electrochemistry

ElectrochemistryElectrochemistryThe Study of the Interchange The Study of the Interchange

of Chemical and Electrical of Chemical and Electrical EnergyEnergy

Galvanic or Voltaic CellsGalvanic or Voltaic CellsThe reactants and products of some The reactants and products of some oxidation-reduction reactions can be oxidation-reduction reactions can be physically separated so that the electron physically separated so that the electron transfer can only occur via a wire. transfer can only occur via a wire. The device or apparatus that is used to The device or apparatus that is used to convert chemical energy to electrical convert chemical energy to electrical energy is called energy is called galvanic (or voltaic) galvanic (or voltaic) cell.cell.

Galvanic or Voltaic CellsGalvanic or Voltaic CellsAs the electrical current passes As the electrical current passes

through the wire, it can be used to through the wire, it can be used to run a device, such as a motor, light run a device, such as a motor, light bulb, voltmeter, etc. As a result, the bulb, voltmeter, etc. As a result, the electrochemical reaction can be electrochemical reaction can be used to provide useful work.used to provide useful work.

Reaction of Zn with CuReaction of Zn with Cu2+2+

If a strip of zinc metal is immersed in If a strip of zinc metal is immersed in a solution of copper(II)sulfate, a reaction a solution of copper(II)sulfate, a reaction will spontaneously occur.will spontaneously occur.Zn(Zn(ss) + Cu) + Cu2+2+((aqaq) ) Zn Zn2+2+((aqaq) + Cu() + Cu(ss))

As the reaction proceeds, some of As the reaction proceeds, some of the metallic zinc dissolves into solution, the metallic zinc dissolves into solution, and the blue copper(II) ion plates out as and the blue copper(II) ion plates out as elemental copper.elemental copper.


As each zinc atom dissolves, it provides two electrons to the copper ions.


As the reaction proceeds, a thin black layer of Cu is formed on the zinc surface. The blue color of the Cu2+ ion fades as it is reduced.


The transfer of electrons occurs directly on the zinc surface. As a result, the movement of electrons cannot be utilized.

Construction of a galvanic cell will allow the electron transfer to occur via a wire. In this way, the electrical current can be used to do work.

Galvanic CellsGalvanic CellsIn the galvanic cell for the reaction, In the galvanic cell for the reaction, the oxidizing reagent (Cuthe oxidizing reagent (Cu2+2+) and the ) and the reducing agent (Zn) are physically reducing agent (Zn) are physically separated into two separated into two half-cellshalf-cells..

The overall net-ionic equation for the The overall net-ionic equation for the reaction is:reaction is:

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

Galvanic CellsGalvanic CellsZn(Zn(ss) + Cu) + Cu2+2+((aqaq) ) Zn Zn2+2+((aqaq) + ) +

Cu(Cu(ss))

The half-reactions are:The half-reactions are:Zn(s) Zn(s) Zn Zn2+2+((aqaq) + 2 e) + 2 e-- (oxidation) (oxidation)CuCu2+2+((aqaq) + 2 e) + 2 e- - Cu( Cu(ss) (reduction)) (reduction)

Galvanic CellsGalvanic CellsIn galvanic cells, the In galvanic cells, the

components of two half-reactions are components of two half-reactions are physically separated into two physically separated into two beakers. The two beakers can then beakers. The two beakers can then be connected by a wire and a be connected by a wire and a salt salt bridge bridge so that the electron transfer so that the electron transfer can occur.can occur.

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

The half-reactions are:The half-reactions are:Zn(s) Zn(s) Zn Zn2+2+((aqaq) + 2 e) + 2 e-- (oxidation) (oxidation)CuCu2+2+((aqaq) + 2 e) + 2 e- - Cu( Cu(ss) (reduction)) (reduction)

One half-cell will contain metallic Zn One half-cell will contain metallic Zn in a solution of zinc ion, and the other in a solution of zinc ion, and the other half-cell will contain metallic copper half-cell will contain metallic copper in a solution of copper (II) ion.in a solution of copper (II) ion.


Zn(s) Zn(s) Zn Zn2+2+((aqaq) + 2 e) + 2 e-- (oxidation) (oxidation)

Oxidation takes place at the Oxidation takes place at the anodeanode, so , so the strip of zinc metal will serve as the strip of zinc metal will serve as the anode. It will be immersed in an the anode. It will be immersed in an aqueous solution of a zinc salt, such aqueous solution of a zinc salt, such as zinc sulfate. The sulfate ions are as zinc sulfate. The sulfate ions are inert, and are just spectator ions.inert, and are just spectator ions.


CuCu2+2+((aqaq) + 2 e) + 2 e- - Cu( Cu(ss) (reduction)) (reduction)

Reduction takes place at the Reduction takes place at the cathodecathode, , so the copper strip will serve as the so the copper strip will serve as the cathode. It will be immersed in an cathode. It will be immersed in an aqueous solution of a copper (II) salt, aqueous solution of a copper (II) salt, such as copper(II)sulfate. The sulfate such as copper(II)sulfate. The sulfate ion is inert, and will serve as a ion is inert, and will serve as a spectator.spectator.

The Salt BridgeThe Salt BridgeThe salt bridge, porous cup, or The salt bridge, porous cup, or

glass frit allows the flow of ions. glass frit allows the flow of ions. This is necessary in order to This is necessary in order to maintain a neutral charge in each maintain a neutral charge in each half-cell.half-cell.

The Salt BridgeThe Salt BridgeThe salt bridge, porous cup, or The salt bridge, porous cup, or

glass frit allows the flow of ions.glass frit allows the flow of ions.


Each combination of half-cells produces a characteristic voltage.

Oxidation at the AnodeOxidation at the Anode

Reduction at the CathodeReduction at the Cathode

Voltage or EMFVoltage or EMFThe voltage of the galvanic cell is also The voltage of the galvanic cell is also called the called the cell potentialcell potential, or the , or the electromotive forceelectromotive force ( emf). It is related ( emf). It is related to the driving force of the reaction. to the driving force of the reaction.

The units of cell potential are volts (V). The units of cell potential are volts (V). A volt is exactly 1 joule of work per A volt is exactly 1 joule of work per coulomb of charge transferred.coulomb of charge transferred.

Cell PotentialCell PotentialThere can be no reduction without There can be no reduction without oxidation, and thus, each galvanic cell oxidation, and thus, each galvanic cell needs two half-cells in order to produce a needs two half-cells in order to produce a voltage. voltage.

Scientists have devised a system to Scientists have devised a system to measure the potential (voltage) of any measure the potential (voltage) of any half-cell relative to a standard half cell. half-cell relative to a standard half cell. The potential of the standard half-cell is The potential of the standard half-cell is set at 0.00 volts.set at 0.00 volts.

The Hydrogen Half-CellThe Hydrogen Half-Cell

The half-cell consists of an inert platinum electrode immersed in 1M strong acid. Hydrogen gas is bubbled over the electrode.

Pt electrode

Standard Reduction Standard Reduction PotentialsPotentials

Each half-cell is connected to a Each half-cell is connected to a standard hydrogen electrode. All cells standard hydrogen electrode. All cells contain solutions which are 1.00M, and contain solutions which are 1.00M, and all gases are at a pressure of 1.00 all gases are at a pressure of 1.00 atmospheres.atmospheres.Since the voltage of the hydrogen Since the voltage of the hydrogen electrode is set at zero, the voltage of electrode is set at zero, the voltage of the galvanic cell represents the the galvanic cell represents the assigned voltage of the other half-cell. assigned voltage of the other half-cell.


The emf of the standard Zn half-cell is 0.76 volts relative to the hydrogen half-cell. In this cell, Zn is being oxidized, and H+ reduced.


The half-reactions are:The half-reactions are:Zn(Zn(ss) ) Zn Zn2+2+((aqaq) + 2e) + 2e--

2H2H++((aqaq) + 2e) + 2e-- H H22((gg))

Since the potential of the Since the potential of the hydrogen reaction is set at zero, the hydrogen reaction is set at zero, the potential for the oxidation of zinc is potential for the oxidation of zinc is the measured value of 0.76 volts.the measured value of 0.76 volts.


All half-reactions are tabulated as All half-reactions are tabulated as reductionsreductions. Since the potential for . Since the potential for the oxidation of zinc is +0.76 volts, the oxidation of zinc is +0.76 volts, the reduction potential is -.76 volts.the reduction potential is -.76 volts.

ZnZn2+2+(aq) + 2e(aq) + 2e-- Zn(s) E Zn(s) Eoo = -0.76V = -0.76V


In this way, the reduction In this way, the reduction potentials for all half-cells are potentials for all half-cells are obtained relative to the hydrogen obtained relative to the hydrogen electrode. The values of reduction electrode. The values of reduction potentials are listed from highest potentials are listed from highest potential to lowest.potential to lowest.

Cell PotentialsCell PotentialsWhen a galvanic cell is When a galvanic cell is

constructed, one half-reaction constructed, one half-reaction is a is a reduction reaction, and the other is reduction reaction, and the other is an oxidation reaction. an oxidation reaction.

When a reduction half-reaction When a reduction half-reaction is reversed to make it an oxidation is reversed to make it an oxidation reaction, the sign on its cell reaction, the sign on its cell potential is reversed.potential is reversed.

Cell PotentialsCell Potentials

The reaction which will occur The reaction which will occur spontaneously is the oxidation and spontaneously is the oxidation and reduction that produces reduction that produces the most the most positive cell potentialpositive cell potential..

Cell Potentials - ProblemCell Potentials - Problem Determine the balanced reaction Determine the balanced reaction

and the standard cell potential for a and the standard cell potential for a galvanic cell with the following half-galvanic cell with the following half-cells:cells:

CrCr22OO772-2- + 14 H + 14 H++ + 6 e + 6 e-- 2 2

CrCr3+3+ + 7 H + 7 H22O O HH22OO22 + 2H + 2H++ + 2 e + 2 e-- 2 H 2 H22OO

Cell Potentials - ProblemCell Potentials - Problem Determine the balanced reaction and the Determine the balanced reaction and the

standard cell potential for a galvanic cell standard cell potential for a galvanic cell with the following half-cells:with the following half-cells:

CrCr22OO772-2- + 14 H + 14 H++ + 6 e + 6 e-- 2 Cr 2 Cr3+3+ + 7 + 7

HH22O O HH22OO22 + 2H + 2H++ + 2 e + 2 e-- 2 H 2 H22OO

1. Look up the standard reduction 1. Look up the standard reduction potentials for potentials for both half-reactions. both half-reactions.



CrCr22OO772-2- + 14 H + 14 H++ + 6 e + 6 e-- 2 Cr2 Cr3+3+ + 7 H + 7 H22O O

1.33V H1.33V H22OO22 + 2H + 2H++ + 2 e + 2 e-- 2 H 2 H22OO 1.78V1.78V

2. Reverse one half-reaction so that the 2. Reverse one half-reaction so that the net cell net cell potential is the largest potential is the largest positive number. positive number.



CrCr22OO772-2- + 14 H + 14 H++ + 6 e + 6 e-- 2 Cr2 Cr3+3+ + 7 H + 7 H22O O


2. Reverse one half-reaction so that the 2. Reverse one half-reaction so that the net cell net cell potential is the largest potential is the largest positive number. positive number.

reverse



2 Cr2 Cr3+3+ + 7 H + 7 H22O O Cr Cr22OO772-2- + 14 H + 14 H++ + 6 e + 6 e-- - -


3. Multiply half-reactions so that the 3. Multiply half-reactions so that the electrons electrons lost = electrons gained. lost = electrons gained. Reduction Reduction potentials are potentials are notnot multiplied. multiplied.


standard cell potential for a galvanic cell with standard cell potential for a galvanic cell with the following half-cells:the following half-cells:


1.33V 1.33V ((HH22OO22 + 2H + 2H++ + 2 e + 2 e-- 2 H 2 H22O O )3)3 1.78V1.78V

3. Multiply half-reactions so that the electrons 3. Multiply half-reactions so that the electrons lost = electrons gained. Reduction potentials lost = electrons gained. Reduction potentials are are notnot multiplied. multiplied.

Cell Potentials - ProblemCell Potentials - Problem Determine the balanced reaction and Determine the balanced reaction and

the standard cell potential for a galvanic the standard cell potential for a galvanic cell with the following half-cells:cell with the following half-cells:


1.33V 3H1.33V 3H22OO22 + 6H + 6H++ + 6 e + 6 e-- 6 H 6 H22O O 1.78V 1.78V

4. Add the two half-reactions and their 4. Add the two half-reactions and their cell cell potentials. potentials.


standard cell potential for a galvanic cell with standard cell potential for a galvanic cell with the following half-cells:the following half-cells:

2 Cr2 Cr3+3+ + 7 H + 7 H22O O Cr Cr22OO772-2- + 14 H + 14 H++ + 6 e + 6 e-- -1.33V -1.33V

3H3H22OO22 + 6H + 6H++ + 6 e + 6 e-- 6 H 6 H22O O 1.78V1.78V

3H3H22OO22 + 2 Cr + 2 Cr3+ 3+ + 1 H+ 1 H22O O Cr Cr22OO772-2- + 8 H + 8 H++

EEoo=0.45V=0.45V

81

Line NotationLine NotationThere is a system of notation, There is a system of notation,

called called line notationline notation, used to describe , used to describe a galvanic cell. For the cell pictured a galvanic cell. For the cell pictured below:below: The anode is

written on the left, the cathode on the right.

Line NotationLine NotationThe anode is written on the left, the cathode on the right.A single vertical line A single vertical line represents a phase represents a phase boundary, and a boundary, and a pair of vertical lines pair of vertical lines indicate a salt indicate a salt bridge or porous bridge or porous disk.disk.

Line NotationLine Notation

The notation for this cell is:Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)


The notation for this cell is: Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)

Note that Zn is written on the left because it is the anode.


The notation for this cell is:Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)

Note that the standard hydrogen electrode is written on the right because it is the cathode.

Cell Potential and Free Cell Potential and Free EnergyEnergy

An electrochemical cell An electrochemical cell produces a voltage as a result of the produces a voltage as a result of the driving force for electron transfer. driving force for electron transfer. As a result, cell potentials are As a result, cell potentials are directly related to ∆G for the directly related to ∆G for the reaction. reaction.


For standard conditions,For standard conditions,∆∆GGoo = -nFE = -nFEoo

where n is the moles of where n is the moles of electrons transferred, electrons transferred, and F = 96,485 coulombs/mol eand F = 96,485 coulombs/mol e- -

(Faraday’s constant), and a volt (Faraday’s constant), and a volt equals 1 joule/coulomb.equals 1 joule/coulomb.


For standard conditions,For standard conditions,∆∆GGoo = -nFE = -nFEoo

A positive cell potential A positive cell potential (spontaneous reaction) yields a (spontaneous reaction) yields a negative value for ∆G.negative value for ∆G.

Cell Potential and Cell Potential and ConcentrationConcentration

Standard conditions dictate that all Standard conditions dictate that all solutions be 1M, and all gases have a solutions be 1M, and all gases have a pressure of 1 atm. Cell potential will pressure of 1 atm. Cell potential will vary with concentration.vary with concentration.

As a galvanic cell produces voltage, As a galvanic cell produces voltage, the concentrations in each half-cell the concentrations in each half-cell change, and the voltage gradually change, and the voltage gradually decreases to zero.decreases to zero.


The cell potential of a non-standard The cell potential of a non-standard cell can be calculated using the cell can be calculated using the Nernst equation:Nernst equation:

E = EE = Eoo--RT RT ln(Q) ln(Q) nnwhere T is temperature in Kelvins andwhere T is temperature in Kelvins andQ is the reaction quotient. Q is the reaction quotient.


For standard temperature For standard temperature (25(25ooC), the equation becomes:C), the equation becomes:

E = EE = Eoo--0.0591 0.0591 log(Q) log(Q) nn

where n is the number of moles of where n is the number of moles of electrons transferred.electrons transferred.

Concentration CellsConcentration Cells

Concentration Concentration cells are cells are electrochemical electrochemical cells with half-cells with half-cells that differ cells that differ only in the only in the concentration of concentration of reactants. reactants.

Concentration CellsConcentration CellsThe two solutions The two solutions would mix if they would mix if they weren’t physically weren’t physically separated. The separated. The transfer of electrons transfer of electrons occurs until the occurs until the concentration in concentration in both beakers is the both beakers is the same.same.

Concentration CellsConcentration CellsThe electrode in the The electrode in the dilute solution dilute solution dissolves, thus dissolves, thus raising the raising the concentration. concentration. Electrons flow to Electrons flow to the concentrated the concentrated beaker, where silver beaker, where silver ion plates out.ion plates out.

Concentration CellsConcentration Cells

1.0M CuSO4

0.1 M CuSO4

Concentration CellsConcentration CellsThe potential of a concentration The potential of a concentration

cell can be calculated.cell can be calculated.

EEcellcell = = -.0591 -.0591 log log [dilute][dilute] nn [conc.] [conc.]

Applications of Applications of Concentration CellsConcentration Cells

Special electrodes have been Special electrodes have been developed to determine the developed to determine the concentration of a variety of ions. The concentration of a variety of ions. The pH probe is an electrode that contains pH probe is an electrode that contains dilute hydrochloric acid. A thin glass dilute hydrochloric acid. A thin glass membrane in the electrode is put in membrane in the electrode is put in contact with a solution of unknown contact with a solution of unknown pH. The difference in potential results pH. The difference in potential results in the measurement of pH.in the measurement of pH.

pH ElectrodespH ElectrodesThe pH electrode is The pH electrode is a half-cell containing a half-cell containing a silver wire, silver a silver wire, silver chloride and dilute chloride and dilute hydrochloric acid. hydrochloric acid. The potential The potential depends upon the depends upon the difference in [Hdifference in [H33OO++] ] inside and outside of inside and outside of the electrode.the electrode.

Cell Potential and KCell Potential and KAt equilibrium, the cell potential is At equilibrium, the cell potential is 0.0 volts, and Q=K. Using the 0.0 volts, and Q=K. Using the Nernst Equation at 25Nernst Equation at 25ooC:C:

E = EE = Eoo--0.0591 0.0591 log(Q) log(Q) nn

0 = E0 = Eoo-- 0.0591 0.0591 log(K) log(K) nn

log K = nElog K = nEoo/0.0591/0.0591

Cell Potential and KCell Potential and K

log K = nElog K = nEoo/0.0591/0.0591

Since redox reactions often have Since redox reactions often have very large equilibrium constants, very large equilibrium constants, measuring cell potential is often the measuring cell potential is often the only way to obtain the value of K.only way to obtain the value of K.

ProblemProblem Write the chemical reaction and calculate Write the chemical reaction and calculate

the equilibrium constant for the following the equilibrium constant for the following galvanic cell under standard conditions.galvanic cell under standard conditions.

Pt(Pt(ss) | Cu) | Cu1+1+((aqaq), Cu), Cu2+2+((aqaq) || Au) || Au3+3+((aqaq) | Au() | Au(ss))1. Write the half-reactions.1. Write the half-reactions.Since the anode is on the left, copper(I) is Since the anode is on the left, copper(I) is oxidized to form copper(II). Gold(III) ion oxidized to form copper(II). Gold(III) ion is reduced to elemental gold at the is reduced to elemental gold at the cathode.cathode.

ProblemProblem Write the chemical reaction and Write the chemical reaction and

calculate the equilibrium constant for calculate the equilibrium constant for the following galvanic cell under the following galvanic cell under standard conditions.standard conditions.Pt(Pt(ss) | Cu) | Cu1+1+((aqaq), Cu), Cu2+2+((aqaq) || Au) || Au3+3+((aqaq) | ) |

Au(Au(ss) ) 1. Write the half-reactions.1. Write the half-reactions.

CuCu1+1+((aqaq) ) Cu Cu2+2+((aqaq) + 1 e) + 1 e--

AuAu3+3+((aqaq) + 3 e) + 3 e-- Au( Au(ss))

ProblemProblemPt(Pt(ss) | Cu) | Cu1+1+((aqaq), Cu), Cu2+2+((aqaq) || Au) || Au3+3+((aqaq) | ) |

Au(Au(ss) ) 2. Look up the reduction potentials.2. Look up the reduction potentials. CuCu1+1+((aqaq) ) Cu Cu2+2+((aqaq) + 1 e) + 1 e- - EEoo = - = -

(0.16V)(0.16V) AuAu3+3+((aqaq) + 3 e) + 3 e-- Au( Au(ss) E) Eoo = = 1.50V 1.50V

ProblemProblemPt(Pt(ss) | Cu) | Cu1+1+((aqaq), Cu), Cu2+2+((aqaq) || Au) || Au3+3+((aqaq) | Au() | Au(ss) )

3. Combine the reactions.3. Combine the reactions. 3[3[CuCu1+1+((aqaq) ) Cu Cu2+2+((aqaq) + 1 e) + 1 e--]] EEoo = - = -(0.16V)(0.16V) AuAu3+3+((aqaq) + 3 e) + 3 e-- Au( Au(ss) ) E Eoo = = 1.50V1.50V

3 Cu3 Cu1+1+((aqaq) + Au) + Au3+3+((aqaq) ) 3 Cu 3 Cu2+2+((aqaq) + Au() + Au(ss))

EEoo = 1.34V = 1.34V

ProblemProblem3 Cu3 Cu1+1+((aqaq) + Au) + Au3+3+((aqaq) ) 3 Cu 3 Cu2+2+((aqaq) + ) +

Au(Au(ss))EEoo = 1.34V = 1.34V

4. Calculate the value of K using E4. Calculate the value of K using Eoo..log K = nElog K = nEoo/0.0591/0.0591

log K = (3) (1.34)/.0591 = 68.0log K = (3) (1.34)/.0591 = 68.0K = 1 x 10K = 1 x 106868

Application - BatteriesApplication - BatteriesA A battery battery is a galvanic cell or a is a galvanic cell or a

group of galvanic cells connected in group of galvanic cells connected in series. They are “storage devices” series. They are “storage devices” for electrochemical energy.for electrochemical energy.

Lead Storage BatteryLead Storage BatteryEach cell of this Each cell of this battery produces battery produces approximately 2 approximately 2 volts. Six are volts. Six are connected in series connected in series to make the typical to make the typical 12V car battery. 12V car battery.

Lead Storage BatteryLead Storage BatteryThe anode is made of lead, and the The anode is made of lead, and the cathode is PbOcathode is PbO22..

Anode: Pb + HSOAnode: Pb + HSO44-1-1 PbSO PbSO44 + H + H++

+ 2e+ 2e--

Cathode: PbOCathode: PbO22 + HSO + HSO44-1 -1 +3 H+3 H++ + 2e + 2e- -

PbSOPbSO44 + 2H + 2H22OO

Lead Storage BatteryLead Storage BatteryNet Reaction: Net Reaction:

Pb(Pb(ss) + PbO) + PbO22((ss) + 2H) + 2H++((aqaq) + 2 HSO) + 2 HSO44--

11((aqaq) )

2 PbSO2 PbSO44((ss) + 2H) + 2H22O(O(ll))

Other BatteriesOther Batteries

1.5V Dry Cell

Mercury Battery

The Potato ClockThe Potato ClockAn An electrochemical electrochemical reaction that reaction that depends upon depends upon salts and acids salts and acids in the potato can in the potato can be used to power be used to power a digital clock. a digital clock.

ElectrolysisElectrolysisAn An electrolytic cellelectrolytic cell uses electricity to uses electricity to produce a non-produce a non-spontaneous spontaneous chemical reaction. chemical reaction. Examples include Examples include the electrolysis of the electrolysis of water to produce water to produce hydrogen and hydrogen and oxygen.oxygen.

ElectrolysisElectrolysisElectrolysis is also used in Electrolysis is also used in electroplatingelectroplating, , in which a metal such as silver is formed in which a metal such as silver is formed on the surface of a less expensive metal.on the surface of a less expensive metal.

Typically, an electrical current is passed Typically, an electrical current is passed through a solution of the ion to be through a solution of the ion to be deposited. The current is expressed in deposited. The current is expressed in amperesamperes (amp), indicated with the (amp), indicated with the symbol A.symbol A.

ElectrolysisElectrolysis1 amp = 1 coulomb/second1 amp = 1 coulomb/second

If the current (in A) is multiplied If the current (in A) is multiplied by the time the current flows (in s), by the time the current flows (in s), the total number of coulombs of the total number of coulombs of charge is obtained.charge is obtained.

Coulombs of charge = amps x secondsCoulombs of charge = amps x seconds

ElectrolysisElectrolysisCoulombs of charge = amps x Coulombs of charge = amps x

secondsseconds

Coulombs can be converted to moles Coulombs can be converted to moles of an element deposited as follows:of an element deposited as follows:

coulombs coulombs moles e moles e-- moles of moles of elementelement

96,485 C/mol e-

divide by n

Problem: ElectrolysisProblem: Electrolysis How many grams of copper will be How many grams of copper will be

plated out if a current of 5.00 amps plated out if a current of 5.00 amps is passed through a Cuis passed through a Cu2+2+ solution for solution for an hour. Assume an excess of Cuan hour. Assume an excess of Cu2+2+. .

Application: CorrosionApplication: Corrosion

The rusting of iron involves oxidation The rusting of iron involves oxidation of iron at the anode, and reduction of of iron at the anode, and reduction of oxygen with water at the cathode.oxygen with water at the cathode.

Application: CorrosionApplication: CorrosionTo help prevent To help prevent corrosion of corrosion of underground underground fuel tanks or the fuel tanks or the hulls of ships, a hulls of ships, a sacrificial anodesacrificial anode of a more of a more reactive metal reactive metal than iron is used.than iron is used.

electrochemistry

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