Ch. 17-Galvanic Cells

5/27/2018 Ch. 17-Galvanic Cells

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Electrochemistry-study of interchange of

chemical and electricalenergy

Generating electric current from spontaneous chemical

reactions and use of current to produce chemicalchange

Labs

#31 The Thermodynamics of the Dissolution of Borax#33 Electrolytic Cells; Avogadros Number

Chemical Equations

Chapter 12


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Oxidation-Reduction Reactions

(Redox Reactions)

Oxidation: (LEO)

Loss of electrons-atoms or ions undergo increasein oxidation state (is oxidized)

Electrons given to another atom which is beingreduced (reducing agentor reductant)

Reduction: (goes GER)

Gain of electrons-atoms/ions undergo decrease inoxidation state (is reduced)

Takes electrons away from another atom which isbeing oxidized (oxidizing agentor oxidant)


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Redox reaction can be written as two half-reactions(one reduction, one oxidation)

To generate current

Separate oxidizing agent from reducing agent Electron transfer occurs through wire

Electron transfer directed through device toprovide useful work

Reactants separated by salt bridge/porouspartition

Each half-reaction has a potential, or voltage,associated with it

Given as reduction half-reactions Read in reverse and change sign on voltage to

get oxidation potentials


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Electrochemical cells

device associated with redox reaction

(galvanic cells, electrolytic cells)

Electrochemical process involves electron

transfer at interface between electrode and

solution

Species undergoing reduction receive

electrons from cathode

Species in solution act as oxidizing agent

Species undergoing oxidation donate electronsto anode

Species in solution act as reducing agenthttp://college.hmco.com/chemistry/shared/media/animations/anodereaction.html

http://college.hmco.com/chemistry/shared/media/animations/cathodereaction.html
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Galvanic Cell (voltaic cell)

Chemical energyelectrical energy

Harnesses energy of spontaneous redox reactions

Physically separate chemicals in 2 half-reactions

Electrons generated by oxidation half-reaction flowthrough electrical conductor before being used inreduction half-reaction

Flow diverted through meters, motors, light bulbs toperform useful work before reaching destination

Current (defined by physicists as flow of positivecharge)always in opposite direction from flow ofelectrons (always from anode to cathode)


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Electrodes are metal strips

Sign of electrodes determined by

Since electrons flow out of anode and into

external circuit, anode is negative

Since electrons flow from external circuit into

cathode, cathode is positive Opposite is true for electrolytic cells

Where reactions occur (The Red Cat ate

An Ox) Oxidation occurs at anode (AN OX)-on left

Reduction occurs at cathode (RED CAT)-on

right


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Porous barrier separate two

compartments

Allows for migration of

positive/negative ions

between half-cells,

completing electriccircuit

Problem with porous

barriers

Inside barriers, ionicsolutions mix

Has effect on operation

of cell

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http://college.hmco.com/chemistry/shared/media/animations/electrochemicalhalf.htmlhttp://college.hmco.com/chemistry/shared/media/animations/electrochemicalhalf.html


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Galvanic cells with salt bridges

Inverted tube contains electrolyte Gel (agar) added to provides

firmness but permits ion flow

Prevents two reacting solutions from mixing

Ions dont react with other ions in cell or with

electrode material Maintains electrical neutrality in system

Provides - ions to equal + ions created at anode(during oxidation) and + ions toreplace - ions being used up at

cathode (during reduction) Anions always migrate toward anode

Cations toward cathode


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Electron Flow


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The ANODE... The CATHODE...

Supplies electronsto external

circuit (wire)

Accepts electronsfrom external

circuit (wire)

Is negativepole of battery Is positivepole of battery

Is site of OXIDATION Is site of REDUCTION

Is written on left- hand sideif

convention is followed

Is written on right-hand sideif

convention is followed

Is half-cell with lowestelectrode

potential

Is half-cell with highestelectrode

potential


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Cell Potential (Ecell) or

Electromotive force (emf)

Force with which electrons flow from - electrode(anode on left) to + electrode (cathode on right)through external wire

Due to PE difference of electrons before/after transfer

In electrochemical cell, electric potential createdbetween two dissimilar metals

Greater tendency or potential of two half-reactions tooccur spontaneously, greater emf of cell

Measured in volts (V-why called cell voltage)

1 V = 1 J/coulomb(of charge transferred)

Measured with voltmeter which draws current throughknown resistance (heat is produced)


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Voltage of voltaic cells All based on spontaneous chemical reactions

G must always be negative

Voltage of voltaic cell is always positive (+EMF) Subtract smaller reduction potential from larger one

Same as EMF = cathodeanode

Under standard conditions, voltage of cell is same

as total voltage of redox reaction

Standard emf of standard cell potential (E0cell)

Under nonstandard conditions, cell voltage

computed by using Nernstequation As galvanic cell operates

Redox reaction of cell approaches equilibrium

Capacity to deliver useful electrical energy decreases

At equilibrium, cell ceases to function (dead battery)


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Standard Reduction

PotentialsCell potentials can be

measured

Half-cell potential cannot4/7/2014 14


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Measuring Potential

Galvanic cell understandard conditions madeusing arbitrary standardhydrogen electrode(SHE) and test half-cell

w/different half reaction Assigned standard electrodepotential of exactly 0.00 V

Half reaction alwayswritten as reduction

Eo

values corresponding toreduction half-reactions withall solutes standardreduction potentials-Eocell

Under ideal conditions whereideal behavior is assumed


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Standard Reduction Potentials All solutions are 1M, gases at 1 atm, T 25oC

Write oxidation/reduction half-reactions for cell Look of reduction potential (Eoreduction) for reduction half-

reaction in table

Half reaction w/higher reduction potential

Look of reduction potential for reverse of oxidation half-

reaction and reverse sign (Eooxidation= -Eoreduction) Half reaction w/lower reduction potential/sign reversed

Add potentials to get overall standard cell potential

Two half reactions are balanced for # electronsexchanged but value of each Eoremains unchanged(intensive property-does not depend on how manytimes reaction occurs)

If sum positive, reaction is spontaneous/runs on own(always positive for electrochemical cells)

If sum negative, energy supplied to make reaction go


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Each half-reactionassociated w/signednumerical value

More positive it is,greater oxidizing powerof redox half-reaction

More negative it is,greater reducing powerof reverse redox half-reaction

Larger differencebetween Ered values,larger Ecell


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(1 atm)


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Line Notation

Anodewritten first on left/cathodeon right Reactants written 1ston each side

Vertical baris boundary between two phases

If both substances in same phase, separated

by comma, not vertical bar Double line represents salt bridge or porous disk

When platinum electrode present, placed at leftand/or right end of cell diagram


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Using the table of standard reductions provided write

the equation for the reaction between the following two

half cells, and determine its voltage

Au (s) | Au 3+(aq) ||Cu 2+ (aq) | Cu (s)

Gold is higher on table and written as found on table

Au 3+(aq) + 3 e -Au (s) Eo= + 1.50 V

Copper is lower so it's written as oxidation (sign reversed)

Cu (s)Cu 2+(aq) + 2 e- Eo= -0.34 V Equation balanced for exchange of electrons (each needs 6)

2 Au 3+(aq) + 6e -2 Au (s) E = + 1.50 V

3 Cu (s)3 Cu 2+(aq) + 6e E = -0.34 V ***Size of values for Eoof reactions not changed ***

Add two half reactions and E0values

2 Au 3+(aq) + 3 Cu (s)2 Au (s) + 3 Cu 2+(aq)

E = + 1.50 V + ( - 0.34 V ) = 1.16 V


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Place the following in order of increasing

strength as oxidizing agents:

-0.44 0.954 2.87 0.22


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Describe a galvanic cell based

on the two half-reactions below:


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Homework:

Read 17.1-17.2, pp.827-837

Q pp. 867-869, #14 a/b/c/f/i/k, 16 a/b/d/f,

26a, 28 (a only), 30b, 32 (b only), 34a, 36b


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Cell Potential,Electrical Work,

and Free Energy


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Complete Description of a

Galvanic Cell Electrical energy delivered by galvanic cell equal to

quantity of useful work obtained as result of cell operation Work, w, measured in relation to amount of charge, q, transferred

between anode/cathode of cell

This quantity, potential difference, E, is defined as E = w/q.

SI unit is joule per coulomb or volt (V)

Cell potential (always positive for galvanic cell)

Direction of electron flow (direction that yields + potential)

Designation of anode/cathode

Nature of each electrode/ions present in compartments Chemically inert conductor (such as Pt) required if only ions are

present (no substance in reaction is conducting solid)


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Electric work and cell potential

Free energy change occurring during chemicalreaction is measure of maximum work thatsystem can perform

Potential (E) = -Work (w) / Charge (q) So w = -qE

Work leaving system has negative charge

Faraday (F) = charge in coulombs per mole of

electrons (96,485 C/mol e-) Then q = nF and w = -nFE n = number of moles of electrons


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G (Gibbs free energy) is measure of spontaneity of

process occurring at constant T/P

emf, E, of redox reaction also indicates whetherreaction is spontaneous

From thermodynamics we know that

G =UTS +(PV) and U = heat + w

Therefore, at constant T/P:G = w

Therefore:G = -nFE and at standard state:G0= -nFE0

(relationship between emf/free energy changes )

Go = Standard Gibbs free energy change (kJ/mol or Joules)

n = moles of electrons exchanged in reaction (mol)

F = Faradays constant, 96,485 coulombs/mole (1 mole of

electrons has a charge of 96,485 coulombs)

Eo= Standard reaction potential (V or Joules/Coulomb)


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Since both n/F are + + value of E leads tovalue ofG,

which indicates spontaneous reaction IfG and E have opposite signs, E

predicts direction of reaction If Eois positive, Gois negative (< 0)-

reaction spontaneous (has positive cellpotentials)

If Eois negative, Gois positive (> 0)-reaction is nonspontaneous (but isspontaneous in reverse direction)


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Relationship between thermodynamics (push

behind electrons) and electrochemistry

Relationship between reaction potential and free

energy for a redox reaction is given by

Emf = potential difference (V) = work (J)

charge (C)

Driving force (emf) is defined in terms of potential difference(in V) between two points in circuit

One coulomb is amount of charge that moves past any given

point in circuit when current of 1 ampere (amp) is supplied for

one second (1 ampere = 1 coulomb/sec)

Faradays law states that during electrolysis, passage of 1

faraday through circuit brings about oxidation of one

equivalent weight of substance at one electrode (anode) and

reduction of one equivalent weight at other electrode

(cathode)


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emf is not converted to work with 100% efficiency Energy always lost as heat, but wmaxuseful for calculating

efficiency of conversion wmax = -qEmax

Relationship to free energy (energy driving reactiondue to movement of charged particles giving rise topotential difference)

wmax= EG

EG = -qEmax= -nFEmax

EG = -qEmax

EG0= -nFE0 When Ecellpositive (spontaneous), EG will be negative

(spontaneous), so there is agreement

Standard cell potential, Eocell, measured and standardelectrode potential of test half-cell determined by using

Eocell= Eocathode- Eoanode Eocathode-standard reduction potential for reaction occurring at

cathode, represents tendency to remove es from electrodesurface

Eoanode-standard reduction potential for reaction occurring atanode and represents its tendency to remove es from anode


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Dependence of CellPotential on

ConcentrationCell voltages at

nonstandardconcentrations (not 1 M)


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Nernst Equation-

way to relate E0at standard conditions and E,

potential at any real condition

Standard state impossible to achieve in

reality

As soon as wire hooked to two half-cells,

reaction proceeds and changes

concentrations of all reactants and

products

Heating or cooling makes reaction deviatefrom standard temperature


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From thermodynamics, recall G = Go+ RTlnQ

If we divide everything bynF

sinceG =nFE, or E =G/(nF)

Ecell= cell potential at non-standard conditions

E0cell= standard reduction potential

R = 8.314 J/molK (the gas constant) F = 96485 coul/mol (Faraday's constant)

T = absolute temperature

n = number of moles of electrons transferred in balanced equation

Q = reaction quotient for reaction aA + bBcC + dD

Expressed in terms of base 10 rather than ln (standardconditions of 298K)

Can be used to find cell potential at any set of conditions

Cells spontaneously discharge until they achieve

equilibrium (at equilibrium, cell is dead)

Calculating Nernst Equation


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Consider the Daniell Cell at 25 C

Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) Find cell potential at following conditions

when [Cu2+] = 1.00 M, [Zn2+] = 1.0109M and when [Cu2+] = 0.10 M, [Zn2+] =0.90 M.

Recallthat standard potential for Daniell

cell is Eo

= +1.10 V Nernst equation used to find potentials at

nonstandard conditions:

When [Cu2+] = 1.00 M, [Zn2+] = 1.0109M

When [Cu2+] = 0.10 M, [Zn2+] = 0.90 M


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Depiction of concentration cell

E E

n

Q 0.0592

log


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Cell in which current flows due only to difference

in concentration of ion in 2 different

compartments of cell Le Chteliers principle used to determine effect

on potential

Shift to left reduces potential

Shift to right increases potential

If concentrations are different, stress is put on

system that will be equalized by electron flow to

allow reduction and oxidation to occur

Voltages typically small

When concentrations in half-cells become equal,

E0cell= 0 and system is at equilibrium


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Calculate the EMF of the cell Zn(s) |

Zn2+(0.024 M) || Zn2+(2.4 M) | Zn(s)

Zn2+(2.4 M) + 2 e = Zn Reduction

Zn = Zn2+(0.024 M) + 2 e Oxidation

Zn2+(2.4 M) = Zn2+(0.024 M), DE = 0.00

Using Nernst equation:

(0.024) DE= 0.00 - 0.0592/2 log (0.024/(2.4) = (-0.296)(-2.0)

= 0.0592 V

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Show that voltage of electric cell is


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Show that voltage of electric cell is

unaffected by multiplying reaction

equation by positive number

Mg | Mg2+|| Ag+| Ag

Mg + 2 Ag+= Mg2++ 2 Ag

DE = DEo0.0592/2 log [Mg2+]/[Ag+]2

2 Mg + 4 Ag+= 2 Mg2++ 4 Ag

DE = DEo0.0592/4 log [Mg2+]2/[Ag+]4

Simplified to the 1stequation, showing cell

potential DE not affected

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Calculation of Equilibrium


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Calculation of EquilibriumConstants for Redox

Reactions The standard reaction potential is related to theequilibrium constant

At equilibrium, Ecell= 0 and Q = K

As cells discharge, concentration changes, Ecell changes.

For a cell at concentrations and conditions other than

standard, a potential can be calculated using the Nernst

equation

If Eois positive, then K > 1 and forward reaction favored

If Eois negative, then K < 1 and reverse reaction is favored

log( ).

K nE

0 0592at 25 C


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The standard cell potential dE for the reaction


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pFe + Zn2+= Zn + Fe2+is -0.353 V. If a piece of

iron is placed in a 1 M Zn2+solution, what is the

equilibrium concentration of Fe2+?

Equilibrium constant Kmay be calculated

using K= 10(nDE)/0.0592

= 10-11.93

= 1.2x10-12

= [Fe2+]/[Zn2+]. Since [Zn2+] = 1 M, it isevident that [Fe2+] = 1.2-12 M

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Homework:

Read 17.3-17.4, pp. 837-846

Q pp. 869-871, #38, 40, 46, 48, 54, 60, 66,

70


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Batteries Portable, self-contained electrochemical power source

(DC) consisting of one or more voltaic cells, connected in

series Greater voltages achieved by using multiple voltaic cells

in single battery (12V)

When connected in series, battery produces

voltage that is sum of emfs of individual cells Higher emfs achieved by using multiple batteries in

series

Electrodes marked + (cathode) and(anode)


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Lead-acid storage battery

As battery discharged, uses up sulfateions/electrodes become coated w/lead sulfate

Reverse reactions regenerate sulfate ion insolution/reduce amount of lead sulfate

contaminating electrode surfaces Each pair produces ~2 volts (6 pairs ofelectrodes used in 12-volt car battery)

When jump starting car, connect ground cable

on dead car to metallic contact away frombattery. Otherwise, could explode

Causes electrolysis of water/production ofH2/O2which could ignite


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C D C ll B tt


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Common Dry Cell Battery

(acid version)

Zinc-anode

Carbon rod in

contact with moist

paste-cathode


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Alkaline dry cell

NH4Cl replaced w/KOH or NaOH

Last longer than acid cells because zinc (anode)corrodes more slowly in basic environment

Cathode (graphite rod) inserted into paste made of

manganese dioxide, water and potassium hydroxide Zn(s) + 2OH-(aq)ZnO(s) + H2O + 2e- (anode)

2MnO2(s) + H2O + 2e-Mn2O3(s) + 2OH-(aq)(cathode)

Total voltage is 1.54 volts Not rechargeable


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Mercury Battery

Fuel Cells


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Fuel Cells Galvanic cells where

reactants continuouslysupplied

Energy normally lost asheat is captured andused to produce anelectric current

Redox reaction Hydrogen oxidized at

anode and oxygenreduced at cathode toform water and electricity

2x as effective as gas, oilor coal-poweredgenerators in convertingchemical energy intoelectricity

Type of Battery Advantages Disadvantages


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Lead acid Rechargeable

Long life

12 V

Heavy

Contain acid

Weather issues

Alkaline/Dry cell InexpensiveNo toxic metals used

Lots of power (1.5 V down to 1.2 V)

Heavier

Lithium (solid state) Lightweight

Higher capacity (3.6-3.9 V)

Rechargeable

Longer lastingLess likely to leak/explode

Expensive

Fuel cells No rechargingNo harmful pollutants

high initial cost

Fuel not readily

available

Nickel-Cadmium (Ni-Cd) Fast/simple charge (rechargeable)High # charges

1.2 V continuously

Toxic metalsExpensive

Mercury More constant voltage (1.35 V)

Longer life

Lighter

More expensive


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Corrosionoxidation of metal

Oxidation of most metals by oxygen isspontaneous redox reactions

Many metals develop thin coating ofmetal oxide on outside that preventsfurther oxidation

Some metals, such as copper, gold,silver and platinum (noble metals), arerelatively difficult to oxidize

C i f I


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Corrosion of Iron Anodic regions

Regions of steel alloy where iron is more easilyoxidized

FeFe2+ + 2e-

Cathodic regions

Areas resistant to oxidation

Electrons flow from anodic regions & react

w/oxygen

O2+ 2H

2O + 4e- 4OH-

Presence of water essential to iron corrosion

Presence of salt accelerates corrosion by increasing

electron conduction from anodic to cathodic regions


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Prevention of Corrosion

Coating w/metal that forms oxide coat to protect metalthat would not develop protective coat

Galvanizing Place sacrificial of more easily oxidized metal on top of metal

to protect Zinc over iron

Alloying Addition of metals that change steels reduction potential.

Nickel and chromium alloyed to iron

Cathodic Protection


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Cathodic Protection

Connection of easily oxidized metals (an

anode) to less easily oxidized metals keeps

less from experiencing corrosion

Anode corrodes-must be replaced periodically

Magnesium as anode to iron pipe

Titanium as anode to steel ships hull


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Electrolysis

Decomposition of substanceby electric current

Galvanic Cell Electrolytic Cell

Cathode + (reduction) - (reduction)Anode - (oxidation) + (oxidation)

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Electrolytic cells

Nonspontaneous reactions

Electrical energy required to induce reaction Two electrodes immersed in electrically conductive sample

Electrical voltage (>1.10 V) applied to them

Voltage increased until electrons flow in opposite direction

(electrolytic)

At cathode-reduction occurs (RED CAT)

At anode-oxidation occurs (AN OX)

Electrical energy is converted into chemical energy

Electrolytic cells are used for electroplating

(a) Standard galvanic cell based on spontaneous


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( ) g

reaction

Zn + Cu2+Zn2++ Cu

(b) Standard electrolytic cell: Power source forcesopposite reaction

Cu + Zn2+Cu2++ Zn

What voltage is necessary to force the


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What voltage is necessary to force the

following electrolysis reaction to occur?

2I-(aq) + Cu2+(aq)I2(s) + Cu(s) Which process would occur at the anode? Cathode?

Assuming the iodine oxidation takes place at a platinum

electrode, what is the direction of electron flow in this

cell?


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Calculating Quantity of Substance


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Calculating Quantity of Substance

Produced or Consumed

To determine quantity of substance either

produced or consumed during electrolysis

given time known current flowed

Write balanced half-reactions involved

Calculate number of moles of electrons that

were transferred

Calculate number of moles of substance thatwas produced/consumed at electrode

Convert moles of substance to desired units of

measure4/7/2014

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A 40.0 amp current flowed through molten iron(III) chloride for 10.0 hours


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(36,000 s). Determine the mass of iron and the volume of chlorine gas

(measured at 25oC and 1 atm) that is produced during this time.

Write half-reactions that take place at anode/cathode

anode (oxidation): 2 Cl- Cl2(g) + 2 e

cathode (reduction) Fe3++ 3 e- Fe(s)

Calculate number of moles of electrons

Calculate moles of iron/chlorine produced using number of moles ofelectrons calculated and stoichiometries from balanced half-reactions. (3

moles electrons produce 1 mole of Fe/2 moles of electrons produce 1

mole of chlorine gas)

Calculate mass of iron using molar mass and calculate volume of

chlorine gas using ideal gas law (PV = nRT).

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Calculating Time Required

To determine quantity of time required to

produce known quantity of substance given

amount of current that flowed

Find quantity of substance produced/consumedin moles

Write balanced half-reaction involved

Calculate number of moles of electrons required Convert moles of electrons into coulombs

Calculate time required

How long must a 20.0 amp current flow


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g p

through a solution of ZnSO4in order to

produce 25.00 g of Zn metal

Convert mass of Zn produced into moles using molar massof Zn

Write the half-reaction for the production of Zn at the

cathode Zn2+(aq) + 2 e- Zn(s)

Calculate moles of e-required to produce moles of Zn usingstoichiometry of the balanced half-reaction (2 moles of

electrons produce 1 mole of zinc)

Convert moles of electrons into coulombs of charge using

Faraday's constant

Calculate time using current and coulombs of charge

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Calculating Current Required

To determine amount of current necessary to

produce known quantity of substance in

given amount of time

Find quantity of substance produced/orconsumed in moles

Write equation for half-reaction taking place

Calculate number of moles of electrons required Convert moles of electrons into coulombs of

charge

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What current is required to produce 400.0 L of


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hydrogen gas, measured at STP, from the electrolysis

of water in 1 hour (3600 s)?

Calculate number of moles of H2 Write equation for half-reaction that takes place. Hydrogen

produced during reduction of water at cathode. Equation for

this half-reaction is 4 e-+ 4 H2O(l) 2 H2(g) + 4 OH-(aq)

Calculate number of moles of electrons (4 mole of e-

required to produce 2 moles of hydrogen gas, or 2 moles of

e-'s for every one mole of hydrogen gas)

Convert moles of electrons into coulombs of charge

Calculate current required

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75

How many grams of copper can be reduced by


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applying a 3.00 A current for 16.2 min to a

solution containing Cu2+ions?

TimecurrentCoulombsmoles e-moles Cug Cu

16.2 min 60 sec 3 C 1 mol e- 1 mol Cu 63.54 g Cu =

1 min 1 sec 96,486 C 2 mol e- 1 mol Cu

0.96 g Cu

Electrolysis can be used to separate mixture of


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Electrolysis can be used to separate mixture of

ions, if reduction potentials are fairly far apart

Remember, metal ion with highestreduction potential is easiest to reduce

Predict order of reduction and which of

following ions will reduce first at cathode ofelectrolytic cell: Ag+, Zn2+, IO3

-


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Electroplating

Deposit neutral metal atoms onelectrode by reducing metal

ions in solution

One metal coated with another

Presence of active electrodethat takes part in electrolysis

reaction

Anode-piece of plating metal

Cathode-object to be plated Plating solution is NiSO4because

SO42-ion does not participate in

plating reaction

How long must a current of 5.00 A be


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applied to a solution of Ag+ to produce

10.5 g silver metal?

10.5 g Ag 1 mol Ag 1 mol e- 96,485 C 1 sec 1 min =

107.868 g Ag 1 mol Ag 1 mol e- 5 C 60 sec

31.3 min

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Electrolysis of Water Requires soluble salt/dilute

acid to serve as electrolyte

Anode

2H2O(l)O2(g) + 4H+(aq) + 4e-

Eoox= -1.23 V

Cathode

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

Eored= -0.83 V

Overall reaction

6H2O(l)2 H2(g) + O2(g) +

4H+(aq) + 4e- Eocell= -2.06 V

If in single container, H+/OH-

combine to yield 4 additional

water molecules

What volume of H2(g) and O2(g) is produced by


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electrolyzing water at a current of 4.00 A for

12.0 minutes (assuming ideal conditions)?

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

Actual ratio is not exactly 2:1 for a variety

of reasons including oxygen solubility.12 min 60 sec 4 C 1 mol e- 1 mol H2 22.4 L =

1 min 1 sec 96,486 C 2 mol e- 1 mol H2

0.334 L H20.167 L O2

El t l i f lt lt


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Electrolysis of molten salts

NaCl Good conductor as

liquid

Melting salt freesions

Makes it electricallyconductive

Ions of oppositecharge migrate tothese electrodesand react

Electrolysis of aqueous solutions


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Electrolysis of aqueous solutions

Aqueous solutions of salts are electrically

conductive and can be electrolyzed For solutions, two possible reactions occur

(water can be both oxidized and reduced)

At cathode If metal ion is very active metal, water will be

reduced (2H2O + 2e-H2+ 2OH-)

If metal ion is inactive or active metal, metal ion will

be reduced At anode

Oxidation of salts anion or ()

Oxidation of water (2H2OO2+ 4H+ + 4e-)

To determine which will occur at anode


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If anion is polyatomic ion, it generally will

not be oxidized

SO42-/NO3

-/ClO4-not oxidized in aqueous

solution

Cl-/Br-/l-will be oxidized in aqueous solution

If anion in one salt is oxidized in aqueouselectrolysis, that same anion in any other

salt will also be oxidized

If solution of NaBr results in Br- being oxidized

to Br2, predict that solutions of KBr, CaBr2,NH4Br and AlBr3will all produce Br2at anode

Commercial


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Commercial

ElectrolyticProcesses

Since metals are easily oxidized,most found as ores, mixtures of

ionic compounds. Au, Ag, and Ptare more difficult to oxidize, so often

found as pure metals.

Abundance of elements on earth:

1st Oxygen

2nd Silicon

3rd Aluminum (very active metal so difficult

and expensive originally to purify)

Production of aluminum from molten-saltl t l i


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electrolysis

(purification of aluminum from bauxite ore)

Hall-Heroult process

Electrorefining (purifying) metals


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Electrorefining (purifying) metals

Copper ore is refined by roasting

Impure copper is anode Small strip pure copper is cathode

During electrolysis, copper is oxidized to

Cu2+at anode and then reduced to copper

metal again at cathode

Impurities such as silver and gold drop to

bottom as sludge which is then salvaged

Metal Plating


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Metal Plating

Electroplating thin layers of decorative metal

on less expensive metal (silver and gold ontoiron, chromium on to car parts for decoration

and resistance to corrosion)

Electrolysis of concentrated aqueous sodium chloride

l i (b i ) d h d d h d id


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solutions (brine) produces hydrogen and hydroxide

ions at cathode and chlorine gas at anode

If electrodes separated byporous membrane, H2,

NaOH, and Cl2 produced

If solution stirred, chlorine

gas reacts with sodiumhydroxide to form sodium

hypochlorite (NaOCl)

solution (bleach)

Electrolysis of moltenNaCl produces sodium

metal and chlorine gas

(Downs cell)

H k


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Homework:

Read 17.5-17.8, pp. 846-866Q pp. 871-872, #74, 76, 78, 80, 84, 88 a (dont forget water)

Do 1 additional exercise and 1 challenge problem

Submit quizzes by email to me:

http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xml

http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xml

http://www.cengage.com/chemistry/book_content/0547125321 zumdahl/ace/launch ace html?folder path=/chemistry
http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace3.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace3.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace3.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace3.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch17_ace1.xml

Documents

Ch. 17-Galvanic Cells