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CHEM 101-GENERAL CHEMISTRY CHAPTER 13

ELECTROCHEMISTRY

INSTR : FİLİZ ALSHANABLEH

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CHAPTER 13 ELECTROCHEMISTRY

• Oxidation – Reduction Reactions

• Galvanic Cells

• Cell Potentials

• Cell Potentials and Equilibrium

• Batteries

• Electrolysis

• Electrolysis and Stoichiometry

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INTRODUCTION

• Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy.

• This conversion takes place in electrochemical cells which are two types: I. Galvanic Cells: electricity is produced by means of spontaneous redox reaction. CHEMICAL ENERGY ELECTRICAL ENERGY II. Electrolytic Cells: electrical energy is used to cause a non- spontaneous reaction to occur. ELECTRICAL ENERGY CHEMICAL ENERGY

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Oxidation-Reduction Reactions

• Reactions that transfer electrons between reactants are known as oxidation-reduction or redox reactions. • Oxidation is the loss of electrons from some chemical

species. • Reduction is the gain of electrons to some chemical

species.

• When copper wire is placed in a silver nitrate solution, a redox reaction occurs. • A reaction is observed to occur because the solution

changes color and crystals form on the surface of the copper wire.

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Oxidation-Reduction and Half-Reactions

• When a clean copper wire is placed into a colorless solution of silver nitrate, it is quickly apparent that a chemical reaction occurs.

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Oxidation-Reduction and Half-Reactions

• The solution’s blue color is indicative of Cu2+ ions in solution. • Cu2+ is formed when a copper atom loses two electrons. • The copper metal is oxidized.

• The crystals forming on the surface of the copper wire are silver metal. • Silver is formed when a silver cation gains an electron. • The silver cation is reduced.

Cu(s) → Cu2+(aq) + 2e−

Ag+(aq) + 1e− → Ag(s)

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Oxidation-Reduction and Half-Reactions

• For the reaction between silver cation and copper metal, two half-reactions are written. • One for the oxidation of copper and one for the reduction

of silver. • Neither half-reaction can occur without the other.

• The half-reactions as written indicate that Ag+ only accepts

one electron whereas Cu loses two electrons. • The electron transfer must balance, so the reduction half-

reaction is multiplied by 2.

Cu(s) → Cu2+(aq) + 2e−

2Ag+(aq) + 2e− → 2Ag(s)

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Oxidation-Reduction and Half-Reactions

• Add the two half-reactions together, the electrons cancel out, leaving the net ionic equation for the redox reaction.

Cu(s) → Cu2+(aq) + 2e−

2Ag+(aq) + 2e− → 2Ag(s)

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

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Oxidation-Reduction and Half-Reactions

• The species undergoing oxidation is referred to as a reducing agent. • The Cu was oxidized and is the reducing agent. • The Cu facilitated the reduction of Ag+ by losing electrons.

• The species undergoing reduction is referred to as an

oxidizing agent. • The Ag+ was reduced and is the oxidizing agent. • The Ag+ facilitated the oxidation of Cu by gaining

electrons.

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Problem 13.5

• Prob. 13.5. For the following oxidaton-reduction reactions, identify the half reactions and label them as oxidation or reduction.

(a) Cu(s) + Ni2+(aq) Ni(s) + Cu2+(aq) oxidation rxn: Cu(s) Cu2+(aq) + 2e-

reduction rxn: Ni2+(aq) + 2e- Ni(s) Cu is oxidized and Cu is reducing agent Ni is reduced and Ni2+ is oxidizing agent (b) 2 Fe3+ (aq) + 3 Ba (s) 3 Ba2+ (aq) + 2 Fe (s) oxidation rxn: Ba(s) Ba2+(aq) + 2e-

reduction rxn: Fe3+(aq) + 3e- Fe(s) Ba is oxidized and Ba is reducing agent Fe is reduced and Fe3+ is oxidizing agent

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Building a Galvanic Cell

• A galvanic cell is any electrochemical cell in which a spontaneous chemical reaction can be used to generate an electric current.

• To harness electricity from a galvanic cell, each half-reaction is prepared separately in half-cells. • Cu metal immersed in Cu2+ solution is one half-cell. • Ag metal immersed in Ag+ solution is the second half-cell.

• Current flows by the migration of ions in solution.

• To transfer current between the half-cells, a salt bridge is used. • The salt bridge contains a strong electrolyte that allows either

cations or anions to migrate into the solution where they are needed to maintain charge neutrality.

• A metal wire cannot transport ions and cannot be used.

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Building a Galvanic Cell

• A salt bridge is crucial to a galvanic cell. The salt bridge allows ions to flow between each half-cell, completing the circuit.

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Terminology for Galvanic Cells

• Electrodes are the electrically conducting sites at which either oxidation or reduction occurs. • The electrode where oxidation occurs is the anode. • The electrode where reduction occurs is the cathode.

• Cell notation - a shorthand notation for the specific chemistry

of an electrochemical cell. • Cell notation lists the metals and ions involved in the

reaction. • A vertical line, |, denotes a phase boundary. • A double vertical line, ||, denotes a salt bridge. • The anode is written on the left, the cathode on the right.

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• General form of cell notation

• For the previous example of copper and silver

• An electrochemical cell is at its standard state when the

electrolyte concentrations are 1 M.

• For half-cells that generate or consume a gas, a partial pressure of 1 atm is required for the standard state.

anode | anode electrolyte || cathode electrolyte | cathode

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

Terminology for Galvanic Cells

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Problem 13.11

• Prob 13.11. The following oxidation-reduction reactions are used in electrochemical cells. Write them using cell notation.

(a) Ni (s) + 2 Ag+(aq)(0.50 M) 2 Ag (s) + Ni2+(aq)(0.20 M) oxidation rxn: Ni (s) Ni2+(aq) + 2e-

reduction rxn: Ag+(aq) + e- Ag(s) Ag|M) (0.50 (aq)Ag||M) (0.20 (aq)Ni|Ni(s) ++2

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Problem 13.12

• Prob 13.12. Write a balanced chemical equation for the overall cell reaction represented as

(a) oxidation rxn: Ag (s) Ag+(aq) + e-

reduction rxn: Sn4+(aq) + 2e- Sn2+(aq) Sn4+(aq) + 2 Ag (s) 2 Ag+ (s) + Sn2+(aq)

Pt(s)|(aq)Sn (aq),Sn|| (aq)Ag|Ag(s) +2+4+

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Cell Potentials

• The build up of charge on the electrode means there is potential for electrical work. • This potential is the cell potential, or electromotive force

(EMF).

• EMF is related to the maximum work obtainable from an electrochemical cell. • wmax = qE • q is the charge, E is the cell potential.

• A voltmeter measures the size of the electrical potential and

also its polarity - the locations of the negative charge (negative pole) and the positive charge (positive pole).

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Measuring Cell Potential

• When a voltmeter is connected to the previously described copper/silver cell, a potential of 0.462 V is measured.

• Connecting the copper half-cell to a reducing iron(III)/iron(II)

half-cell, a cell potential of 0.434 V is measured. • Connecting the iron(III)/iron(II) half-cell to the silver half-cell

results in a cell potential of 0.028.

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Measuring Cell Potential

• Measurement of standard cell voltages for various combinations of the same half-reactions suggests that a characteristic potential can be associated with a particular half-reaction.

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Measuring Cell Potential

• The standard hydrogen electrode (SHE) is the choice for the standard component in cell potential measurements.

• The cell is constructed of a platinum wire or foil as the electrode. • The electrode is immersed in a

1 M HCl solution through which H2 gas with a pressure of 1 atm is bubbled.

• The SHE has been chosen as the reference point for the scale of standard reduction potentials, and assigned a potential of exactly zero volts.

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Measuring Cell Potential

• For the Standard Hydrogen Electrode • The half-reaction for the SHE is: 2 H+(aq) + 2 e– → H2(g).

• The half-cell notation is: Pt(s) | H2(g, 1 atm) | H+ (1 M).

• The half-cell is assigned a potential of exactly zero volts.

• The cell potential is attributed to the other half-reaction.

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Standard Reduction Potentials

• To compare the oxidation-reduction trends of species used in electrochemistry, all half-cell potentials are written as reductions. • A table of standard reduction potentials lists the potential

of any half-reaction when connected to a SHE.

• All materials are 1 M for aqueous species and 1 atm partial pressure for gases.

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Standard Reduction Potentials

• Standard reduction potentials for several half-reactions involved in the cells discussed in the text.

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Standard Reduction Potentials

• Although the half-reactions are listed as reductions in the table, one half-reaction in any electrochemical cell must be an oxidation and, therefore, reversed from what appears in the table. • The cell potential sign must be changed when writing the

half-reaction as an oxidation.

• Some half-reactions have positive potentials, whereas others have negative potentials.

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Standard Reduction Potentials

• The tendency for the chemicals involved in a half-reaction to be an oxidation or reduction depends on the value of the reduction potential. • A large, positive value for the standard reduction potential

implies the substance is reduced readily and a good oxidizing agent.

• A large, negative value for the standard reduction potential implies the substance is oxidized readily and a good reducing agent.

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Standard Reduction Potentials

• For a galvanic cell, • The half-reaction with the more positive reduction potential

will be the cathode. • The half-reaction with the more negative reduction potential

will be the anode.

• The standard reduction potential for any pair of half-reactions, Eo

cell, is calculated from the standard reduction potentials for the cathode and anode. Eo

cell = Eored + Eo

ox • Eo

red is the standard reduction potential for the cathode and Eo

ox is the standard reduction potential for the anode.

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Standard Reduction Potentials

• For standard reduction potentials arranged horizontally, the anode and cathode for a galvanic cell can easily be determined. The reduction potential furthest to the left is the anode.

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Problem 13.20 • Prob 13.20. Four voltaic cell are set up. In each, one half contains a standard

hydrogen electrode. The second half cell is one of the following: (a) In which of the voltaic cell does the hydrogen electrode serve as the cathode ? (b) Which voltaic cell produces the highest voltage? Which produces the lowest ? (a) For hydrogen to serve as Cathode Eo

red ( H+| H2) > Eored ( X)

since Eored ( Cr3+ | Cr) , Eo

red ( Fe2+ | Fe) , Eored ( Mg2+ | Mg) < Eo

red ( H+| H2) Hydrogen will be cathode (reduced) and others will be anode (oxidized)

(b) Eocell = Eo

red + Eoox

Eocell = Eo

red ( H+| H2) + Eoox ( Mg | Mg2+) = 0.00 + (+2.37) = 2.37 V the highest

Eocell = Eo

red ( H+| H2) + Eoox ( Cr | Cr2+) = 0.00 + (+0.74) = 0.74 V

Eocell = Eo

red ( H+| H2) + Eoox ( Fe | Fe3+) = 0.00 + (+0.41) = 0.41 V

Eocell = Eo

red ( Cu2+| Cu) + Eoox (H2 | H+) = 0.337 + 0.00 = 0.337 V the lowest

Cr3+ (aq) | Cr(s) , Fe2+ (aq) | Fe(s) , Cu2+ (aq) | Cu (s) , Mg2+ (aq) | Mg (s)

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Problem 13.24 • Prob 13.24. Consider for the reduction of the following metal ions to the metal:

Among the metal ions and metals that make up these half reactions: a. Which metal ion is the weakest oxidizing agent? b. Which metal ion is the strongest oxidizing agent? c. Which metal is the strongest reducing agent? d. Which metal is the weakest reducing agent? e. Will Sn(s) reduce Cu2+ (aq) to Cu(s)? f. Will Ag(s) reduce Co2+ (aq) to Co(s)? g. Which metal ions in the list can be reduced by Sn(s)? h. What metals can be oxidized by Ag+(aq)? ANSWERS a. With the lowest Eo

red value Zn2+ is the weakest oxidizing agent.

b. With the highest Eored value Au+ is the strongest oxidizing agent.

c. With the highest Eoox value Zn is the strongest reducing agent.

d. With the lowest Eoox value Au is the weakest reducing agent.

e. Yes, since Eored ( Cu2+ | Cu) > Eo

red ( Sn2+ | Sn) , Sn will reduce Cu2+ to Cu. f. No, since Eo

red ( Co2+ | Co) < Eored ( Ag+ | Ag) , Ag will not reduce Co2+ to Co.

h. Except Au (Eored ( Au+ | Au) > Eo

red ( Ag+ | Ag)) , all other metals; Zn, Co, Sn, Cu will be oxidized by Ag +.

Sn2+ (aq) , Au+(aq), Zn2+ (aq) , Co2+ (aq) , Ag+(aq), Cu2+ (aq)

Eored ( Zn2+ | Zn) = - 0.763 V

Eored ( Co2+ | Co) = - 0.28 V

Eored ( Sn2+ | Sn) = - 0.14 V

Eored ( Cu2+ | Cu) = + 0.337V

Eored ( Ag+ | Ag) = + 0.799 V

Eored ( Au+ | Au) = + 1.68 V

---------------------------------------------------------------------------------------- Eo

ox ( Au | Au +) = - 1.68 V Eo

ox ( Ag | Ag +) = - 0.799 V Eo

ox ( Cu | Cu2+) = - 0.337V Eo

ox ( Sn | Sn2+) = + 0.14 V Eo

ox ( Co | Co2+) = + 0.28 V Eo

ox ( Zn | Zn2+) = + 0.763 V

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Problem 13.25 & 26 • Prob 13.25. Calculate cell potentials of the following cells:

a. Ga(s) | Ga3+(aq) ║ Ag+(aq) | Ag(s) Eo

red ( Ag+ | Ag) = + 0.799 V Eo

red ( Ga3+ | Ga) = - 0.53 V

oxidation half-rxn : Ga(s) Ga3+(aq) + 3e- Eoox ( Ga | Ga3+) = + 0.53 V

reduction half-rxn : Ag+(aq) + e- Ag(s) Eored ( Ag+ | Ag) = + 0.799 V

Eocell = Eo

red + Eoox

= 0.799+ 0.53 Eo

cell = 1.33 V

• Prob 13.26. Calculate cell potentials of the following cells: b. Pt | Fe2+ (aq), Fe3+(aq) ║ MnO4

- (aq), H+ (aq) , Mn 2+ (aq) | Pt Eo

red (Fe3+ | Fe2+ ) = + 0.771 V Eo

red (MnO4- | Mn 2+ ) = 1.507 V

oxidation half-rxn : Fe2+ (aq) Fe3+(aq) + e- Eoox (Fe2+ | Fe3+ ) = - 0.771 V

reduction half-rxn : MnO4- (aq) + 8 H+ (aq) + 5e- Mn 2+ (aq) + +H2O Eo

red (MnO4- | Mn 2+ ) = 1.507 V

Eocell = Eo

red + Eoox

= - 0.771+ 1.507 Eo

cell = 0.736 V

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Nonstandard Conditions

• The cell potential at nonstandard conditions is calculated using the Nernst equation.

• Q is the reaction quotient, and n is the number of electrons transferred

in the reaction.

• For rxn; aA + bB cC + dD

• In Q calculations only concentration of (aq) state and partial pressure of (g) state included.

( ) Qn

EE o log0591.0−=

[ ] [ ][ ] [ ]ba

dc

BADCQ =

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Problem 13.29

• Prob.13.29. Calculate the cell potentials of the following cells at 25 oC. a. 2 Ag+ (aq) (0.50M) + Ni(s) 2 Ag(s) + Ni2+(aq) (0.20M) oxidation half-rxn : Ni(s) Ni 2+(aq) + 2e- Eo

ox ( Ni | Ni2+) = + 0.25 V reduction half-rxn : Ag+(aq) + e- Ag(s) Eo

red ( Ag+ | Ag) = + 0.799 V Eo

cell = Eored + Eo

ox

= 0.799 + 0.25 = 1.049 V n = 2

( ) Qn

EE o log0591.0−=

[ ][ ]

[ ][ ]

8.050.020.0

2

1

2

12

===+

+

AgNiQ

( ) VQn

EE o 052.1)8.0log(2

)0591.0(049.1log0591.0=−=−=

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Electrolytic Cells

Electrolytic Cells: electrical energy is used to cause a non-spontaneous reaction to occur.

• Electrolysis is the process of passing an electric current through an

ionic solution or molten salt to produce a chemical reaction.

• Electrolytic cells are divided into two categories based on the nature of the electrodes used. • Passive electrolysis: the electrodes are chemically inert materials

that simply provide a path for electrons.

• Active electrolysis: the electrodes are part of the electrolytic reaction.

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Electrolysis

• Electrolysis changes the polarity of the electrodes in a system. • For reduction, electrons are forced to the cathode. The cathode

becomes the negative electrode. • For oxidation, electrons are pulled from the anode. The anode

becomes the positive electrode.

• In electrolysis, an external source of current drives a redox reaction that would otherwise not be spontaneous. The flow of ions through the solution completes the circuit.

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Electroplating

• The process of depositing a thin coat of metal on another metal by using electrolysis is electroplating. • In some cases, the thin coating is cosmetic, or to provide some

vital functionality for the coated piece, such as corrosion resistance or desirable conductive properties.

• Silver is plated onto electrical devices because silver is a good conductor and resistant to corrosion. • Ag+ (aq) + e- Ag(s) Rxn occur at cathode.

• To collect 1 mol of Ag 1mol of e- is required

• Metal ions reduce at the surface of cathode; • M+x(aq) + Xe- M (s) • 1 mol M = X mol e-

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Electrolysis and Stoichiometry

• For electroplating, it can be vitally important to use carefully controlled amounts of materials. • Controlling the flow of electrons (current) in an

electroplating operation provides a method to accurately limit the amount of material deposited.

• Electroplating is often used to prevent galvanic corrosion in an electrical apparatus in places where different metals come into contact with one another.

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Current and Charge

• When current is measured in an electric circuit, the observation is the flow of charge for a period of time. • The unit of current, the ampere (A), is defined as one

coulomb per second: 1 A = 1 C s-1.

• If a known current flows through a circuit for a known time, the charge can be easily calculated.

Charge = current × time Q = I × t

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Current and Charge

• Using Faraday’s constant, F = 96,485 C mol-1 and the calculated charge, the number of moles of electrons that pass through the circuit can be calculated.

• If the number of electrons required to reduce each metal cation is known, the number of moles of material plated can be calculated.

• Electricity use is often measured in terms of power. The SI unit for power is the watt (1 watt = 1 J s-1)

• Electrical utilities normally determine consumption in kilowatt-hours, kWh (1 kWh = 3.60 × 106 J)

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Calculations Using Masses of Substances in Electrolysis

• A knowledge of current, how long the current flows, stoichiometry, and the number of electrons required to reduce a metal cation are used to answer the following questions. • How much material is plated given a specific current for an

allotted time or electrical energy expenditure?

• How long must a given current to pass through the cell to yield a desired mass of plated material?

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Example Problem 13.75

• Prob.13.75. If current of 15 A is run through an electrolysis cell for 2.0 hours, how many moles of electrons have moved?

Charge = Current x time Q = 15 x (2.0 x 3600) = 1.08 x105 C 1 mol e- = 96,500 C 96,500 C 1 mol e- 1.08 x105 C X = 1.12 mol e-

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Example Problem 13.79

• Prob.13.79. In a copper plating experiment in which copper metal is deposited from copper(II) ion solution, the system is run for 2.6 hours at a current of 12.0 A. What mass of copper depozited?

copper(II) ion solution Cu2+(aq) Cu2+(aq) + 2e- Cu(s)

Charge = Current x time = 12.0 x (2.6 x 3600) = 1.12 x105 C 96,500 C 1 mol e- 1.12 x105 C X = 1.16 mol e- 2 mol e- : 1mol Cu 1.16 mol e- X = 0.58 mol Cu = 0.58 (63.55) = 37.0 g Cu