Chapter 18Electrochemistry

Chemistry II

Redox Reaction

• one or more elements change oxidation numberall single displacement, and combustion,some synthesis and decomposition

Redox Reaction

• always have both oxidation and reductionsplit reaction into oxidation half-reaction and a reduction

half-reaction

• aka e- transfer reactionshalf-reactions include e-

oxidizing agent is reactant molecule that causes oxidationcontains element reduced

reducing agent is reactant molecule that causes reductioncontains the element oxidized

Oxidation & Reduction

• oxidation:ox number of an element increaseselement loses e-

compound adds Ocompound loses Hhalf-reaction has e- as products

• reduction:ox number of an element decreaseselement gains e-

compound loses Ocompound gains Hhalf-reactions have e- as reactants

Rules for Assigning Oxidation States

• rules are in order of priority1. free elements have an oxidation state = 0

Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)

2. monatomic ions have an oxidation state equal to their charge

Na = +1 and Cl = -1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0

Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0

Rules for Assigning Oxidation States

3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion

N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1

4. (a) Group I metals have an oxidation state of +1 in all their compounds

Na = +1 in NaCl

(b) Group II metals have an oxidation state of +2 in all their compounds

Mg = +2 in MgCl2

Rules for Assigning Oxidation States5. in their compounds, nonmetals have oxidation states according to the

table below nonmetals higher on the table take priority

Nonmetal Oxidation State Example

F -1 CF4

H +1 CH4

O -2 CO2

Group 7A -1 CCl4

Group 6A -2 CS2

Group 5A -3 NH3

Oxidation and Reduction

• oxidation occurs when an atom’s oxidation state increases during a reaction

• reduction occurs when an atom’s oxidation state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2

oxidationreduction

Oxidation–Reduction

• oxidation and reduction must occur simultaneously if an atom loses electrons another atom must take them

• reactant that reduces an element in another reactant = reducing agent the reducing agent contains the element that is oxidized

• reactant that oxidizes element in another reactant = oxidizing agent the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agent

Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ → 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O

Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ S + 2 NO + 4 H2O

MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O

+1 -2 +5 -2 +1 0 +2 -2 +1 -2

ox agred ag

+4 -2 +1 -1 +2 -1 0 +1 -2

oxidationreduction

oxidation

reduction

red agox ag

Common Oxidizing Agents

Oxidizing Agent Product when Reduced

O2 O-2

H2O2 H2O

F2, Cl2, Br2, I2 F-, Cl-, Br-, I-

ClO3- (BrO3

-, IO3-) Cl-, (Br-, I-)

H2SO4 (conc) SO2 or S or H2S

SO3-2 S2O3

-2, or S or H2S

HNO3 (conc) or NO3-1 NO2, or NO, or N2O, or N2, or NH3

MnO4- (base) MnO2

MnO4- (acid) Mn+2

CrO4-2 (base) Cr(OH)3

Cr2O7-2 (acid) Cr+3

Common Reducing Agents

Reducing Agent Product when Oxidized

H2 H+

H2O2 O2

I- I2

NH3, N2H4 N2

S-2, H2S S

SO3-2 SO4

-2

NO2- NO3

-

C (as coke or charcoal) CO or CO2

Fe+2 (acid) Fe+3

Cr+2 Cr+3

Sn+2 Sn+4

metals metal ions

Balancing Redox Reactions1. assign oxidation numbers

a) determine element oxidized and element reduced

2. write ox. & red. half-reactions, including e-

a) ox. electrons on right, red. electrons on left of arrow

3. balance half-reactions by massa) first balance elements other than H and Ob) add H2O where need Oc) add H+ where need Hd) neutralize H+ with OH- in base

4. balance half-reactions by chargea) balance charge by adjusting e-

5. balance e- between half-reactions

6. add half-reactions

7. check

Ex 18.3 – Balance the equation:I(aq) + MnO4

(aq) I2(aq) + MnO2(s) in basic solution

Assign Oxidation States

I(aq) + MnO4

(aq) I2(aq) + MnO2(s)

Separate into half-reactions

ox:

red:

Assign Oxidation States

Separate into half-reactions

ox: I(aq) I2(aq)

red: MnO4(aq) MnO2(s)

Ex 18.3 – Balance the equation:I(aq) + MnO4

(aq) I2(aq) + MnO2(s) in basic solution

Balance half-reactions by mass

ox: I(aq) I2(aq)

red: MnO4(aq) MnO2(s)

Balance half-reactions by mass

ox: 2 I(aq) I2(aq)

red: MnO4(aq) MnO2(s)

Balance half-reactions by mass

then O by adding H2O

ox: 2 I(aq) I2(aq)

red: MnO4(aq) MnO2(s) + 2 H2O(l)

Balance half-reactions by mass

then H by adding H+

ox: 2 I(aq) I2(aq)

red: 4 H+(aq) + MnO4

(aq) MnO2(s) + 2 H2O(l)

Balance half-reactions by mass

in base, neutralize the H+ with OH-

ox: 2 I(aq) I2(aq)

red: 4 H+(aq) + MnO4

(aq) MnO2(s) + 2 H2O(l)

4 H+(aq) + 4 OH

(aq) + MnO4(aq) MnO2(s) + 2 H2O(l) + 4 OH

(aq)

4 H2O(aq) + MnO4(aq) MnO2(s) + 2 H2O(l) + 4 OH

(aq)

MnO4(aq) + 2 H2O(l) MnO2(s) + 4 OH

(aq)

Ex 18.3 – Balance the equation:I(aq) + MnO4

(aq) I2(aq) + MnO2(s) in basic solution

Balance Half-reactions by charge

ox: 2 I(aq) I2(aq) + 2 e

red: MnO4(aq) + 2 H2O(l) + 3 e MnO2(s) + 4 OH

(aq)

Balance electrons between half-reactions

ox: 2 I(aq) I2(aq) + 2 e } x3

red: MnO4(aq) + 2 H2O(l) + 3 e MnO2(s) + 4 OH

(aq) }x2

ox: 6 I(aq) 3 I2(aq) + 6 e

red: 2 MnO4(aq) + 4 H2O(l) + 6 e 2 MnO2(s) + 8 OH

(aq)

Ex 18.3 – Balance the equation:I(aq) + MnO4

(aq) I2(aq) + MnO2(s) in basic solution

Add the Half-reactions

ox: 6 I(aq) 3 I2(aq) + 6 e

red: 2 MnO4(aq) + 4 H2O(l) + 6 e 2 MnO2(s) + 8 OH

(aq)

tot: 6 I(aq)+ 2 MnO4

(aq) + 4 H2O(l) 3 I2(aq)+ 2 MnO2(s) + 8 OH

(aq)

Check ReactantCount Element

ProductCount

6 I 6

2 Mn 2

12 O 12

8 H 8

8 charge 8

Practice - Balance the Equation H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O

Practice - Balance the Equation H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O+1 -1 +1 -1 +1 +6 -2 +1 +6 -2 0 +1 -2

oxidationreduction

ox: 2 I- I2 + 2e-

red: H2O2 + 2e- + 2 H+ 2 H2O

tot 2 I- + H2O2 + 2 H+ I2 + 2 H2O

H2O2 + 2 KI + H2SO4 K2SO4 + I2 + 2 H2O

Practice - Balance the EquationClO3

- + Cl- Cl2 (in acid)

Practice - Balance the Equation

ClO3- + Cl- → Cl2 (in acid)

+5 -2 -1 0

oxidationreduction

ox: 2 Cl- → Cl2 + 2 e- } x5red: 2 ClO3

- + 10 e- + 12 H+ → Cl2 + 6 H2O} x1

tot 10 Cl- + 2 ClO3- + 12 H+ → 6 Cl2 + 6 H2O

ClO3- + 5 Cl- + 6 H+ → 3 Cl2

+ 3 H2O

Electrical Current

• current of a liquid in a stream, = amount of water that passes by in a given period of time

• electric current = amount of electric charge that passes a point in a given period of time

whether as e- flowing through a wire or ions flowing through a solution

Redox Reactions & Current

• redox reactions involve the transfer of e- from one substance to another

• therefore, redox reactions have the potential to generate an electric current

• in order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring

Electric Current Flowing Directly Between Atoms

Electric Current Flowing Indirectly Between Atoms

Electrochemical Cells

• electrochemistry is the study of redox reactions that produce or require an electric current

• the conversion between chemical energy and electrical energy is carried out in an electrochemical cell

• spontaneous redox reactions take place in a voltaic cellaka galvanic cells

• nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

Electrochemical Cells

• redox reactions kept separatehalf-cells

• e- flow in a wire along and ion flow in solution constitutes an electric circuit

• requires a conductive metal or graphite electrode to allow the transfer of e-

through external circuit

• ion exchange between the two halves of the systemelectrolyte

Electrodes

• Anodeelectrode where oxidation occursanions attracted to itconnected to positive end of battery in electrolytic cell loses weight in electrolytic cell

• Cathodeelectrode where reduction occurscations attracted to itconnected to negative end of battery in electrolytic cellgains weight in electrolytic cell

electrode where plating takes place in electroplating

Voltaic Cell

the salt bridge is required to complete the circuit and maintain charge balance

Current and Voltage

• # e- that flow through the system per second is the currentunit = Ampere1 A of current = 1 Coulomb of charge per second1 A = 6.242 x 1018 e-/sec.Electrode surface area dictates the number of e- that

can flow

Current and Voltage

• the difference in potential energy between the reactants and products is the potential differenceunit = Volt1 V of force = 1 J of energy/Coulomb of charge the voltage needed to drive electrons through the

external circuitamount of force pushing the electrons through the wire

is called the electromotive force, emf

Cell Potential

• the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential

• cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode

• the cell potential under standard conditions is called the standard emf, E°cell

25°C, 1 atm for gases, 1 M concentration of solutionEcell = EOX + ERED

Cell Notation

• shorthand description of Voltaic cell

electrode | electrolyte || electrolyte | electrode

• oxidation half-cell on left, reduction half-cell on the right

• single | = phase barrier, double line || = salt bridge

if multiple electrolytes in same phase, a comma is used rather than |

often use an inert electrode

Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s)

Standard Reduction Potential

• a half-reaction with a strong tendency to occur has a large + half-cell potential

• two half-cells are connected, e- will flow so that the half-reaction with the stronger tendency will occur

• we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction

• select as a standard half-reaction the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 vstandard hydrogen electrode, SHE

Half-Cell Potentials

• SHE reduction potential is defined to be exactly 0 v

• half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red

• half-reactions with a stronger tendency toward oxidation than the SHE have a value for E°red

• E°cell = E°oxidation + E°reduction

E°oxidation = E°reduction

when adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half-reactions to balance the equation

When E°cell > 0 reaction may be spontaneous

Ex 18.4 – Calculate Ecell for the reaction at 25CAl(s) + NO3

−(aq) + 4 H+

(aq) Al3+(aq) + NO(g) + 2 H2O(l)

Separate the reaction into the oxidation and reduction half-reactions

ox: Al(s) Al3+(aq) + 3 e−

red: NO3−

(aq) + 4 H+(aq) + 3 e− NO(g) + 2 H2O(l)

find the E for each half-reaction and sum to get Ecell

Eox = −Ered = +1.66 v

Ered = +0.96 v

Ecell = Eox + Ered

Ecell = (+1.66 v) + (+0.96 v) = +2.62 v

Ex 18.4a – Predict if the following reaction is spontaneous under standard conditions

Fe(s) + Mg2+(aq) Fe2+

(aq) + Mg(s)

Separate the reaction into the oxidation and reduction half-reactions

ox: Fe(s) Fe2+(aq) + 2 e−

red: Mg2+(aq) + 2 e− Mg(s)

look up the E half-reactions

Ecell = Eox + Ered = +0.45 + -2.37 = -1.92

since Ecell = -ve the reaction is NOT spontaneous as written

[Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written]

the reaction is spontaneous in the reverse direction

Mg(s) + Fe2+(aq) Mg2+

(aq) + Fe(s)

ox: Mg(s) Mg2+(aq) + 2 e−

red: Fe2+(aq) + 2 e− Fe(s)

sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode

Practice - Sketch and Label the Voltaic CellFe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) , Write the Half-

Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.

ox: Fe(s) Fe2+(aq) + 2 e− E = +0.45 V

red: Pb2+(aq) + 2 e− Pb(s) E = −0.13 V

tot: Pb2+(aq) + Fe(s) Fe2+(aq) + Pb(s) E = +0.32 V

Spontaneous

Predicting Whether a Metal Will Dissolve in an Acid

• acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H+

(aq)

•metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid

All have +ve Eox

Ecell = Eox + 0 = +ve

E°cell, ΔG° and K

• for a spontaneous reaction one that proceeds in the forward direction with the

chemicals in their standard statesΔG° < 1 (negative)E° > 1 (positive)K > 1

• ΔG° = −RTlnK = −nFE°cell

n is the number of electronsF = Faraday’s Constant = 96,485 C/mol e−

Example 18.6- Calculate ΔG° for the reactionI2(s) + 2 Br−

(aq) → Br2(l) + 2 I−(aq)

since G° is +, the reaction is not spontaneous in the forward direction under standard conditions

Answer:

Solve:

Concept Plan:

Relationships:

I2(s) + 2 Br−(aq) → Br2(l) + 2 I−

(aq)

G, (J)

Given:

Find:

E°ox, E°red E°cell G°redoxcell EEE

cellFEG n

ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 v

red: I2(l) + 2 e− → 2 I−(aq) E° = +0.54 v

tot: I2(l) + 2Br−(aq) → 2I−

(aq) + Br2(l) E° = −0.55 v

cellFEG n

J 101.1G

55.0485,96 mol 2G

5

C

J

mol

C

ee

E°cell, ΔG° and K

• ΔG° = −RTlnK = −nFE°cell

E°cell = RT x ln K

nF

R = 8.314 J/mol.K

lnK = 2.303log K

F = 96,485 C/mol e-

E°cell = 0.0592 logK

n

125.11 102.310

5.11V 0592.0

mol 2V 34.0log

K

eK

Example 18.7- Calculate K at 25°C for the reactionCu(s) + 2 H+

(aq) → H2(g) + Cu2+(aq)

since < 1, the position of equilibrium lies far to the left under standard conditions

Answer:

Solve:

Concept Plan:

Relationships:

Cu(s) + 2 H+(aq) → H2(g) + Cu2+

(aq)

Given:

Find:

E°ox, E°red E°cell redoxcell EEE K

nlog

V 0592.0Ecell

ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 v

red: 2 H+(aq) + 2 e− → H2(aq) E° = +0.00 v

tot: Cu(s) + 2H+(aq) → Cu2+

(aq) + H2(g) E° = −0.34 v

Kn

logV 0592.0

Ecell

Nonstandard Conditions - the Nernst Equation

• Relationship between Ecell (nonstandard) and E°cell (standard)

ΔG = ΔG° + RT ln Q

• Subs. ΔG° = -nFE°cell into above eqn.

-nFEcell = -nFE°cell + -RTlnQ (divide by -nF)

Ecell = E°cell - (RT/nF) log Q

R = 8.314 J/mol.K, (RT/nF)lnQ = (0.0592/n)logQ

Ecell = E° - (0.0592/n) logQ

Called the Nernst equation

Nonstandard Conditions - the Nernst Equation

Ecell = E°cell - (0.0592/n) log Q at 25°C

1. when Q = 1 (std. conditions) Ecell = E°cell

2. At equilibrium,Q = K,

Ecell = E°cell - (0.0592/n) log K and (0.0592/n) log K = E°cell

Ecell = 0

• Potential reaches zero as concentrations approach equilibrium

• Used to calculate E when concentrations not 1 M

E at Nonstandard Conditions

Reactant conc. > standard conditionsProduct conc. < standard conditions … reaction shifts right

V 41.1E

.0]1[.0]2[

]010.0[log

6

V 0592.0V 34.1E

][H][MnO

]Cu[log

V 0592.0EE

cell

83

3

cell

834

32

cellcell

n

Example 18.8- Calculate Ecell at 25°C for the reaction3 Cu(s) + 2 MnO4

−(aq) + 8 H+

(aq) → 2 MnO2(s) + 3 Cu2+(aq) + 4 H2O(l)

units are correct, Ecell > E°cell as expected because [MnO4

−] > 1 M and [Cu2+] < 1 M

Check:

Solve:

Concept Plan:

Relationships:

3 Cu(s) + 2 MnO4−

(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+

(aq) + 4 H2O(l)

[Cu2+] = 0.010 M, [MnO4−] = 2.0 M, [H+] = 1.0 M

Ecell

Given:

Find:

E°ox, E°red E°cell Ecell

redoxcell EEE Q

nlog

V 0592.0EE cellcell

ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = −0.34 v

red: MnO4−

(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = +1.68 v

tot: 3 Cu(s) + 2 MnO4−

(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+

(aq) + 4 H2O(l)) E° = +1.34 v

Qn

logV 0592.0

EE cellcell

Concentration Cells

• it is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different

• the difference in energy is due to the entropic difference in the solutions the more concentrated solution has lower entropy than the less

concentrated

• electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution oxidation of the electrode in the less concentrated solution will

increase the ion concentration in the solution – the less concentrated solution has the anode

reduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode

when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow

Concentration Cell

Cu2+(aq) + 2e- → Cu(s) 0.34VCu(s) → Cu2+(aq) + 2e- -0.34V

Cu2+(aq) + Cu(s) → Cu(s) + Cu2+(aq)

E°cell = E°red + E°ox = 0V

Cu(s) Cu2+(aq) (1 M) Cu2+

(aq) (1 M) Cu(s)

Concentration Cell

when the cell concentrations are different, e- flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)

Cu(s) Cu2+(aq) (0.010 M) Cu2+

(aq) (2.0 M) Cu(s)

Cell potential Ecell calculated using Nernst eqn.

Ecell = E° - (0.0592/n) logQ

Ecell = E° - (0.0592/n) log([OX]/[RED])

= 0.068V

LeClanche’ Acidic Dry Cell

• electrolyte in paste form ZnCl2 + NH4Cl

or MgBr2

• anode = Zn (or Mg)Zn(s) → Zn2+(aq) + 2 e-

• cathode = graphite rod

• MnO2 is reduced

2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-

→ 2 NH4OH(aq) + 2 Mn(O)OH(s)

• cell voltage = 1.5 v• expensive, nonrechargeable, heavy, easily corroded

Alkaline Dry Cell• same basic cell as acidic dry cell, except electrolyte is alkaline

KOH paste

• anode = Zn (or Mg)

Zn(s) → Zn2+(aq) + 2 e-

• cathode = brass rod

• MnO2 is reduced:

2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-

→ 2 NH4OH(aq) + 2 Mn(O)OH(s)

• cell voltage = 1.54 v

• longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc

Lead Storage Battery

• 6 cells in series

• electrolyte = 30% H2SO4

• anode = Pb

Pb(s) + SO42-(aq) → PbSO4(s) + 2 e-

• cathode = Pb coated with PbO2

• PbO2 is reduced:

PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e-

→ PbSO4(s) + 2 H2O(l)

• cell voltage = 2.09 v

• rechargeable, heavy

NiCad Battery

• electrolyte is concentrated KOH solution• anode = Cd

Cd(s) + 2 OH-(aq) → Cd(OH)2(s) + 2 e- E0 = 0.81 v

• cathode = Ni coated with NiO2

• NiO2 is reduced:

NiO2(s) + 2 H2O(l) + 2 e- → Ni(OH)2(s) + 2OH- E0 = 0.49 v

• cell voltage = 1.30 v

• rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown

Ni-MH Battery

• electrolyte is concentrated KOH solution• anode = metal alloy with dissolved hydrogen

oxidation of H from H0 to H+

M∙H(s) + OH-(aq) → M(s) + H2O(l) + e- E° = 0.89 v

• cathode = Ni coated with NiO2

• NiO2 is reduced:

NiO2(s) + 2 H2O(l) + 2 e- → Ni(OH)2(s) + 2OH- E0 = 0.49 v

• cell voltage = 1.30 v

• rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad

Lithium Ion Battery

• electrolyte is concentrated KOH solution

• anode = graphite impregnated with Li ions

• cathode = Li - transition metal oxide reduction of transition metal

• work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode

• rechargeable, long life, very light, more environmentally friendly, greater energy density

Fuel Cells

• like batteries in which reactants are constantly being added so it never runs down!

• Anode and Cathode both Pt coated metal

• Electrolyte is OH– solution

• Anode Reaction:

2 H2 + 4 OH–

→ 4 H2O(l) + 4 e-

• Cathode Reaction:

O2 + 4 H2O + 4 e- → 4 OH–

Electrolytic Cell

• uses electrical energy to overcome the energy barrier and cause a non-spontaneous reactionmust be DC source

• the + terminal of the battery = anode

• the - terminal of the battery = cathode

• cations attracted to the cathode, anions to the anode

• cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized

• some electrolysis reactions require more voltage than Etot, called the overvoltage

electroplating

In electroplating, the work piece is the cathode.

Cations are reduced at cathode and plate to the surface of the work piece.The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution

Electrochemical Cells• in all electrochemical cells, oxidation occurs at the anode, reduction

occurs at the cathode

• in voltaic cells, anode is the source of electrons and has a (−) charge cathode draws electrons and has a (+) charge

• in electrolytic cells electrons are drawn off the anode, so it must have a place to

release the electrons, the + terminal of the battery electrons are forced toward the anode, so it must have a source of

electrons, the − terminal of the battery

Electrolysis • electrolysis is the process of using electricity

to break a compound apart

• electrolysis is done in an electrolytic cell

• electrolytic cells can be used to separate elements from their compounds generate H2 from water for fuel cells recover metals from their ores

Electrolysis of Water

Electrolysis of Pure Compounds

• must be in molten (liquid) state

• electrodes normally graphite

• cations are reduced at the cathode to metal element

• anions oxidized at anode to nonmetal element

Electrolysis of NaCl(l)

Mixtures of Ions

• when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode least negative or most positive E°red

• when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode least negative or most positive E°ox

Electrolysis of Aqueous Solutions• Complicated by more than one possible oxidation and reduction• possible cathode reactions

reduction of cation to metal reduction of water to H2

2 H2O + 2 e- → H2 + 2 OH- E° = -0.83 v @ stand. cond.E° = -0.41 v @ pH 7

• possible anode reactions oxidation of anion to element oxidation of H2O to O2

2 H2O → O2 + 4e- + 4H+ E° = -1.23 v @ stand. cond.E° = -0.82 v @ pH 7

oxidation of electrodeparticularly Cugraphite doesn’t oxidize

• half-reactions that lead to least negative Etot will occur unless overvoltage changes the conditions

Electrolysis of NaI(aq)

with Inert Electrodes

possible oxidations2 I- → I2 + 2 e- E° = −0.54 v2 H2O → O2 + 4e- + 4H+ E° = −0.82 v

possible reductionsNa+ + e- → Na0 E° = −2.71 v2 H2O + 2 e- → H2 + 2 OH- E° = −0.41 v

possible oxidations2 I- → I2 + 2 e- E° = −0.54 v2 H2O → O2 + 4e- + 4H+ E° = −0.82 v

possible reductionsNa+ + e- → Na0 E° = −2.71 v2 H2O + 2 e- → H2 + 2 OH- E° = −0.41 v

overall reaction2 I−

(aq) + 2 H2O(l) → I2(aq) + H2(g) + 2 OH-(aq)

Faraday’s Law

• the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runscharge that flows through the cell = current x time

Example 18.10- Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction

Au3+(aq) + 3 e− → Au(s)

units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−

Check:

Solve:

Concept Plan:

Relationships:

3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min

mass Au, g

Given:

Find:

s 1

C 5.5

Au g 6.5

Au mol 1

g 196.97

mol 3

Au mol 1

C 96,485

mol 1

s 1

C 5.5

min 1

s 60min 25

e

e

t(s), amp charge (C) mol e− mol Au g Au

C 6,4859

mol 1 ee mol 3

Au mol 1

Au mol 1

g 196.97

Corrosion

• corrosion is the spontaneous oxidation of a metal by chemicals in the environment

• since many materials we use are active metals, corrosion can be a very big problem

Rusting

• rust is hydrated iron(III) oxide• moisture must be present

water is a reactant required for flow between cathode and anode

• electrolytes promote rustingenhances current flow

• acids promote rusting lower pH = lower E°red

Preventing Corrosion

• one way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment paint some metals, like Al, form an oxide that strongly attaches to the

metal surface, preventing the rest from corroding

• another method to protect one metal is to attach it to a more reactive metal that is cheap sacrificial electrode

galvanized nails

Sacrificial Anode