ElectrochemistElectrochemistryry

Electrochemistry Terminology Electrochemistry Terminology #1#1

OxidationOxidation – A process in which an element attains a more positive oxidation state

Na(s) Na+ + e-

ReductionReduction – A process in which an element attains a more negative oxidation state

Cl2 + 2e- 2Cl-

Electrochemistry Terminology Electrochemistry Terminology #2#2

GGainain EElectronslectrons = = RReductioneduction

An old memory device for An old memory device for oxidation and reduction goes oxidation and reduction goes like this…like this… LEOLEO says says GERGER

LLoseose EElectronslectrons = = OOxidationxidation

Electrochemistry Terminology Electrochemistry Terminology #3#3

Oxidizing agentOxidizing agentThe substance that is reduced is

the oxidizing agent Reducing agentReducing agent

The substance that is oxidized is the reducing agent

Electrochemistry Terminology Electrochemistry Terminology #4#4

Anode Anode The electrode

where oxidation occurs

CathodeCathodeThe electrode

where reduction occurs

Memory Memory device:device:

RedReductionuctionat the at the

CatCathodehode

Table of Table of Reduction Reduction PotentialsPotentials

Measured Measured against against

the the StandardStandardHydrogenHydrogenElectrodeElectrode

Measuring Measuring Standard Standard Electrode Electrode PotentialPotential

Potentials are measured against a Potentials are measured against a hydrogen ion reduction reaction, which hydrogen ion reduction reaction, which is arbitrarily assigned a potential of is arbitrarily assigned a potential of zero zero voltsvolts..

Galvanic (Electrochemical) Galvanic (Electrochemical) CellsCells

Spontaneous redox processes

have:A positive cell potential, E0

A negative free energy change, (-G)

Zn - Cu Zn - Cu Galvanic Galvanic

CellCell

Zn2+ + 2e- Zn E = -0.76V

Cu2+ + 2e- Cu E = +0.34V

From a From a table of table of reduction reduction potentials:potentials:

Zn - Cu Zn - Cu Galvanic Galvanic

CellCell

Cu2+ + 2e- Cu E = +0.34V

The less positive, The less positive, or more negative or more negative reduction reduction potential potential becomes the becomes the oxidation…oxidation…

Zn Zn2+ + 2e- E = +0.76V

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Line Line NotationNotation

Zn(Zn(ss) | Zn) | Zn2+2+((aqaq) || Cu) || Cu2+2+((aqaq) | Cu() | Cu(ss))

An abbreviated An abbreviated representation representation of an of an electrochemical electrochemical cellcell

AnodeAnodesolutionsolution

AnodeAnodematerialmaterial

CathodeCathodesolutionsolution

CathodeCathodematerialmaterial|| ||||||

Calculating Calculating GG00 for a Cell for a Cell

).)()((Coulomb

Joules

emol

coulombsemolG 1014859620

GG00 = -nFE = -nFE00

nn = moles of electrons in balanced redox equation = moles of electrons in balanced redox equation

FF = Faraday constant = 96,485 coulombs/mol e = Faraday constant = 96,485 coulombs/mol e--

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

kJJoulesG 2122122670

The Nernst EquationThe Nernst Equation

)ln(0 QnF

RTEE

Standard potentials assume a Standard potentials assume a concentration of 1 M. The Nernst concentration of 1 M. The Nernst equation allows us to calculate potential equation allows us to calculate potential when the two cells are not 1.0 M.when the two cells are not 1.0 M.

RR = 8.31 J/(mol = 8.31 J/(molK) K)

TT = Temperature in K = Temperature in K

nn = moles of electrons in balanced redox equation = moles of electrons in balanced redox equation

FF = Faraday constant = 96,485 coulombs/mol e = Faraday constant = 96,485 coulombs/mol e--

Nernst Equation SimplifiedNernst Equation Simplified

)log(0591.00 Qn

EE

At 25 At 25 C (298 K) the Nernst Equation C (298 K) the Nernst Equation is simplified this way:is simplified this way:

Equilibrium Constants and Cell Equilibrium Constants and Cell PotentialPotential

At equilibrium, forward and reverse reactions occur at equal rates, therefore:1.1. The battery is “dead”

2. The cell potential, E, is zero volts

)log(0591.0

0 0 Kn

Evolts

Modifying the Nernst Equation (at 25 C):

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Calculating an Equilibrium Calculating an Equilibrium Constant from a Cell PotentialConstant from a Cell Potential

)log(2

0591.010.10 Kvolts

)log(0591.0

)2)(10.1(K

)log(2.37 K

372.37 1058.110 xK

ConcentratioConcentration Celln Cell

Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

Both sides have the same

components but at different

concentrations.

??????

ConcentratioConcentration Celln Cell

Both sides have the same

components but at different

concentrations.

The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentrationZn2+ (1.0M) + 2e- Zn (reduction) Zn Zn2+ (0.10M) + 2e-

(oxidation)

??????

CathodeCathodeAnodeAnode

Zn2+ (1.0M) Zn2+

(0.10M)

Concentration CellConcentration Cell

)log(0591.00 Qn

EE

Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C).

Both sides have the same

components but at different

concentrations.

??????

CathodeCathodeAnodeAnode

Zn2+ (1.0M) Zn2+

(0.10M)

ConcentratioConcentration Celln Cell

VoltsE 0.00 )0.1(

)10.0(Q2n

VoltsE 030.0)0.1

10.0log(

2

0591.00.0

Nernst CalculationsNernst Calculations

Zn2+ (1.0M) Zn2+

(0.10M) )log(

0591.00 Qn

EE

ElectrolytElectrolytic ic

ProcesseProcessess

A negative cell potential, (-E0)

A positive free energy change, (+G)

Electrolytic processes are NOT spontaneous. They have:

Electrolysis of Electrolysis of WaterWater

eHOOH 442 22 OHHeOH 4244 22

In acidic solutionIn acidic solution

Anode rxn:Anode rxn:

Cathode rxn:Cathode rxn:-1.23 V-1.23 V

-0.83 V-0.83 V

-2.06 V-2.06 V222 22 OHOH

Electroplating Electroplating of Silverof Silver

Anode reactionAnode reaction::

Ag Ag Ag Ag++ + e + e--

Electroplating requirementsElectroplating requirements::1. Solution of the plating metal1. Solution of the plating metal

3. Cathode with the object to be plated3. Cathode with the object to be plated2. Anode made of the plating metal2. Anode made of the plating metal

4. Source of current4. Source of current

Cathode reactionCathode reaction::

AgAg++ + e + e-- Ag Ag

Solving an Electroplating Solving an Electroplating ProblemProblem

Q: How many seconds will it take to Q: How many seconds will it take to plate out 5.0 grams of silver from a plate out 5.0 grams of silver from a solution of AgNOsolution of AgNO33 using a 20.0 Ampere using a 20.0 Ampere current?current?

5.0 g

AgAg++ + e + e- - Ag Ag

1 mol Ag

107.87 g

1 mol e-

1 mol Ag

96 485 C

1 mol e-

1 s

20.0 C

= 2.2 x 10= 2.2 x 1022 s s