062 mol

we need 2 mol e’s for each Cu so mol Cu = 0.031 mol

2. If we have 100 ml of solution (1 M in each ion) obtain the final conc of the ions and the cell potential after 10 min.

Zn + Cu2+ 2e→ Zn2+ + Cu

0.1 - 0.031 0.1 + 0.031 mol can be used, ml cancels below

0.062 mol e’s will plate 0.062/2 mol Cu = 0.031 x 63.54 = 1.97 g

E = 1.10 - 0.0592/2 log ( 0.131/0.069) =1.09 V a spontaneous reaction will always produce a decrease in E over time.

E = Eo - 0.0592/2 log ([Zn2+] / [Ag+]2 )

OX Zn → Zn2+ + 2 e- + 0.76 V

RED x 2 Ag+ + 1 e- → Ag + 0.80 V

Zn + 2 Ag+ → Zn 2+ + 2 Ag + 1.56 V

10 amps x 360 s = 3600 C 3600 C/ 96,485 C/mol = 0.0373 mol e’s

For 0.0373 mol of e’s we gain 0.0185 mol Zn2+ and lose 0.0373 mol Ag+ . From salt bridge anode gets 0.0373 mol NO3

-, cathode 0.0373 K+

in millimoles [ Zn2+] = (100 + 18.5)mmol / 1000 ml = 0.1185 M

[Ag+] = (100 - 37.3)mmol / 1000 ml = 0.0627 M

E = Eo -0.0592/2 log [Zn2+ ]/[Ag+]2 = 1.55 V

the e’s will run until Ag+ is used up after 100 mmol flow = 965 seconds.

<Zn\Zn2+ (0.1 M) \\ Ag+ (0.1 M) \Ag>

Start with 1 L in each beaker, a KNO3 salt bridge. After 10 amps flow for 6 min what do we have?

Batteries derive electrical energy from chemistry spontaneous e flow - to + , forced uphill e flow + to - . anode is where OX occurs

anode Pb + H2SO4 → PbSO4 (S) + 2 e- + 2 H+ ~ 0. 54 V

cathode PbO2 (s) + H2SO4 + 2 e- + 2 H+ → PbSO4 (s) + 2 H2O ~ 1.46 V

NET Pb + PbO2 + 2 H2SO4 → 2 PbSO4 + 2 H2O 2.0 V

Ni/Cd battery < Cd\Cd(OH)2\\Ni(OH)2\ NiO2>

anode Cd (s) → Cd(OH)2 (s) + 2 e- + 0.81 V

cathode NiO2 (s) + 2e- + H2O → Ni(OH)2 (s) + 0.49 V

Cd + NiO2 → Cd(OH)2 + Ni(OH)2 + 1.30 V

Were you wondering wny the Cd half reaction has Eo = + 0.81 V ?

Cd2+ + 2e- → Cd Eo1/2 = - 0.40 V Cd(OH)2 has a Ksp = 2.5 x 10-14

in 1 M base Cd2+ = Ksp/[OH]2 = 2.5 x 10-14

Eredn = - 0.40 -0.0592/2 log {1/[Cd2+]} = - 0.81 V

Eox then is +0.81 V in base.

Dry cell battery and Alkaline battery (non-rechargeable)

Zn(s) + 2 Cl- + 2 NH3 + 2 MnO2 (s) + H2O → Mn2O3 (s) + Zn(NH3)2Cl2 (s) + 2 OH-

LeClanche’ Zn (s) + 2 MnO2 (s) + 2OH- + H2O → Mn2O3 (s) + Zn(OH)4 2-

Lithium Ion

Battery -new tech

Anode: Cgr Lix → Cgr + X Li+ + x e- Li+/Li

Cathode: Li1-XCoO2 + X Li+ + x e- → LiCoO2 CoIV/CoIII

Li+ moves from graphite to holes in LiCoO2 /electrons move to maintain neutrality

∆E0 = 3.7 V

electrolyte is

nonaqueous media

that allows Li+

transport between

graphite and a metal

oxide

Electro-refining of Copper

Crude Cu bars are used as the anode and a pure Cu wire as the cathode

A 0.1 V potential is applied

Metals more active than Cu dissolve Al, Fe, Ni, Co, Zn but do not plate out at cathode

Metals less active drop off as “anode sludge” Ag, Au, Pt - these pay the cost of the

process !

Cu → Cu2+ + 2e- Cu2+ + 2 e- → Cu

net reaction just involves transport of Cu from anode to cathode

anode oxid 2 Cl-→ Cl2 (g) + 2 e- cathode redn 2H2O + 2 e- → 2 OH- + H2 (g)

Eo1/2 = -1.358 V Eo

1/2 = -0.828 V Ecell = -2.19 V

Cl2 production in a diaphragm cell

Cl2 Production via Chlor-Alkali Process

Anode 2 Cl-→ Cl2 + 2e- -1.36 V Cathode Na+ + 1 e- → Na(Hg) - 1.86 V

Mercury serves as a liquid electrode, the amalgam stabilizes Nao but even with 99% recovery of the Hg there are environmental consequences.