REVIEW of Electrochemistry

Lecture 04

I. Introduction A. Terms

Electrochemistry - deals with the relationship of chemical reactions and electricity.

Redox Reactions - Chemical reactions which involve a transfer of electrons.

Electrochemical Cell - a system consisting of electrodes that dip into an electrolyte in which a reaction either generates or uses an electric current.

Cell 1: Voltaic Cell - An electrochemical cell in which a spontaneous reaction generates an electric current.

Cell 2: Electrolytic Cell - An electrochemical cell in which an electric current forces a reaction to occur.

I. Introduction B. review - Redox Reactions

- Two General Classes of Chemical Reactions: Redox & Non-Redox

- Redox reactions = ones in which electrons are transferred & two processes are associated with redox reactions:

1) Electrolysis: Electricity forces a chemical reaction to occur. Example:

2 H2O + electricity 2 H2(g) + O2(g)

2) A Voltaic Cell (battery): spontaneous redox reaction generates electrical potential. Example (1 M Cu+2 and Zn+2)

Zno + Cu+2 Zn+2 + Cuo + electricity (1.1 volts)

B. Review - Redox Reactions

- Oxidation = loss of electrons. Know- Reduction = gain of electrons. Know- Both must occur together, but we frequently break them up into

two half reactions.

Example: Zno + Cu+2 -----) Zn+2 + Cuo

Oxidn: Zno -----) Zn+2 + 2 e-

Redn: Cu+2 + 2 e- -----) Cuo

a) Oxidizing Agent causes oxidn. & is reduced (Cu2+) b) Reducing Agent causes redn. & is oxidized (Zno)

- Classes of Redox Reactions: Combination, Decomposition, Displacement, Combustion

B. Review - Oxidation Numbers (ON)

Definition: The charge that an atom (or group of atoms) would have if it were ionic

Usefulness:

1) Naming variably charged compounds. Example: FeCl3 = Iron(III)chloride2) Determining if a reaction is redox.3) Balancing a redox reaction.

Rules for Determining ON: Note - priority of rules are important; lowest # rule takes precedent when a conflict occurs. Know Rules

B. Review - Oxidation Numbers (ON). Rules

1) The sum of ONs must add up to give the charge

2) The ON of a neutral element by itself is 0

3) Group I, II, III elements in ionic form are +1, +2 & +3 respectively (H -1 when combined with Ia, IIa, IIIa)

4) F is -1

5) O is -2 (O is -1 when found as a peroxide, O2-2)

6) Cl, Br, I are -1 (when together, most electronegative rules)

B. Review of ON

Examples: Determine the ON of single N in each:(Treat as a Quiz & Complete in < 5 minutes)

N2 N3-1 NH3 N2O NO NF3 N2O4 NO3

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

- 1) Which one would be most reactive and why?

- 2) Ethanol/Nitric Acid experience in the lab.

- 3) HClO4 - What is ON of Cl? Comments on perchloric acid.

B. Review – ON Examples

- The top two are redox:

1) Mg + Cl2 MgCl2

2) CH4 + 2 O2 CO2 + 2 H2O

3) HCl + NaOH H2O + NaCl

- In any redox reaction you should be able to determine:The ON for each elementWhich reagent has been oxidizedWhich reagent has been reducedWhich reagent is the oxidizing agentWhich reagent is the reducing agentWhat are the two half reactions for eachHow many moles of e- are transferred in each half rxn

II. Balancing Redox Rxns A. acidic media

- Many are difficult and require a systematic method: Note that you must first determine if the reaction occurs in acid or base.

- Steps: 1-3 for acidic media & Steps 1-6 for basic media1. Assign ON’s & split into two half reactions

2. Complete & balance each half reaction:a. balance atom undergoing redox

b. balance O by adding H2O’sc. balance H by adding H+’sd. balance ionic charge by adding e-’s

3. Combine half reactions and finish (if in acidic media):a. multiply each half reaction by integer to cancel e-’s.b. add & simplify by canceling same species on both

sides.

II. Balancing Redox Rxns B. Basic media

4. If in basic media add OH- equal to # H+ to both sides.

5. Convert H+ & OH- on same side to H2O & simplify by canceling H2O if needed.

6. Check result by noting if following are both balanced: elements & charge.

II. Balancing Redox Rxns C. Example 1 in acid

Zn + NO3- + H+ -----) Zn+2 + NH4

+ + H2O

1) Zn -----) Zn2+ 0 ---) +2 on Zn (Oxidn.)

NO3- -----) NH4

+ +5 ---) -3 on N (Redn.)

2) Zn -----) Zn2+ + 2 e- (balance redox atom)

NO3- -----) NH4

+ + 3 H2O (balance O with H2O)

10 H+ + NO3- -----) NH4

+ + 3 H2O (balance H with H+ )

10 H+ + NO3- + 8 e------) NH4

+ + 3 H2O (bal. Charge w e-)

3) 4 [Zn -----) Zn2+ + 2 e-] (8e- lost)

1 [10 H+ + NO3- + 8 e------) NH4

+ + 3 H2O] (8e- gained)

4 Zn + 10 H+ + NO3- + 8 e----) 4 Zn2+ + 8 e- + NH4

+ + 3 H2O

4 Zn + 10 H+ + NO3- -----) 4 Zn2+ + NH4

+ + 3 H2O

II. Balancing Redox Rxns C. Example 2 (basic)Pb(OH)3

- + ClO- -----) PbO2 + Cl-

1) ClO- -----) Cl- +1 ---) -1 on Cl (rdn)Pb(OH)3

- -----) PbO2 +2 ---) +4 on Pb (oxn)

2) 2 H+ + ClO- + 2 e- -----) Cl- + H2OPb(OH)3

- -----) PbO2 + H2O + H+ + 2e-

3) 2H+ + ClO- + Pb(OH)3- + 2e- -----) Cl- + 2H2O + H+ + PbO2 + 2e-

H+ + ClO- + Pb(OH)3- -----) Cl- + 2H2O + PbO2

4) OH- + H+ + ClO- + Pb(OH)3- ---) Cl- + 2H2O + PbO2 + OH-

5) H2O + ClO- + Pb(OH)3- ---) Cl- + 2 H2O + PbO2 + OH-

6) ClO- + Pb(OH)3- Cl- + H2O + PbO2 + OH-

II. Balancing Redox Rxns C. Example 3MnO4

- + S2O32- -----) Mn2+ + SO4

2- (acidic)

1) a) MnO4- ----) Mn2+ [+7 ---) +2 Redn]

b) S2O32- ----) SO4

2- [+4 ---) +6 per S Oxn]

2) a) 5e- + 8H+ + MnO4- ----) Mn2+ + 4H2O (x 8)

b) 1S2O32- -----) 2SO4

2-

5H2O + S2O32- -----) 2SO4

2- + 10H+ 5H2O + S2O3

2- -----) 2SO42- + 10H+ + 8e- (x 5)

3)

40e- + 64H+ + 8MnO4- + 25H2O + 5S2O3

2- ---) 8Mn2+ + 32H2O + 10SO42- + 50H+ + 40e-

14 H+ + 8 MnO4- + 5 S2O3

2- ----) 8 Mn2+ + 7 H2O + 10 SO42-

II. Balancing Redox Rxns C. Example 4MnO4

- + I- -----) MnO2 + IO3- (basic)

1) MnO4- -----) MnO2 +7 ---) +4 Mn

I- -----) IO3- -1 ----) +5 I

2) 4H+ + MnO4- + 3e- -----) MnO2 + 2H2O

3H2O + I- -----) IO3- + 6H+ + 6e-

3) 2 [4H+ + MnO4- + 3e- -----) MnO2 + 2H2O]

1 [3H2O + I- -----) IO3- + 6H+ + 6e-]

8H+ + 2MnO4- + 6e- + 3H2O + I- ----) 2MnO2 + 4H2O + IO3

- + 6H+ + 6e-

2H+ + 2MnO4- + I- ----) 2MnO2 + H2O + IO3

-

4) 2OH- + 2H+ + 2MnO4- + I- ----) 2MnO2 + H2O + IO3

- + 2OH-

5) 2H2O + 2MnO4- + I- ----) 2MnO2 + H2O + IO3

- + 2OH-

H2O + 2MnO4- + I- ----) 2MnO2 + IO3

- + 2OH-

6) Double check balancing of atoms & charge.

III. Voltaic Cells

A. Introduction- A voltaic cell (battery) is an electrochemical cell in which a

spontaneous redox reaction generates an electric current.

- It consists of two half cells which are connected:

Oxidation half cell consists of electrode (anode = - ) & electrolyte.

Reduction half cell consists of electrode (cathode = + ) & electrolyte.

Internal connection of a salt bridge that allows the flow of ions but prevents mixing of electrolytes from two half cells.

External connection of anode to the cathode.

Electrons travel from anode to cathode.

B. Example

Zn + Cu2+ -----) Zn2+ + Cu Anode - Zn ---) Zn2+ + 2e- Cathode + Cu2+ + 2e- ---) Cu

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

C. Voltaic Cell Shorthand Notation same example: Zn+2 & Cu2+ at 1.0 M

Zn + Cu2+ -----) Zn2+ + Cu

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

Anode (-) (oxidation) Zn ---) Zn2+ + 2e- Cathode (+) (reduction) Cu2+ + 2e- ---) Cu

Know the following Conventions:

Anode written on leftCathode written on right| = phase boundary|| = salt bridgeZn = anode electrodeCu = cathode electrode, = separates different ions in the same half cellM = Molar concentration of ions

This battery develops 1.1 volts at 25 oC

C. Voltaic Cell Notation (ions at 0.1 M)

Cd + 2Ag+ -----) Cd2+ + 2Ag Anode - Cd ---) Cd2+ + 2e- Cathode + 2Ag+ + 2e- ---) 2Ag

Cd | Cd2+ (0.1 M) || Ag+ (0.1M) | Ag (this cell develops 1.2v)

C. Voltaic Cell NotationAdditional Examples

Pt | H2 | H+|| (Half Cell)Hydrogen electrode as anode (electrode on left)H2 -----) 2H+ + 2 e-

Three phases: Pt solid, H2 gas, H+ solution

||Fe3+, Fe2+ | Pt Cathode Reaction (electrode on right) Fe3+ + e- -----) Fe2+

||Cd2+ | Cd Cathode Reaction (electrode on right)Cd2+ + 2e- ------) Cd

Tl | Tl+ || Sn2+ | Sn 2 Tl + Sn2+ -----) 2 Tl+ + Sn2 Tl ----) 2 Tl+ + 2 e- (anode)Sn2+ + 2e- ---) Sn (cathode)

D. Work & Voltage of a Cell

• Much of the early work on cell physics & chemistry came from Michael Faraday - 1791 to 1867

• He was responsible for uncovering ONs, terminology (anode, cathode, etc), benzene, electromagnetism, Faraday’s Laws & Faraday’s constant.

• 1 Faraday = Charge on 1 mole of electrons = 96,500 c.

D. Work & Voltage of a Cell

Work done in a redox rxn = ∆G = - n x F x Ecell – n= moles of e- transferred– Ecell = voltage of the cell

– F = Charge of 1 mole of e- = 1 Faraday = 96,485 coulombs

Notes:o Units: Joules = moles x (coul/mole) x volts = coul x volt = Jo 1 coulomb x 1 volt = 1 joule o Since batteries are spontaneous redox rxns the sign is -: ∆G = - nFEcell

Free Energy, ΔG = maximum useful work done ΔG = - nFEcell

At 1M concentrations, 1 atm pressure & 25 oC: ΔG = ΔGo

ΔGo = - nFEocell

E. Standard Cell Voltages - some notes

- Can measure the [cell voltage or potential or electromotive force (EMF)] using a voltmeter; and can calculate theoretical voltages from equations & tables.

- The voltage (E) developed by a cell is the sum of the oxidation and reduction potentials: Ecell = Eoxidation + Ereduction (note: you supply signs)

- Note: can’t separate the two half reactions in the lab; so, how calculate E r & O? By setting an arbitrary standard and comparing all other half cell reactions to this standard.

- Arbitrary Standard is the hydrogen half cell reduction under standard conditions (1 M, 1 atm, 25oC) ; a voltage of 0.000 was assigned.

E. Standard Cell Voltages H2 Electrode

|| H+ (1.0 M) | H2 (1.0 atm) | Pt Eo = 0.0000 volts at 25 oC

2H+ + 2e- H2 (1 M HCl & 1 atm H2)

E. Standard Cell Voltages- If we couple the hydrogen electrode under standard conditions with another half

reaction and measure the Ecell , then we can determine the E for the unknown half reaction.

Eocell = Eo

half reaction + Eohydrogen electrode = Eo

half reaction + 0.000

Eocell = Eo

half reaction under standard conditions

- Example: Zn | Zn2+ (1.00M) || H+ (1.00M) | H2 (1.00atm) | Pt

Zn + 2H+ Zn2+ + H2 Eocell measured as 0.76 volts

Eocell = Eo

Zn + EoH2 = Eo

Zn + 0.000; EoZn = Eo

cell = 0.76 v

Notes: 1) Convention: Half reactions are written as reductions. See Table 19.1, pg 786; if need half reaction as oxidn, then reverse the sign.

2) Eo1/2 value is the same regardless of #e- transferred = constant.

E. Standard Cell Voltages – Calculate Eocell

Eo cell = Eo reduction + Eo oxidation

Use Table 19.1 & must change sign for oxidation. Example: I2 + 2Li -----) 2Li+ + 2I-

Table values: 2Li+ + 2e- ---) 2Li Eo = -3.04 vI2 + 2e- -----) 2I- Eo = 0.54 v

What is Eocell? Eo

cell = 3.58 v (reverse Li1/2 rxn)

Li Iodine battery – can reverse the rxn by adding electricity to recharge battery

E. Standard Cell Voltages

Use of Standard Cell Voltages (table 19.1, pg 786)

1) Can tell strength of reducing/oxidizing agents:Reaction with largest + volts wants to go most

2) Can determine the Eocell from Standard ½ Cell

Voltages: Eo

cell = Eo ½ reduction + Eo

½ oxidation

3) Can tell which way a redox rxn wants to go: Will spontaneously go such that Ecell is largest

4) Can be used to calculate ΔG: ΔGo = - nFEocell

5) Can be used to calculate Keq: Eo = (2.30 RT / nF) Log K

6) Used to calculate Ecell not under std conditions.

F. Calculations Involving Eo

1) Can tell strength of reducing/oxidizing agents:Reaction with largest + Eo

1/2 value wants to go mostMg2+ + 2e- ---) Mg Eo

1/2 = -2.4 F2 + 2e- ---) 2F- Eo

1/2 = 2.9 F2 wants to be reduced more than Mg+2

2) Can determine the Eocell from Standard ½ Cell Voltages:

Given: Al | Al3+ || Cr3+ | Cr (Direction mandated by convention)Al ---) Al3+ + 3e- Eo = 1.66 v Cr3+ + 3e- ----) Cr Eo = - 0.74 VEo

cell = Eoox + Eo

red = 1.66 + (-0.74) = 0.92 V

3) Can tell which way a redox reaction will spontaneously go.Goes such that Eo

cell is the largest positive value.

Zn2+ + Cu ---) Zn + Cu2+ Eo = - 1.10 v = Not Spontaneous Zn + Cu2+ ---) Zn2+ + Cu Eo = + 1.10 v = Spontaneous

F. Calculations Involving Eo

4) Free Energy from Eo & Eo from ΔGo

ΔGo = - nFEocell Eo

cell = -ΔGo/nF

Example: Calculate the ΔGo in J for: Zn | Zn2+ (1.0M) || Ag+(1.0M) | Ag

Zn + 2Ag+ ----) 2Ag + Zn2+

Zno -----) Zn+2 + 2e- Eo1/2 = +0.76 v

2Ag+1 + 2e- -----) 2Ago Eo1/2 = +0.80 v

Eo = 0.76 + 0.80 = 1.56 v

Note: 2 m of e- transferred, but Eo1/2 = 0.80v for Ag (Eo

cell is a constant)

ΔGo = - nFEocell = - 2.00m x 9.65x104c/m x 1.56v = - 3.01x105 J

Example: Calculate Eo for a rxn in which ΔGo = -409 kJ & n=2

Eo = -ΔGo/nF = - (-409x103J) / (2m x 9.65x104c/m) = 2.12 J/c = 2.12 v(1 coulomb x 1 volt = 1 joule; 1 volt = 1 joule/coulomb)

F. Calculations Involving Eo 5) Get Keq from Eo

ΔGo = - nFEo & ΔGo = - RTLn K (from last chapter)

- nFEo = - RT Ln K -nFEo = - 2.30 RT Log K

Eo = (2.30 RT / nF) Log K @ 298 K (25 oC) 2.30RT/F = 0.0592 V

Eo = (0.0592) Log Keq (true at 25 oC)

n

Example: Calc K at 25oC for Zn + Cu2+ Zn2+ + Cu Eo = 1.10 V

Eo = (0.0592) Log K 1.10 = (0.0592) Log K n 2

Log K = 37.2 K = 1037.2 K = 1.58 x 1037

F. Calculations Involving Eo

6) Cell EMF for Nonstandard Conditions (not 1 M or not 25 0C)

From: ΔG = - nFE & ΔG = ΔGo + RT Ln Q (last chapter)Get: E = Eo - RT/nF Ln Q = Eo - 2.30RT/nF Log Q

E = Eo - 0.0592/n Log Q at 25 oC = Nerst Equation (know)

- Can now calculate E at various temperatures & concentrations.- n = # moles of electrons in the balanced redox reaction- Q = same form as Keq except initial M used - not equilibrium concentrations.

Example: Calculate E at 25 oC for Zn | Zn2+(1.00x10-5) || Cu2+(0.100) | Cu

Zn (s) + Cu2+ (------) Zn2+ + Cu (s) Eo = 1.10 V

Q = [Zn2+]/[Cu2+] = 1.00x10-5 M / 0.100 M = 1.00x10-4

E = Eo - 0.0592/n Log Q = 1.10 - 0.0592/2 Log (1.00x10-4)

E = 1.10 - (-0.12) = 1.22 V

IV. Electrolytic Cells, Examples

- Place inert electrodes in a solution and force a non-spontaneous redox rxn to occur by running electricity through the solution with a battery = electrolysis.

1. Molten NaCl - at cathode: 2Na+ + 2e- ---) 2Na- at anode: 2Cl- ---) Cl2 + 2e-

2NaCl(s) + elec. -----) 2Na(s) + Cl2 (g)

2. Aqueous NaCl - at cathode: 2H2O + 2e- ---) H2 + 2OH-

- at anode: 2Cl- ---) Cl2 + 2e-

2H2O + 2Cl- + elec. -----) H2 + 2OH- + Cl2

3. Water + non reacting Electrolyte (like Na2SO4)

- at cathode: 4H2O + 4e- -----) 2H2(g) + 4OH-

- at anode: 2H2O -----) 4e- + 4H+ + 1O2(g)

2H2O + elec. ----) 2H2(g) + 1O2(g)

IV. Electrolytic CellsElectroplating - process of depositing a metal on an electrode. Can be

used for analysis, beauty, or for protective coating.

- Can be used for quantitative analysis:

Example: Analysis of Cu in brass (mix of Sn, Pb, Cu)

Dissolve brass in HNO3; add two Pt electrodes to the HNO3 solution; apply > +0.34 volts. Cu forms at the cathode: Cu2+ + 2e ----) Cuo

Sn+2 Eo = -0.14 v Pb+2 Eo = -0.13 v Cu+2 Eo = +0.34 v

Weigh the cathode before and after the process to quantitatively determine the amount of Cu in the brass.

- Note: 1. Can predict what will form at electrodes from knowing the ions in the solution & from a table of Eo values; the metal ion with the largest + reduction potential will plate out on the cathode first.

2. Can also determine amount of Cu by measuring current (amps) and time necessary to plate out all of the Cu (see next section).

IV. Electrolytic Cells - Stoichiometry

- Note that electrolysis can be used for stoichiometric calculations.

- Equations and values to know: coulombs = amperes x seconds

c = a x sec a= c/sec 1F = Charge on 1 mole of e- = 96500 c

Example: How many g of Cu are present in the previous example if a current of 0.852 amp is needed for 600 seconds in order to remove all of the blue color?

1Cu+2 + 2e- ------) 1Cu(s)

c = a x s = 0.852 a x 600 s = 511 coulombs

511 c x 1 mole e- x 1 mole Cu x 63.5 g Cu = 0.168 g Cu 96500 c 2 mole e- 1 mole Cu

- Note: Can also weigh a Pt cathode before & after; would gain 0.168 g.

V. More Applications of Electrochemistry

pH Electrode: Ecell = Ereference electrode + Eindicator electrode

V. Few More Applications of Electrochemistry

pH electrode: AgCl(s) + H+Inside (---) Ag(s) + H+

outside + Cl-

[Cl-] = 1 M n = 1 [H+Inside ] = constant

Ecell = Eo - (0.0592/n) Log Q = Eo - 0.0592 Log {[H+out ]x[Cl-] / [H+

Ins ]}

Ecell = K’ - 0.0592 Log [H+out ] = K’ + 0.0592 pH

Ecell = 0.0592 pH + K’ [1) Eqn a pH meter uses; 2) a plot of voltage vs pH will be a straight line with a slope of 0.0592; 3) when calibrating a pH meter, you are setting the slope = 0.0592 and the Y intercept = K’ ]

Other Ion Specific Electrodes - Example: Fluoride Specific Electrodetypical F- ISE (ion-selective electrode)

F- F- LaF3 Crystal - Potential developed across membrane

Et = Constant - 0.0592 Log F- (at 25oC)

F- F- F- F- Note the potential interferences from the F lab.

V. Few More Applications of Electrochemistry

Electrochemical Detectors for Liquid Chromatography- Control the voltage of an electrode system in an LC mobile phase. If compound of interest goes past the cell & can be oxidized at the selected voltage, then will get a current which is proportional to the concentration of the compound. Can detect pg amounts of chemicals.

V. More Applications of Electrochemistry

Batteries Example: Car Battery

Pb + HSO4- -----) PbSO4 + H+ + 2e-

PbO2 + 3H+ + HSO4- + 2e- -----) PbSO4 + 2H2O

PbO2 + Pb + 2H+ + 2HSO4- (-----) 2 PbSO4 + 2H2O

One cell develops 2 V; so, a six cell battery connected in series will develop 12 Volts