tro chapter 18.1 & 18.3 18.9 electrochemistry spring 2015v4.6(1) (1)

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Chapter 18: Electrochemistry

Mid-term Exam 3F March 27Redox (Tro Ch. 4.9 & 18.2) and Electrochemistry(Tro Ch. 18.1, 18.3-18.9)

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Electrochemistry Electrochemical reactions are redox reactions occurring in

an electrochemical cell. The oxidation and reduction half-reactions are physically separated

into half-cells. Electrons flow in an external circuit.

There are two kinds electrochemical cells: Non-spontaneous chemical reactions occur in electrolytic cells. Spontaneous chemical reactions occur in voltaic or galvanic cells.

Electrolytic cells require input of electrical energy to force the reaction to occur.

Voltaic/galvanic cells release energy from the chemical reaction as electrical energy.

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Electrolytic Cells

Electrical energy is used to force non-spontaneous chemical reactions to occur.

The process is called electrolysis. Two examples of commercial electrolytic

reactions are: The electroplating of jewelry and auto parts. The electrolysis of chemical compounds to make other useful compounds.

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Electrolysis of Molten Potassium Chloride

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Electrical Conduction and Electrochemistry

Metals are good conductors of electrical currents. In metallic conduction there is electron flow with no atomic

translational motion of the metal atoms in the electrode. Solid metals, can therefore, make good electrodes and

“wires”, (if they don’t participate in the redox reaction!) Electrodes are connected to each other by an external

circuit (conducting “wires”), allowing electron flow between electrodes.

Inert electrodes do not chemically react with the liquids or products of the electrochemical reaction but do conduct electrons.

Graphite (a form of carbon) and platinum are two commonly used inert electrodes.

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Electrolytic Cells In all electrolytic cells, electrons are forced

to flow from the positive electrode (anode) to the negative electrode (cathode) by an external source of electrical energy.

In ionic or electrolytic conduction ionic motion transports oxidizable or reducible species to the electrodes. Positively charged ions, cations, move toward the negative

electrode. Negatively charged ions, anions, move toward the positive

electrode.

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Electrodes The following naming convention is used for

either electrolytic or voltaic cells: The cathode is the electrode at which

reduction occurs. The cathode is negative in electrolytic cells and positive

in voltaic cells.

The anode is the electrode at which oxidation occurs.

The anode is positive in electrolytic cells and negative in voltaic cells.

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Electrolysis of Molten Potassium Chloride The non-spontaneous redox reaction that occurs is:

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Electrolysis of Potassium Chloride in Aqueous Solution

2 H2O + 2e- → H2 (g) + 2 OH-

cathode reaction

Battery, a source of direct currente- e-

aqueous KCl

2Cl- → Cl2 (g) + 2e- anode reaction

cathode (-) anode (-)

H2 gas Cl2 gas

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Electrolysis of Potassium Chloride in Aqueous Solution

Electrolysis of an aqueous solution of KCl results in production of hydrogen gas (H2(g)) at one electrode (the cathode). The solution becomes basic near this electrode. The reduction half-rxn occurs at this electrode, the cathode. the cathode.

Gaseous chlorine (Cl2(g)) is produced at the other electrode. the oxidation half-rxn occurs at this electrode, the anode.

These empirical (i.e., experimental) observations lead us to the non- spontaneous reaction:

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Electrolysis of Aqueous Potassium Sulfate

e- e-

cathode (-) anode (+)

H2 gas O2 gas

2H2O + 2e- → H2(g) + 2OH-

cathode reaction

2H2O → O2 (g) + 4H+ + 4e-

anode reaction

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In this electrolysis, hydrogen gas is produced at one electrode (the cathode). The solution becomes basic near this electrode. Reduction half-rxn occurrs at this electrode.

Gaseous oxygen is produced at the other electrode (the anode). The solution becomes acidic near this electrode. the oxidation half-rxn occurs at this electrode.

These experimental facts lead us to the following electrode reactions:

Electrolysis of Aqueous Potassium Sulfate

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The Electrolysis of Aqueous Potassium Sulfate

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Electrolytic Cells

In all electrolytic cells: the most easily reduced species is reduced and, the most easily oxidized species is oxidized.

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Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Faraday’s Law - The amount of substance undergoing

chemical reaction at each electrode during electrolysis is directly proportional to the amount of electricity (or number of electrons) that passes through the electrolytic cell.

A Faraday is the amount of electricity that reduces one equivalent (= 1 mole of electrons) of a species at the cathode and must, therefore also oxidize one equivalent of a species at the anode.

1 Faraday of electricity = 6.022 x 1023 e-

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Counting Electrons: Coulometry and Faraday’s Law of Electrolysis A coulomb (C) is the amount of charge that passes a

given point when a current of one ampere (A) flows for one second.

1 ampere (A) = 1 coulomb (C)/second 1 Faraday (F) = 96,487 coulombs/mol e-

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Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Faraday’s Law states that during electrolysis, one

Faraday of electricity (96,487 coulombs) reduces and oxidizes, respectively, one equivalent of the oxidizing agent and the reducing agent. A Faraday corresponds to the passage of one mole of

electrons through the electrolytic cell. One reducing equivalent ==> gain of 1 mol of e-. One oxidizing equivalent ==> loss of 1 mol of e-.

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Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Example: Calculate the mass of palladium produced by the

reduction of palladium(II) (Pd2+) ions during the passage of 3.20 amperes of current through an aqueous solution of palladium (II) sulfate for 30.0 minutes.

Cathode: Pd2+ + 2e- Pd0

1 mol 2 mol 1 mol

2(96,487 C) 106 g

3.20 amp = 3.20 C/s

?g Pd = 30.0 min x (60 s/min) x (3.20 C/s) x (1 mol e-/96,487 C)

x 1 mol Pd/2 mol e- x 106 g/mol Pd

= 3.16 g Pd

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Counting Electrons: Coulometry and Faraday’s Law of Electrolysis Example: Calculate the volume of oxygen

(measured at STP) produced by the oxidation of water in the previous example.

Anode: 2H2O O2(g) + 4H+ + 4e-

2 mol 1 mol 4 mole 4(96,487C)

22.4 LSTP

?L O2(g) = 30.0 min x (60 s/min) x 3.20 C/s x (1 mol e-/96,487 C)

x (1 mol O2(g)/4 mol e-) x (22.4 L O2(g)/mol)

= 0.334 L O2(g)

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Voltaic or Galvanic Cells Electrochemical cells in which a spontaneous chemical

reaction (∆G < 0) produces electrical energy. Half-cells are physically separated so that electrons are

forced to travel through external wires. The key concept: the potential difference between the

two half-cells drives the external electron flow.

Auto batteries Flashlight batteries Computer, calculator, cell phone batteries

Batteries are common examples of voltaic cells.

The electrical energy produced by these spontaneous reactions can be converted to useful work via the electron flow in the external circuit.

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The Standard Zinc-Copper Voltaic Cell

The cell’s initial voltage is 1.10 volts at 25 0C.

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The Construction of Simple Voltaic Cells Voltaic cells consist of two half-cells:

one containing a substance that can be oxidized and the other containing a substance that can be reduced.

in electrical contact with each other.

A simple voltaic cell: Two half cells, each containing a metal electrode

immersed in a solution of its ions. A wire connecting the two half-cells. A salt bridge to complete the circuit, allowing ion diffusion

to maintain charge neutrality.

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The Standard Zinc-Copper Voltaic Cell

Cell components for the Zn-Cu cell are: A metallic Cu strip immersed in 1.0 M copper (II) sulfate.

A metallic Zn strip immersed in 1.0 M zinc (II) sulfate.

A wire and a salt bridge to complete circuit.

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The Zinc-Copper Voltaic Cell

In all voltaic cells, electrons flow spontaneously from the negative electrode where oxidation occurs (the anode) to the positive electrode where reduction occurs (the cathode).

Anode half-rxn: Zn(s) Zn2+(aq) + 2e-

Cathode half-rxn: Cu2+(aq) + 2e- Cu(s)

Overall rxn: Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu(s)

A spontaneous rxn: E0cell = +1.10 V under standard conditions

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The Zinc-Copper Cell A commonly used short-hand notation for voltaic

cells (using the Zn-Cu cell as an example):

Zn(s) | Zn2+(aq)(1.0 M) || Cu2+(aq)(1.0 M) | Cu(s)

Soluble, oxidized species

cathodesalt bridgeanode

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The Copper - Silver Cell

Cell components: A Cu strip immersed in 1.0 M copper (II) sulfate.

A Ag strip immersed in 1.0 M silver (I) nitrate.

A wire and a salt bridge to complete the circuit.

The initial cell voltage is 0.46 V at 25 0C.

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The Copper - Silver Cell

Compare the Zn-Cu cell to the Cu-Ag cell

Anode reaction: Cu(s) Cu2+ + 2e-

Cathode reaction: 2(Ag+ + e- Ag(s))Overall cell reaction: Cu(s) + 2Ag+ Cu2+ + 2Ag(s)

The Cu electrode is the cathode in the Zn-Cu cell.

The Cu electrode is the anode in the Cu-Ag cell. In the Zn-Cu cell, Zn is more active than Cu, i.e., Zn “wants” to get oxidized more than does Cu. In the Cu-Ag cell, Cu “wants” to get oxidized more than does Ag.

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The Copper - Silver Cell The preceding two experimental observations

demonstrate that Cu2+ is a stronger oxidizing agent than Zn2+. Cu2+ oxidizes metallic Zn(s) to Zn2+.

Similarly, Ag+ is is a stronger oxidizing agent than Cu2+. Ag+ oxidizes metallic Cu(s) to Cu2+.

Strength as oxidizing agent: Zn2+ < Cu2+ < Ag+

Strength as reducing agent: Zn(s) > Cu(s) > Ag(s)

A measure of the relative “activities” of the metal.

More active metal is more easily oxidized.

So order of “activities is: Zn > Cu> Ag

These species can be, thus, be arranged in order of increasing oxidizing or reducing strengths:

Combine the SHE with other half-cells to quantitatively measure relative oxidizing and reducing strengths.

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Establish a reference electrode (half-cell) against which all other half-reactions can be compared.

The reference half-cell is the Standard Hydrogen Electrode (SHE): 1 M H+ (pH 0) /1 atm H2, 25 0C. The SHE is assigned an arbitrary voltage of 0.000000… V

The Standard Hydrogen Electrode (SHE)

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The Zinc-SHE Cell For this cell the component are:

1. A Zn strip (electrode) immersed in 1.0 M zinc (II) sulfate.

2. the Standard Hydrogen Electrode.

3. A wire and a salt bridge to complete the circuit.

Under standard conditions, Zn(s) reduces H+.

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The Zinc-SHE Cell

The cathode is the SHE. 2H+ is reduced to H2.

The anode is Zn(s) metal rod. Zn(s) metal is oxidized to Zn2+ ions.

The overall (spontaneous) cell reaction under standard conditions: Zn(s) metal reduces H+.

Anode rxn: Zn(s) Zn2+(aq) + 2e- E0 = +0.763 V

Cathode rxn: 2H+(aq) + 2e- H2(g) E0 = 0.000 V

Overall rxn: Zn(s) + 2H+(aq) Zn2+

(aq) + H2(g) E0cell = +0.763 V

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The Copper-SHE Cell The two half-cells are:

1. A Cu strip immersed in 1.0 M copper (II) sulfate.

2. The SHE.

Under standard conditions H2 reduces Cu2+.

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The Copper-SHE Cell

In this cell the SHE is the anode H2 is oxidized to 2H+.

The Cu(s) is the cathode. The Cu2+ ions are reduced to Cu metal.

Anode rxn: H2(g) 2H+(aq) + 2e- E0 = 0.000 V

Cathode rxn: Cu2+(aq) + 2e- Cu(s) E0 = +0.337 V

Overall rxn: H2(g) + Cu2+(aq) 2H+

(aq) + Cu(s) E0cell = +0.337 V

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Uses of Standard Reduction Potentials Half-cells that force the SHE to act as an anode (i.e., H2 gets oxidized)

are assigned positive standard reduction potentials. Half-cells that force the SHE to act as the cathode (i.e., H+ gets

reduced) are assigned negative standard reduction potentials. Standard reduction potentials, E0 (V) are:

for the half-cells measured relative to the SHE. a quantitative measure of the tendencies of half-reactions to

occur in the direction of reduction. Standard potentials are tabulated for the half reactions written in

the direction of reduction, i.e. the Table lists standard reduction potentials.

For the previous two examples, the standard reduction potentials are:

Zn2+(aq) + 2e- Zn(s) E0 = -0.763 V

Cu2+(aq) + 2e- Cu(s) E0 = +0.337 V

Midterm Exam 3 information

Covers material through today’s lecture. Special office hours:

W March 25 11 AM (right after class period). Th March 26 3 PM. Both office hours in my office (BSE 4.324).

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Uses of Standard Reduction Potentials

Can use standard reduction potentials to predict whether an electrochemical (redox) reaction will occur spontaneously under standard conditions.

Example: Will silver ions, Ag+, oxidize metallic Zn to Zn2+ ions, or will Zn2+ ions oxidize metallic Ag to Ag+ ions (under standard conditions)?

Use the following sequence to determine which direction is the spontaneous electrochemical (redox) reaction:

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Uses of Standard Electrode Potentials Choose the appropriate half-reactions from a table of

standard reduction potentials. Write the equation for the half-reaction with the more

positive E0 value first, along with its E0 value. Write the equation for the other half-reaction in the reverse

direction (i.e., as an oxidation), and reverse the sign of the tabulated E0.

Multiply the half rxns by appropriate coefficients to balance the electrons. DON’T MULTIPLY THE E0 values!

Add the half-rxns and corresponding E0 values to obtain the overall rxn and its potential, E0

cell.

This procedure always produces an equation for the reaction with positive E0

cell, which indicates that the forward reaction is spontaneous.

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Ag+ + e- Ag(s) E0 =+0.799 V

Zn(s) Zn2+ + 2e- E0 =-(-0.763 V)

2( )

2Ag+ + Zn(s) 2Ag(s) + Zn2+ Eocell = +1.562 V

The positive E0cell tells us:

the spontaneous (forward) reaction is oxidation of Zn(s) by Ag+.

Zn(s) is a better reducing agent than Ag(s). Conversely, Ag+ is a better oxidizing agent than Zn2+.

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Uses of Standard Reduction Potentials Example: Will MnO4

- oxidize Fe2+ to Fe3+, or will Fe3+

oxidize Mn2+ to MnO4- in acidic solution?

Thus, the answer is, yes, MnO4- will oxidize Fe2+ ions to Fe3+,

and MnO4- will be reduced to Mn2+ in acidic solution.

Reduction: MnO4- + 8H+ 5e- Mn2+ + 4H2O E0 = +1.51 V

Oxidation: 5(Fe2+ Fe3+ + e-) -E0 = -(+0.77 V)

Cell rxn: MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+ E0

cell = +0.74 V

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Example: Will nitric acid, HNO3, oxidize arsenous acid, H3AsO3,to arsenic acid, H3AsO4, in acidic solution? The reduction product of HNO3 is NO in this reaction.

Reduction: 2(NO3- + 4H+ + 3e- NO + 2H2O) E0 = +0.96 V

Oxidation: 3(H3AsO3 + H2O H3AsO4 + 2H+ + 2e-) -E0 = -(+0.58 V)

Cell rxn: 2NO3- + 2H+ + 3H3AsO3 2NO + H2O + 3H3AsO4 E0

cell = +0.38 V

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Corrosion: Oxidation of Metal by O2. Metallic corrosion consists of oxidation of the metal by O2,

aided by diffusion through surface moisture and acidification by dissolved CO2 or acidic pollutants.

pit

Overall rust reaction: 4Fe + 3O2 + xH2O(l) → 2Fe2O3 · xH2O

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Corrosion Protection

Examples of corrosion protection:1 Coat the corrosion-prone metal surface with with a thin

layer of a less active (less easily oxidized) metal.

“Tin cans” are actually made of steel coated with a thin layer of tin.

Tin is harder to oxidize than is iron (i.e., Sn(s) is less active than Fe(s).

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Corrosion Protection2. Connect the metal to a “sacrificial anode”, a piece of a

more active (i.e., more easily oxidized) metal.

• Ships hulls often have Zn exterior “sacrificial anodes”.

Fe2+ + 2e- Fe(s) E0 = -0.41V

Zn2+ + 2e- Zn(s) E0 = -0.76V

Mg2+ + 2e- Mg(s) E0 = -2.39V

Ship’s hull showing lighter colored Zn blocks, which function as sacrificial anodes.

• The Mg or Zn are more active than iron and will, therefore, be preferentially oxidized by O2.

• Mg and Zn are “sacrificed” to protect iron.

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Mg bar


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Corrosion Protection3. Coating or mixing with a more active (more

easily oxidized) metal.

Stainless steel is an alloy of Cr (~10%) and Fe. The Cr is more easily oxidized than is Fe, and the Cr2O3 forms a protective coating on the grains of Fe.

Galvanizing, either dipping steel in molten zinc or electrochemical reduction of Zn2+ on the steel surface), provides a more active metal on the exterior.

The thin coat of Zn functions as a sacrificial anode inhibiting the “rusting” of iron underneath. The oxidized Zn also forms an additional hard protective layer of zinc oxide (ZnO).

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4. Paint or coat with a polymeric material such as plastic or ceramic.

Steel bathtubs coated with porcelain, a ceramic material akin to glass.

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5. Allow a protective film to form naturally.

4Al(s) + 3O2 2Al2O3

Al2O3 forms a hard, transparent film on aluminum surfaces.

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Effect of Non-Standard Conditions on Electrode Potentials Standard electrode potentials are determined at:

1.00 M solution concentrations. 1.00 atm for gases. Pure liquids or solids (activities = 1). 25o C (298 K).

Electrode and cell potentials under non-standard conditions can be calculated using the Nernst equation.

Non-standard potentials are often more useful for “real-world” conditions.

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The Nernst Equation

E = E0 -2.303RT

nFlogQ

• E is the electrode (half-cell) potential under the non-standard conditions. • E0 is the standard electrode (half-cell) potential (from the table).

• R is the universal gas constant, 8.314 J/mol-K (not the same units as for gas

law).

• T is absolute temperature (Kelvins).

• n is number of electrons transferred in the half-reaction listed in the table.

• 1 V = 1J/C

• F is the Faraday (96,487 C/mol e-)

• Q is the reaction quotient.

The Nernst equation for electrochemical half-reactions under non-standard conditions:

x (1J/C·V) = 96,487 J/V · mol e- (J V-1mol -1).

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The Nernst Equation Substitution of the values of the constants into the

Nernst equation for half-reactions at 25o C gives:

MEMORIZE this form of the Nernst equation for non-standard half-cells at 25 oC!

2.303RT

F

2.303 x (8.314J/mol-K) x 298K

96,487 J/V-mol e-= = 0.0592

E = E0 -0.0592

nlogQ

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The Nernst Equation

Can write Nernst equations for half-reactions, for example:

The corresponding Nernst equation is:

E = E0 - 0.0592 1

log [Cu+]

[Cu2+]

Cu2+ + e- Cu+ E0 = +0.153 V

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The Nernst Equation Substituting E0 into the above expression gives:

If [Cu2+] and [Cu+] are both 1.0 M, then E = E0 (= 0.153 V) because the concentration ratio, Q = 1 and, therefore, logQ equals zero.

E = 0.153 - 0.0592 1

log[Cu+]

[Cu2+]

E = 0.153 - 0.0592 1

log 11

= 0.153

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The Nernst Equation Example: Calculate the potential for the Cu2+/Cu+

electrode at 250C when the concentration of Cu+ ions is three times that of Cu2+ ions.

Cu2+ + e- Cu+

[Cu+]

[Cu2+]Q = = 3

E = +0.153 - 0.0592 1

log 3

E = +0.153 - 0.0282 = +0.125 V

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The Nernst Equation Example: Calculate the potential for the Cu2+/Cu+

electrode at 25 0C when the Cu+ ion concentration is 1/3 of the Cu2+ ion concentration.

Cu2+ + e- Cu+

[Cu+]

[Cu2+]Q = = 1/3

E = 0.153 - 0.0592 1

log 0.333

E = 0.153 + 0.0282 = +0.181 V

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The Nernst Equation Example: Calculate the reduction potential for a

hydrogen electrode in which the [H+] is 1.0 x 10-3 M and the H2 pressure is 0.50 atmosphere.

0.50

(1.0 x 10-3)2Q = = 5.0 x 105

2H+ + 2e- H2

E = 0 - 0.0592 2

log 5.0 x 105

E = 0 - 0.168 = -0.168 V

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The Nernst Equation

The Nernst equation also applies to overall redox reactions occurring in a voltaic cell.

For an overall cell reaction, substitute Ecell and E0cell in

place of E and E0, respectively:

The Nernst equation can, therefore, be used to calculate the cell potential, Ecell, and, thus, spontaneity of a reaction

in a voltaic cell that consists of two non-standard electrodes.

Ecell = E0cell -

0.0592 n

logQ

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The Nernst Equation

Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu(s)

A spontaneous rxn: E0cell = +1.10 V under

standard conditions

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The Nernst Equation

Cu2+ + 2e- Cu(s) Eo = +0.340 V

Zn2+ + 2e- Zn(s) Eo = -0.763 V

Cu2+ + Zn(s) Cu(s) + Zn2+ Eo

cell = + 1.10 V

Reverse the zinc half-reaction and add:

Example: Starting from the standard Zn|Zn2+||Cu2+|Cu voltaic cell, calculate the cell potential, Ecell, when [Cu2+] is changed to 3 M.

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The Nernst Equation

0.0592 log([Zn2+]/[Cu2+])

0.0592 log 0.333

n

2

Ecell = 1.10 - (-0.477)

Ecell = +1.58 V

Calculate the cell potential, Ecell, when [Cu2+] is changed to 3 M.

Ecell = E0cell -

Ecell = 1.10 -

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The Nernst EquationCan even construct a voltaic cell consisting of the same half reaction at different starting concentrations!

If both half-cells contained either 1 M Cu2+ or 0.1 M Cu2+ :

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The Nernst Equation

0.0592 log (0.10/1.00)

0.0592 log 0.10

n

2

Ecell = 0 - (-0.030)

Ecell = +0.030 V

Cu|Cu2+(0.10M)||Cu2+(1.00M)|Cu

E0cell = 0, no net rxn

Ecell = E0cell -

Ecell = 0 -

For unequal [Cu2+]:

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The Nernst Equation

Example: Calculate the initial potential, Ecell, of a voltaic cell consisting of an Fe3+/Fe2+ half-cell connected to a Sn4+/Sn2+ half-cell with initial concentrations: [Fe3+] = 1.0 x 10-2 M and [Fe2+] = 0.1 M [Sn4+] = 1.0 M and [Sn2+] = 0.10 M.

Ecell = E0cell -

0.0592 n

logQ

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The Nernst Equation

Calculate the E0cell by the usual procedure.

Fe3+ + e- Fe2+ E0 = +0.771 V

Sn2+ Sn4+ + 2e- -E0 = -(0.15 V)

2Fe3+ + Sn2+ 2Fe2+ + Sn4+ Eocell = +0.62 V

2( )

Make sure the overall redox equation is balanced!

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The Nernst Equation

Substitute the ion concentrations into Q to calculate Ecell.

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The Nernst Equation

When the redox reaction is at equilibrium, Ecell = 0, (no net reaction or electron flow) so from the Nernst equation:

expressing constants explicitly:

E0cell = RT

or rearrange to: nFE0cell = RTlnK

From previous chapters the relationship between ∆G0 and the equilibrium constant, K, is:

∆G0rxn = -RTlnK

Equating the two expressions for RTlnK yields:

Therefore, ∆G0rxn = -nFE0

cell

0.0592 logK

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Relationship of E0cell to ∆G0 and K

MEMORIZE!

MEMORIZE!E0cell = n

nFlnK

Tro, Chemistry: A Molecular Approach 80


For a spontaneous redox reaction with the reactants and products in their standard states: ∆G° < 0 (negative) E° > 0 (positive) K > 1

∆G° = −RTlnK = −nFE°cell

n is the number of electrons transferred in the balanced reaction.

F = 96,485 C/mol e- or J/V-mol e-

Example: Calculate K at 25 0C for oxidation of Zn by Ag+:

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First calculate the standard cell potential E0cell:

Ag+ + e- --> Ag0 E0 =+0.799 V

Zn --> Zn2+ + 2e- E0 =-(-0.763 V)

2( )

2Ag+ + Zn0 --> 2Ag0 + Zn2+ Eocell = +1.562 V

[Zn2+]

[Ag+]2

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Use the value of E0cell (1.562 V) and n (= 2) to

obtain K:

K = 5.9 x 1052 =

∆G0rxn = -nFE0

cell

= (2 mol e-)(96,487J/V-mol e-)(1.562V)

= -301.4 kJ/mol rxn

0.0592 logKE0cell = n 0.0592

logK = E0celln

0.0592 logK = 2(1.562)

= 52.7

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Now calculate Ecell and ∆G starting with [Ag+] = 0.30 M and [Zn2+] = 0.50 M.

[Zn2+]

[Ag+]2Q =

(0.50)

(0.30)2= = 5.6

Ecell = Eocell - 0.0592 log Q

n

Ecell = 1.562 - 0.0592 log 5.62

Ecell = 1.562 - 0.022 = 1.540 V

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• So the forward reaction is still spontaneous under these non-standard conditions, but somewhat less so than under standard conditions.

∆Grxn = -nFEcell

= -(2 mol e-)(96,487 J/V-mol e-)(1.540 V)

= 2.97 x 105 J/mol of rxn

∆Grxn = -297 kJ/mol

Now calculate ∆Grxn under these non-standard conditions:

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Primary Voltaic Cells (Non-Rechargeable Batteries) As a voltaic cell discharges, its reagents are consumed. Once the reagents are consumed (or the concentrations

reach equilibrium), further chemical reaction is impossible, and therefore, current can no longer flow.

In primary voltaic cells, the electrodes and electrolytes cannot be regenerated by reversing current flow through the cell. Primary cells are, thus, not rechargable.

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The Alkaline Dry Cell (Non-Rechargeable)

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The Alkaline Dry Cell (Non-Rechargeable) Half reactions for an alkaline dry cell are:

Anode half-rxn: Zn0(s) + 2OH- --> Zn(OH)2(s) + 2e-

Cathode half-rxn: 2MnO2(s) + 2H2O + 2e- --> 2OH- + 2MnO(OH)(s)

Cell rxn: Zn0(s) + 2MnO2(s) + 2H2O --> Zn(OH)2(s) + 2MnO(OH)(s)

E0cell = 1.5V

All the reactants and products are solids or water (activity = 1), so the cell voltage remains nearly constant until most of the reactants are consumed.

Alkaline dry cells contain KOH, which precipitates the redox products, i.e., keeps them solids, so activity = 1.

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Secondary Voltaic Cells

Secondary cells are reversible and, therefore, rechargeable.

The electrodes in a secondary cell can be regenerated (recharged) by application of an external voltage. Application of an external voltage switches the cell from

voltaic to electrolytic.

One example of a secondary voltaic cell is the lead acid storage or car battery.

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The Lead Acid Storage (Car) Battery

In the lead storage battery the electrodes are two sets of lead alloy grids (plates).

Holes in one of the grids (cathodes) are filled with lead(IV) oxide, PbO2.

The holes in the other grids (anodes) are filled with “spongy” lead (Pb0).

The electrolyte is dilute sulfuric acid.

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Pb0

(s) + SO42- -->

PbSO4(s) + 2e-

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

PbSO4(s) + 4H+ + 2H2O

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Lead Acid Storage (Car) Battery

electrolyte = 30% H2SO4

Anode half-rxn: Pb(s) + SO42-(aq) → PbSO4(s) + 2 e-

Cathode half-rxn: PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e- → PbSO4(s) + 2 H2O(l)

Cell rxn: Pb0(s) + PbO2(s) + 4H+ + 2SO4

2- --> 2PbSO4(s) + 2H2O Ecell ~ 2V

A comproportionation reaction: two Pb oxidation states converting to one. PbSO4(s) coats both anodes and cathodes.

Battery voltage = ~2 V/cell x 6 cells connected in series = ~12 V

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During charging (or recharging), Pb2+ in PbSO4(s) (coating both electrodes) is either reoxidized to PbO2(s) or re-reduced to Pb0.

The concentration of the H2SO4 decreases as the cell discharges.

Recharging the cell regenerates the H2SO4.

, e.g., LiCoO2 or LMnO4

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The Hydrogen-Oxygen Fuel Cell

The overall reaction is the combination of hydrogen and oxygen to form water. The cell provides a drinking water supply for the astronauts

as well as the electricity for the lights, computers, etc. on board.

Fuel cells are very efficient. Energy conversion rates of 60-70% are common.

Anode rxn: 2(H2(g) + 2OH-(aq) --> 2H2O + 2e-)

Cathode rxn: O2(g) + 2H2O + 4e- --> 4OH-(aq)

Cell rxn: 2H2(g) + O2(g) --> 2H2O(l)

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Fuel cells are batteries that must have their reactants continuously supplied in the presence of appropriate catalysts.

A hydrogen-oxygen fuel cell is the most common. H2 is oxidized at the anode.

O2 is reduced at the cathode.

Reagents are very light, so saves weight. A potentially “green” alternative the internal

combustion engine for transportation vehicles. However, a source of energy is needed to generate

the H2.

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115

Biological Redox Chemistry and Reduction Potentials Living organisms derive their energy from redox reactions. Respiration is essentially combustion of the the food we

eat, for example:

∆G0 = -2878 kJ/mol

Glucose “combustion”:

In this biological “combustion” of organic molecules, the electrons are passed through several intermediary protein molecules called redox enzymes rather than directly from organic molecules to dioxygen.

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Biological Redox Chemistry and Reduction Potentials

• E0’ (pH 7) of some biologically accessible molecules and ions.

Biological redox reactions are often analyzed in terms of their half-reaction reduction potentials rather than free energies.

• Energy can be derived from reduction of molecules with higher E0’ by any molecule with a lower E0’.

glucose

reductants

glucoseCO2

O2 H2O

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After studying Tro Chapter 18 text and lecture slides and working the problems, you should be able to:1. Explain the difference between electrolytic and voltaic cells.2. Diagram electrolytic and galvanic cells, identify cathodes and anodes and

their half-reactions.3. For a given amount of current passed, calculate grams and moles of

substances oxidized or reduced in an electrochemical cell.4. Not responsible for commercial applications of electrolysis. 5. Determine relative strengths of oxidizing and reducing agents from

electrochemical reactions in voltaic cells or given half-reaction E0 values.6. Calculate E0

cell given half-reaction E0 values.

7. Describe and diagram the process of iron corrosion (rusting).8. Explain various methods for preventing corrosion of metals based on metal

“activities”, i.e. relative ease of oxidation or reduction.9. Use the Nernst equation (memorize) to calculate Ecell under non-standard

conditions given half-reaction E0 values, initial concentrations and the Nernst equation.

10. Calculate values of K and ∆Grxn from E0cell using the Nernst equation and ∆G =

-nFEcell (memorize).

11. Describe and calculate Eo for a concentration cell.

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After studying Tro Chapter 18 text and lecture slides and working the problems, you should be able to:

12. Identify components in an alkaline dry cell given the half reactions and a diagram of the cell. Show direction of current flow, and explain why voltage stays fairly constant during discharge.

13. Draw a diagram of a lead acid car battery given the half reactions, and describe the discharge and recharge cycles.

14. Describe and draw a diagram of a H2/O2 fuel cell.

15. Given E0’ values, describe how energy can be derived from biological redox reactions.

tro chapter 18.1 & 18.3 18.9 electrochemistry spring 2015v4.6(1) (1)

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voltaic cells

positive electrode anode

negative electrode cathode

voltaicgalvanic cells

electrolytic cellsin

redox reactions

kinds electrochemical

spontaneous redox reaction