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Electrochemistry Notes for Chem 1001S
These are notes and outlines of the material covered in class. They are not meant as a substitute for reading the text
or attending lectures.
1. Balancing Redox reactions – method differs from that in text
1. first balance elements other than H and O
2. Assign oxidation states and set up skeletal OX and RED half reactions including number of electrons gained or
lost.
SHORT CUT – if balanced half reactions not required
– otherwise do 4 and 5 for each half rxn first.
3. Multiply each half reaction by integers such that electrons gained = electrons lost and add them.
The full equation contains no electrons at this point.
4.
In acid add H+ and in base OH- to balance charge.
5. Add H2O to balance either H or O and then check the other to complete.
There are many examples to work in Chapter 5. Work them to perfect your skills.
A few tips on balancing equations
1. Always balance the charge.
2. Never have H+ and OH- in the same redox equation
3. Never add H2 or O2 unless they are actually present and involved in the reaction.
4. ALWAYS check H,O and charge balance when finished.
5. Become familiar with the most common oxidizing and reducing agents and their oxidation numbers.
6. Memorize the O2/H2O half cell as students often mess this one up.
7. Begin by asking what was oxidized and what was reduced.
8. Make sure you haven’t combined two oxidations or two reductions or have the electrons on the wrong side.
ASSIGNING OX NUMBERS
1. H has ox state +1 except in H2 (0) or metal hydrides (-1) , simple cations like Na+ (+1) Mg2+ (+2) etc.
2. O has oxidation state -2 except in peroxides (-1) as in H2O2 or O2 or O3 (0)
3. Work out other elements such that the sum of the oxidation numbers adds up to the net charge on the species
Examples N in : NH3 and NH4+ -3 N2H4 -2 N3- (-1/3) HNO2 (+3) NO3- (+5) KNO3 (+5)
Cl in Cl- or KCl or CCl4 (-1) Cl2O (+1) HClO or ClO- (+1) KClO3 (+5)
LEWIS STRUCURES AND OXIDATION NUMBERS
If you have studied Lewis structures in Chem 1000 you may note that oxidation numbers derive from one of the three
arbitrary ways of assigning electrons to atoms in molecules.
1. For assigning octet – count both electrons in a bond for each atom i.e count them twice.
2. For formal charge – divide the bonding electrons equally between the two atoms.
3. For ox number – award both bonding electrons to the more electronegative element. (For homonuclear divide
equally)
The peculiar oxidation number of -1/3 in N3- is seen to arise because of the different ox numbers of each N in a
given Lewis structure which average to -1/3. In N=N=N – each terminal N has 2 lone pairs giving 6 total when
bonding electrons are divided equally (no difference in electronegativity of N vs. N) OX # = -1 each. The central N
has no lone pairs and thus 4 electrons to call its own OX # = +1 On avg we have (-1+1-1)/3
If you write the other resonance form with one single and one triple bond. The single bonded N has 3 lone pairs and
thus 7 electrons (OX #= -2), the triply bonded N has one lone pair and thus 2+3 = 5 (OX # =0) and the central N has
no lone pairs and thus 1 + 3 = 4 electrons (OX # = +1) On avg we have (-2+0+1)/3 = -1/3.
E vs. G.
∆G = -nEF where n = number of electrons and F = Faraday constant = 96,485 Coulombs/mol
1 Coulomb volt = joule so E in volts X F gives Joules/mol.
Note that E is an intensive quantity , not on a per mol or gram basis. E has the same value at any point in a solution.
G is an extensive quantity - on a per mol basis.
An analogy is heat Q vs. temperature T . Heat is extensive, temperature intensive.
Almost everything we have said about G and Go applies to E and Eo
E is a measure of spontaneity + goes forward 0 at equilibrium - goes in reverse
While for G
- goes forward 0 at equilibrium + goes in reverse because ∆G = -nEF
The naught refers to standard concentrations (1 molar ) or pressures (1 atm) just as with G.
At any other conditions E is obtained from the Nernst equation which can be derived as follows:
∆G = ∆Go + RT lnQ
-nEF = -nEoF + RTlnQ
E = Eo – RT/nF ln Q which at 298 and using logs becomes E = Eo – 0.0592/n log Q
At equilibrium E = 0 and thus Eo = 0.0592/n log K
by analogy to ∆Go = - 5.705 log K at 298
Standard Half Cell Potentials and their use.
Eo1/2 are half cell potentials measured relative to the standard hydrogen electrode (SHE).
The potential for 2H+ + 2e- = H2 (g) at 1 M H+ and 1 atm H2 = 0
This definition also sets ∆Gof of H+ (aq) = 0 and forms the basis for the free energies of ions in solution.
An electrochemical cell consists of two half reactions one where oxidation occurs (ANODE) and the other where
reduction occurs (CATHODE). The cell potential is obtained as follows.
TEXT METHOD : E (cell) = E cathode – E anode where both E’s are reduction potentials given in tables.
PREFERRED METHOD E (cell) = EOX + ERED where EOX = the oxidation potential obtained by reversing the
reaction and changing the sign of E for the half cell where oxidation occurs.
EXAMPLE :
Zn + Cu2+ Cu + Zn2+
given Zn2+ + 2e- Zn
Cu2+ + 2e- Cu
Eo = -0.76 V
Eo = + 0.34 V
A. Text method : note that Zn is oxidized and Cu reduced in the reaction E (cell) = 0.34 – (-0.76) = +1.10 V
B. Preferred method : reverse the Zn half reaction to Zn Zn2+ + 2e- E (ox) = +0.76 V
Now add the two half reaction and the E’s = 0.34 + 0.76 = +1.10 V
Why I prefer method B. When combining reactions either using G or E one ADDS. If a reaction is reversed one must
change the sign of E or G for the reverse direction. In the preferred method one can see that the electrons drop out for
a net reaction as they must. The text method is a rule which makes two mistakes that cancel out. You use a reduction
potential for an oxidation half reaction and then you subtract instead of add potentials. This is an E-chem for
dummies approach.
LATIMER DIAGRAMS or Reduction Potential Diagams . ( see Chap 22.2 for example)
Reduction potential data can be displayed as in Table 20.1 listing potentials in order from high to low. (About as
sensible as phone book listings in order of phone number). Such tables give only a few potentials and one must
search for the one you want. They provide no sense of the redox chemistry of a given element.
Latimer diagrams (download a copy from the 1001S website) display reduction potentials by element showing the
range of stable oxidation states, the relative stability of them, and the proper standard states in acid and base solution.
(Continuing the phone book analogy, not only do you get the number by looking up the name, but you also get the
next door neighbor’s as well).
For example Mn(IV) is shown as the insoluble MnO2 (s) , weak acids are protonated in 1 M H+ (HNO2 ) and not in 1
M base (NO2-) etc.
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 = {Eo1/2 ( 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. ∆G = -nEF
. ∆Go ½ = -{n1 E1 F + n2 E2 F } = - (1)(-0.41)F + - (2)(-0.91)F = +2.23 F
Eo1/2 = - ∆Go ½ /nF = - (+2.23) F/3F = -0.74 V
You can check your skill at this using any of the skipped step values shown in the Latimer diagrams.
.
PREDICTING CHEMICAL REACTIONS USING Eo1/2 ‘s
Combining any oxidation half reaction with any reduction half reaction gives a net chemical reaction. If the net E >0
the reaction is spontaneous. Because E is an intensive quantity you do not need to concern yourself with numbers of
electrons or balancing to decide on spontaneity noting that Eo ‘s refer to standard conditions of 1M all reactants,
products and H+ if that is present in the net balanced reaction. Specific conditions can influence the outcome.
Also the Eo does not consider kinetics which is typically slow when gases like H2 and O2 are involved. Kinetic
factors lead to an overpotential for H2 and O2 reactions at electrode surfaces.
There are four reactions which must be examined to decide on thermodynamic stability of a chemical species in
aqueous solution in air. These largely determine what form you find elements on earth.
1. Oxidation by water
2. Reduction by water
3. Oxidation by air (O2)
4. Disproportionation
These will usually be pH dependent so be careful to note whether you are in acid or base. A half cell involving a
difference in the number of H’s or O’s between redox pairs will be pH dependent ! Cl-/Cl2 NO ClO3-/ClO4- YES
In 1 M H+
1. Water acts as an oxidation agent by having the H in it get reduced. 2H+ + 2e- H2 + 2e- Eo1/2 = 0.0 V
Only species with oxidation potentials greater than 0.0 V will liberate H2 from 1 M H+. Eg. Mno
2. Water acts as a reducing agent by getting oxidized 2 H2O O2 + 4e- 4H+ Eo1/2 = - 1.23 V
Only species with reduction potentials greater than + 1.23 V will liberate O2 from water. Eg. MnO43. Oxygen oxidizes by getting reduced to water
O2 + 4H+ + 4e- 2 H2O Eo1/2 = + 1.23 V
( Note 3 = reverse of 2. All species with oxidation potentials greater than -1.23 V will oxidize in air.
Thus Fe2+ and Sn2+ do while Co2+ and Ni2+ don’t.
4. Disproportionation The species oxidizes and reduces itself. Such species are unstable unless kinetics are slow and
usually are not commonly found in solution for very long.
Examples : H2O2 , MnO42- Cl2 in base but not in acid , Cu+ , Mn3+
Example 1.
balanced
Mn3+ + 1e- Mn2+ E (red) = +1.5 V
Mn3+ MnO2 (s) + 1e- E (OX) = -1.0 V
2 Mn3+ Mn2+ + MnO2 E = + 0.5 V spontaneous
2 Mn3+ + 2 H2O Mn2+ + MnO2 (s) + 4 H+
Example 2.
2-
MnO4
3 MnO42balanced 3 MnO42Example 3 in base
Balanced
MnO42- + 2 e- MnO2 E (red) = + 2.26 V
MnO4- + 1e- E (ox) = -0.56 V
2 MnO4- + MnO2 (s) E = + 1.7 V spontaneous
+ 4 H+ 2 MnO4- + MnO2 (s) + 2 H2O
½ Cl2 (g) + 1 e- ½ Cl2 (g) OClCl2 (g) OCl2 OH- + Cl2 (g) Cl+ 1 e+ ClOCl-
E = +1.36 V
E = -0.40 V
E = + 0.96 V spontaneous in 1 M OH+ Cl- + H2O balanced in base
Note in 1 M acid the balanced reaction is Cl2 (g) + H2O HOCl
(not spontaneous at stnd conditions but see below)
+ Cl- + H+
and Eo = +1.36 – 1.63 = -0.27 V
NERNST EXERCISES
The Nernst equation is used to correct for non-standard conditions in a half reaction and in an electrochemical cell.
For equilibrium conditions (when e-flow ceases) set E = 0 and Eo = +0.0592 log K at 298 K.
EX1. E1/2 for the SHE at pH = 7 .
Write the balanced reaction
Write the Nernst eqn
Put in the correct value of n and Q and Eo
Plug in 1 atm PH2 and 10-7 M H+
2 H+ + 2 e- H2 (g)
E = Eo - 0.0592/n log Q
E = 0.0 V – 0.0592/2 log { PH2 / [H+]2}
E = 0.0 – 0.0592 log (1/[10-7] = - 0.414 V
Ex2 – Obtain the value of E1/2 for O2/H2O in 1M OH- using the value for 1 M H+.
Ex 3. If Cl2 gas is bubbled through neutral water the water takes up 0.09 moles of the gas per litre of solution. The
saturated solution becomes acidic. Obtain the concentrations of all relevant species.
Cl2 disproportionates . The balanced reaction in acid is
Cl2 (g) + H2O HOCl
+ Cl- + H+
And the Eo = -0.27 V ( using the 1 M acid reference state)
Let X = the concentration of Cl- . From stoichiometry [H+] = [OCl-] = [Cl-] = X
Nernst E = Eo -0.0592 log {[OCl-][Cl-][H+] / PCl2} = 0 at equilibrium
Eo = -0.27 = 0.0592 log {X3/1atm }
X = 0.03
At equilibrium we have 0.03 M HOCl, Cl-, and H+. Of the 0.09 moles of Cl2 gas taken up by a litre of water 0.06
are present as Cl2 (aq) and .03 each as Cl- and OCl-.
NOTE : You must be consistent in the choice of Eo , reference states, n, and Q in your Nernst expression.
Example A : For 2H+ + 2e- H2 (g) Eo is 0.0 , n = 2 and Q = PH2 / [H+]2
For H+ + 1 e- 1/2 H2 (g) Eo is 0.0 , n = 1 and Q = (PH2)1/2 / [H+]
Both give the same result
Example B: At pH = 4 using the standard base reference potentials you balance the half cell as
2e- + H2O H2 (g) + 2 OH- E = -0.828 -0.0592/2 log { PH2 [OH-]2 }
and plug in 1 atm of hydrogen and OH- = 10-10 to give E1/2 = -0.236 V
OR you could use the standard 1 M H+ reference and balance the half cell below and plug in [H+] = 10-4 M
2H+ + 2e- H2 (g) E = 0.0 – 0.0592/2 log{PH2 / [H+]2} = -0.236 V
Both give the same result
EXAMPLE 4. What would be the concentration of Cu+ remaining in solution at equilibrium following
disproportionation of a 0.2 M solution of cuprous chloride.
2 Cu+ Cuo (s) + Cu2+ ∆Eo = +0.521 – 0.153 = + 0.368 V = 0.0592 log [Cu2+]/[Cu+]2
Since Eo is positive reaction goes largely to right giving (0.1-x/2 ) M [Cu2+] and x M [Cu+]
K = 10+ 0.368/.0592 = 1.6 x 106 = (0.1)/ x2
x = 2.4 x 10-4 M
ELECTROCHEMICAL CELLS
There are two kinds of electrochemical cells.
The voltaic (AKA galvanic cell or battery or fuel cell) is a cell in which electrons spontaneously move from anode
( LEO =lose electrons oxidation) to cathode (GER = gain electrons reduction). These cells output the free energy of
chemical reactions into useful electrical work.
Electrolytic cells are ones where an external voltage is applied to drive a nonspontaneous reaction uphill.
Electroplating of metals, production of Chlorine or electrolysis of water to give H2 + O2 are examples. These cells
input energy to do work on chemicals producing higher energy chemicals.
The basic principles are the same for both. The anode is always the electrode at which oxidation occurs. The only
difference is the sign of the electrodes and the direction of electron flow.
In a voltaic cell the anode is negative and electrons flow downhill from – anode to + cathode.
In an electrolytic cell the anode is positive and electrons are forced to flow uphill from + anode to – cathode.
A shorthand notation for cells is < anode(red form\anode)(ox form)// cathode(ox form)/ cathode(red form> with the
spontaneous direction of e-flow being left to right.
Thus < Zn/Zn2+ (0.01 M) // Cu2+ (0.1 M)/Cu> describes a cell combining the zinc and copper half cells at nonstandard concentrations. E(cell) = E0 -0.0592/2 log [Cu2+]/[Zn2+] = 1.34 – 0.0592/2 log (0.1)/0.01) = 1.31 V
Note that the reaction Zn + Cu2+ Zn2+ + Cu becomes less favorable (less positive E) compared to 1 M .
ELECTRON FLOW, STOICHIOMETRY AND VOLTAGE CHANGE.
The Faraday constant corresponds to the charge of a mole of electrons in Coulombs. You can treat electrons just as
you would a stoichiometric reagent in a chemical reaction. Note that 1 amp = 1 Coulomb per second.
Example 1 . How long must a 10 amp current pass through a solution of Cu2+ before 6.35 grams of copper is plated
out? 6.35 grams of Cu = 0.1 mole, we need 2 electrons for each copper. Thus we need 0.2 F
0.2 (96,485) = 1.93 x 105 C. = amp X time = 10 t t = 1.93 x 104 seconds.
Example 2. Let us consider the < Zn/Zn2+ (0.01 M) // Cu2+ (0.1 M)/Cu> above with a KNO3 salt bridge. We will
assume a litre of each of the two solutions. What do we have after passing 2 amp current for 20 minutes?
2 amp x 20 minutes x 60 s/min = 2.4 x 103 C or 2.4 x 103 C / 96,485C/mol = 0.0248 mol of electrons
In the cell reaction
Zn + Cu2+ initial
0.1
change
- 0.0124
final
0.0876
Zn2+ + Cu
0.01
+0.0124
0.0224 and E(cell) = 1.10 -0.0592/2 log {[.0224]/[0.0876] = 1.08
As you see even large changes have only a small effect on the voltage until one of the concentrations gets near zero.
Assuming an excess of the zinc metal, the cell would go dead when essentially all the Cu2+ was used up. This would
require 0.2 moles of electrons or 1.927 x 104 C or a 2 amp current for 9648 seconds or 160.8 minutes or 2.68 hrs.
At equilibrium E = 0 and the remaining conc of Cu2+ would be :
log K = n∆Eo /0.0592 = 2(1.04 V)/0.0592 = 35.1 K = 10+35.. = [Zn2+]/ [Cu2+] = 0.11/ [Cu2+]
thus [Cu2+] = 10-36 M ( You see why we say redox rxns tend to go all the way. )
One thing to note is that the cell voltage changes very little until one of the reagents is almost used up.
APPLICATIONS We will briefly discuss applications to analytical chemistry, batteries, and the chemical industry.
Some of these are discussed in Chapters 20 -22.
Batteries Pb acid battery, Ni/Cd, dry cell
Analytical Chemistry; 1. Titration of halides using the Ag/Ag+ electrode.
2. Determining Ksp for AgCl or Kf for Ag(NH3)2+
Electrolytic cells: Electrolysis of water, brine, Cl2 production, Al, Na, Cu refining.