Chapter 12

Electrochemical cells and

electrode potentials

Reduction and Oxidation Reactions

• Predict what might happen when a piece of copper wire in a solution of 2% AgNO3.

• If you try this experiment, you will initially see that the copper is a shiny copper color and the solution is clear. In less than one hour the solution is light blue and the wire is covered with shiny silver needles. What happened?

• Copper metal became copper ions in solution and silver ions became silver metal.

• Cu(s) + Ag+(aq) Cu2+

(aq) + Ag(s) (unbalanced)

• The Cu(s) loses electrons to become Cu2+(aq) ions and the Ag+

(aq) ions gain electrons to become Ag(s).

• Reactions that involve the exchange of electrons are called reduction and oxidation (redox) reactions.

• When a chemical species loses electrons we say that it is oxidized, and when a chemical species gains electrons we say that it is reduced.

• The Cu(s) loses electrons to be oxidized to Cu2+(aq).

The Ag+(aq) gain electrons to be reduced to Ag(s).

• We can break the reaction into the following two half reactions. – Cu(s) Cu2+

(aq) + 2e-

– Ag+(aq) + e- Ag(s)

• What would you predict if you placed a piece of Ag metal in a solution of Cu2+?

• Since we observed that the reaction of Ag+ and Cu is spontaneous, we would not expect the reverse reaction to be spontaneous.

• So no reaction occurs between Ag metal and Cu2+. Or we call it nonspontaneous reaction

Terminology

• Reduction: gaining of electrons (decrease in the oxidation

state

• Oxidation: loss of electrons (increase in the oxidation

state)

• Reducing agent (reductant): species that donates

electrons to reduce another reagent.

(The reducing agent get oxidized.)

• Oxidizing agent (oxidant): species that accepts electrons

to oxidize another species.

(The oxidizing agent gets reduced.)

Oxidation-reduction reaction (redox reaction): a reaction in which electrons are transferred from one reactant to another. • For example, the reduction of cerium(IV) by iron(II):

Ce4+ + Fe2+ Ce3+ + Fe3+

• The reduction half-reaction is given by...

Ce4+ + e- Ce3+

• The oxidation half-reaction is given by...

Fe2+ e- + Fe3+

• The half-reactions are the overall reaction broken down into

oxidation and reduction steps. • Half-reactions cannot occur independently, but are used

conceptually to simplify understanding and balancing the equations

Rules for Balancing Oxidation-Reduction Reactions

• Write out half-reaction "skeletons."

• Balance the half-reactions by adding H+, OH- or H2O as needed, maintaining electrical neutrality.

• Combine the two half-reactions such that the number of electrons transferred in each cancels out when combined.

• For example, consider the following reaction of the peroxydisulfate ion with manganese ion:

• S2O82- + Mn2+ SO4

2- + MnO4-

1. The reduction step is:

S2O8

2- + 2 e- 2 SO42-

(Each sulfur atoms goes from +7 to +6 oxidation state.)

2. The oxidation step is:

Mn2+ + 4 H2O 5 e- + MnO4- + 8 H+

(Manganese(II) loses 5 electrons, going from +2 to +7.)

3. In combining the two equations, the oxidation step must be multiplied by "2," and the reduction step must be multiplied by "5" to cancel out the electrons:

2x[Mn2+ + 4 H2O 5 e- + MnO4- + 8 H+]

5x[S2O82- + 2 e- 2 SO4

2-]

4. Adding these two equations together:

S2O82- + 2 Mn2+ + 8 H2O 10 SO4

2- + 2 MnO4- + 16 H+

5. Note that the half-reactions are charge-balanced before adding them together.

Consider the next example:

Cr2O72- + I- Cr3+ + I3

-

• The reduction step is given by...

Cr2O72- + 6 e- + 14 H+ 2 Cr3+ + 7 H2O

(Cr(IV) in the dichromate on is reduced to Cr(III).)

• The oxidation step is give by...

3 I- 2 e- + I3-

• Multiplying the oxidation half-reaction by 3x and adding the two half-reactions together:

Cr2O72- + 9 I- + 14 H+ 2 Cr3+ + 3 I3

- + 7 H2O

Oxidation-Reduction Reactions in Electrochemical Cells

• It is possible to separate the half-reactions of an oxidation-reduction reaction in an electrochemical cell.

• Consider the following reaction:

2 Ag+ + Cu(s) 2 Ag(s) + Cu2+

• The reduction half-reaction is given by...

Ag+ + e- Ag(s)

• The oxidation half-reaction is given by...

Cu(s) 2 e- + Cu2+

Redox reactions can be "separated" in a galvanic cell (also

called a voltaic cell or battery):

Charactristics of the of the galvanic cell (battery)

• The left container (half cell) contains of 0.020 M CuSO4.

• The right cell contains 0.0200 M AgNO3.

• A Cu electrode is immersed in the CuSO4 solution.

• An Ag electrode in immersed in the AgNO3 solution.

• The two solutions "communicate" via a salt bridge which consists of a saturated KCl solution in a tube

• with glass frits in both ends.

• At the anode, oxidation takes place:

Cu(s) 2 e- + Cu2+

• At the cathode, reduction takes place:

2 Ag+ + 2 e- 2 Ag(s)

• Chloride ions move into the CuSO4 solution to maintain electrical neutrality.

• Potassium ions move into the AgNO3 solution to maintain electrical neutrality.

• The volt meter reads 0.412 V at the instant the connection is made between the two electrodes. (This represents the difference is voltage ( Ecell) between the two electrodes.

a. The copper electrode has an initial voltage of 0.2867 V.

b. The silver electrode has an initial voltage of 0.6984 V.

c. Voltage Difference = Voltagecathode - Voltageanode

Voltage Difference = 0.6984 - 0.2867 = + 0.412 V

• This initial voltage drops as the reaction proceeds toward equilibrium as soon as connection is made. At equilibrium, the voltage read zero volts.

• The potential difference (voltage) between the anode and cathode is a measure of the tendency of the reaction to proceed from nonequilibrium to equilibrium

Operation of an electrolytic cell

• An external voltage source with a voltage larger than that of the battery is connected to the galvanic cell - positive pole to the silver electrode; negative pole to the copper electrode.

• The external voltage source reversed the direction of electron flow (and reverses the direction of the reactions at each electrode): – The silver electrode switches from the cathode to

the anode (the site of oxidation).

– The copper electrode switches from the anode to the cathode (the site of reduction.

– The voltage meter will read "negative" if still connected in the original fashion.

– The battery will "recharge.

Schematic Representation of Cells

• The copper(II) sulfate/silver nitrate systems described above would be symbolized:

• Cu|CuSO4 (0.02 M)||AgNO3 (0.02 M)|Ag – Each vertical line ("|") represents a phase

boundary or interface where a potential develops.

– Each double vertical line ("||") represents two phase boundaries (e.g., the salt bridge).

• The direction of electron flow is from left to

right: Cu--->Ag.

The original, initial voltage for the cell

• The original, initial voltage for the cell is given by...

• Ecell = Eright - Eleft = Ecathode - Eanode = EAg - ECu

• The potentials (absolute voltages) at the two electrodes cannot be determined experimentally.

• Only the differences between electrodes (via a volt meter) can be measured.

• Potentials at electrodes are assigned relative values, based upon comparison to a standard.

• By convention, the standard hydrogen electrode (SHE) is used as the "agreed upon" reference half-cell against which all others are compared.

• The voltage or cell potential is related to the free energy of the reaction driving the cell:

G = -nFEcell = -2.303RTlog(Keq)

Standard Hydrogen Electrode (SHE)

• The SHE (standard hydrogen electrode) is the reference point for

determining relative electrode potentials.

• The SHE is symbolized:

Pt, H2 ( hydrogen = 1.00 atm)|H+ (aH+ = 1.00)||Ag+ (aAg+ = 1.00)|Ag)

The platinum electrode is a platinized platinum electrode (platinum

coated with finely divided platinum called platinum black).

The aqueous acid solution has an activity of 1.00 (i.e., approximately

1.00 M hydrogen ion) and is saturated with hydrogen gas, bubbled in

at 1.00 atmosphere.

The SHE is connected via the salt bridge and connecting wire to the

other half-cell of the battery.

The half-cell reaction is given by: 2 H1+ + 2 e- H2(g)

• The SHE acts as either the anode or cathode, depending upon whether electrons are given up to the other half-cell or taken in.

• By convention, the SHE is assigned an absolute potential (voltage) = 0.00 volts at ALL temperatures.

• All other half-cells (and half-reactions) are measured relative to the SHE.

• By definition, the electrode potential is the potential of a cell with the standard hydrogen electrode acting as the anode and the other half-cell acting as the cathode.

– For example, consider the cell formed with the SHE as one half-cell and an Ag/Ag+ half-cell on the other side consisting of a silver electrode and an aAg+ = 1.00 silver nitrate solution.

• At the anode (oxidation): SHE

2 H+ + 2 e- H2(g)

• At the cathode (reduction): Ag/Ag+

Ag+ + e- Ag(s)

• As the instant the connection between half-cells is made, the experimentally observed difference in potential between the two electrodes measured by the volt meter (i.e., the battery voltage) is +0.799 V.

• Electrons flow from the SHE to the Ag electrode.

• By convention, the sign of the voltage is positive if electrons leaves the SHE; negative if electrons are taken in.

•

• Consider what happens if the Ag/Ag1+ half-cell is replaced with a Cd/Cd2+ half-cell (aCd+2+ = 1.00) attached to an SHE.

• 1) At the anode (oxidation): Cd/Cd2+

• Cd Cd2+ + 2 e1-

• 2) At the cathode (reduction): SHE

• 2 H1+ + 2 e1- H2(g)

• 3) At the instant the connection between half-cells is made, the experimentally observed difference in potential between the two electrodes measured by the volt meter (i.e., the battery voltage) is a negative voltage (-0.403 V).

• 4) Electrons flow to the SHE from the Cd electrode.

• 5) By convention, the sign of the voltage is negative if electrons are taken in by the SHE. This means that the half-cell reaction opposite the SHE is more reducing than the SHE.

• 6) A voltage greater than 0.403 V would have to be applied to reverse the flow of electrons.

Accepted IUPAC Conventions and Electrode Potentials

• Half-reactions are always written as reductions (i.e., electrodes potentials are by definitions reduction potentials).

• The sign of the electrode potential is determined relative to the SHE. – Positive (+) means electrons flow out of the SHE to the other

electrode (i.e., the SHE acts an the anode).

– Negative (-) means electrons flow to the SHE from the other electrode (i.e., the SHE becomes the cathode).

• The sign of the electrode potential signifies whether the net reaction of the battery is spontaneous "to the right" or "to the left." For example: – The Ag/Ag+ half-reaction has an electrode potential of +0.799 V,

meaning electrons flow to the Ag electrode.

– The Cd/Cd2+ half-reaction has an electrode potential of -0.403 V, meaning electrons flow to the SHE (and the reactions occur opposite to the directions written above).

• The effect of species concentration of electrode potentials is

described by the Nernst equation. For the reversible half-

reaction:

aA + bB + ... ne1- cC + dD + ...

Nernst equation

where...

E = the electrode potential

Eo= the standard electrode potential (i.e., the potential observed

when species in the half-cell are at a = 1.00 or pressure = 1.00 atm)

R = the gas constant (8.314 JK-1mol-1)

T = Kelvin temperature

n = number of moles of electrons in balanced half reaction

F = Faraday's constant (96,485 coulombs/mole; the charge on a

mole of electrons).

• Note that the standard electrode potential (Eo) is

measured under standard conditions (a = 1, pressure

= 1 atm).

• The Nernst equation corrects for nonstandard

concentrations.

• 2.303RT/nF simplifies to 0.0592/n.

• E = Eo if all species are at a = 1 and pressure = 1 atm.

Calculating battery voltages

• The battery voltage is given by:

Ecell = Eright - Eleft = Ecathode - Eanode

• Each half-cell electrode potential is

calculated, and the difference between

electrode determined.

Applications of electrode potentials

• Redox titrations utilized to measure analytes

via oxidation-reduction titrations will be

discussed. This will include:

• Equilibrium constants for redox reactions.

• Titrations curves - shapes, endpoints,

calculations, etc.

• Indicators for oxidation-reduction reactions.

Measuring Equilibrium Constants with Redox

Reactions • Consider the reaction:

2 Ag+ + Cu(s) 2 Ag(s) + Cu2+

• with associated half-reactions:

2 Ag+ + 2 e- 2 Ag(s) (reduction step)

Cu(s) 2 e- + Cu2+ (oxidation step)

• The equilibrium constant for this reaction is given by: •

•The reaction above is the same as the galvanic cell...

Cu|Cu2+ (?? M)||Ag+ (?? M)|Ag

• When this galvanic cell completely discharges to reach equilibrium, Ecell = Eright - Eleft = Ecathode - Eanode = EAg - ECu = 0

which means: EAg = ECu

• Substituting for each from the Nernst equation:

Note that Keq is so large,

the equilibrium lies heavily

toward the right

Oxidation-Reduction Titrations

Oxidation-Reduction Titrations

• When a oxidation-reduciton (redox) reaction is used to measure

an analyte, the titrations usually follow the electrode potential

as a function of titrant (or analyte) concentration.

• Since the electrode potential is a log function of the

concentrations, it behaves as a "p" function.

• We will discuss the redox titrations in terms of a

specific reaction and look at the calculations for

each of the different regions of the titration curve.

This would include:

• Initially (before any titrant was added)

• Before the equivalence point

• At the equivalence point

• After the equivalence point

• Consider titration of iron(II) with cerium(IV)

Fe2+ + Ce4+ Fe3+ + Ce3+

• Oxidation half-reaction:

Fe2+ Fe3+ + e-

• Reduction half-reaction:

e- + Ce4+ Ce3+ – The reactions must be fully reversible (i.e., the system must be at

equilibrium at all times through the titration).

– To be fully reversible means that:

Esystem = ECe = EFe = Eindicator and Ecell = 0 • The equilibrium concentration ratios of the oxidized and reduced forms

of the two species are such that their attraction for electrons are identical.

• The Nernst equation applies.

– Data from the titration can be used to calculate the titration curve using the Nernst equation for either the cerium(IV) or the iron(II) half-reactions.

• In practice: – Before equivalence, the Fe(III) and Fe(II) concentrations are used to

calculate the Esystem.

– After equivalence, Ce(III) and Ce(IV) concentrations are used to calculate the Esystem.

– At equivalence, simplifying assumptions are made based upon the stoichiometric relationships to calculate the Esystem.

• Example: 50.00 mL of 0.05000 M Fe2+ is titrated with 0.1000 M

Ce4+ in a medium containing 1.0 M H2SO4

• Derivation/calculation of the titration curve

1. Initial region: no Ce(IV) has been added

• Since the amount of Fe3+ cannot be calculated (and is

essentially zero), and the amount of Ce3+ is zero, the potential

cannot be calculated.

• Some amount of reaction must occur before numbers can be

plugged into the Nernst equation.

2. Before equivalence

assume 5.00 mL of Ce4+ solution have been added

3. At equivalence

• All the iron(II) is converted to iron(III); all the cerium(IV) is converted to cerium(III)

– Note that at the equivalence point: [Fe3+] = [Ce3+]

[Fe2+] = [Ce4+] ==> a very small amout!!

– The system potential can be calculated by combining the Nernst equations for both species as follows:

4. Beyond the equivalence point

• assume 25.10 mL of cerium(IV) have been added – The iron(II) is completely titrated. b. The cerium(III) and

cerium(IV) concentrations must be used to to calculate the potential of the system.

– Note that the number of moles of cerium(III) is equal to the number of moles of iron(II) originally present in the sample.

– Cerium(IV) present is due to the excess cerium(IV) added past equivalence.

– The system potential is calculated as follows:

Substituting into the Nernst equation:

Summary Of Redox Titration Curve

1. Titration begins beyond Eo for reactant

2. 50% titrated E = Eo for reactant

3. EP

200% titrated E = Eo2

Oxidation-Reduction Indicators • Two types of indicators are commonly used:

1. Specific indicators.

2. Oxidation-reduction indicators ("true" redox indicators).

• Specific indicators are species that react with one of the titration species to produce a color change.

• "True" redox indicators respond to the system potential to produce a color change. – Redox indicators are more versatile than specific indicators, since

they depend on the system potential, not the specific reaction.

– The half-reaction for the redox indicator is:

In(oxid) + n e- In(red)

• Like pH indicators, the color of redox indicators must change by about 10-fold to be seen. This means to see a full color transition, ratio of the oxidized and reduced indicator species must change from 1:10 to 10:1. – If these ratios are plugged into the Nernst equations for various

"n" values, the change in potential at the endpoint should correspond to EIn

o +/- 0.0592/n to get an appropriate color change. • For n = 1, the range around the endpoint is EIn

o +/- 0.0592/1 = +/- 0.0592 V (meaning the equivalence point must span this "window" for the indicator to work).

• For n = 2, the range around the endpoint is +/- 0.0296 V.

• For n = 3, the range around the endpoint is +/- 0.0197 V

– In practice, the equivalence point of a titration is first calculated, then a redox indicator whose EIn

o most closely matches the equivalence point is chosen.

– Consideration must also be given to the titration "break" at endpoint. If the standard electrode potentials of the analyte and titrant are too close to each other (<~0.40 V), it is nearly impossible to titrate the species. (The reaction is not complete enough to give a clear endpoint transition, and an indicator can not be "fit" to the titration.)

• Examples of Specific Indicators

• Starch-I3- complex

• In the presence of I3-, starch forms a deep blue/black color due

to the formation of a complex between the starch and the I3-.

I2(s) + I-(aq) I3-(aq)

I3-(aq) + Starch I3

-(aq)

.Starch (blue color)

• The presence/absence of I3-(aq) in solution can be used as an

indicator. For example, consider the following titration of copper(II):

2 Cu2+ + 5 I- 2 CuI(s) + I3-

– If excess iodide is added, the reaction is driven to completion, and

the I31- is back titrated with S2O3

2- :

I3

- + 2 S2O32- 3 I- + S4O6

2- (tetrathionate ion)

– I3- appears in solution as a yellowish color, and is

visible to approximately 5 x 10-6 M concentration.

– During back-titration, as the yellow disappears, starch may be added to produce a blue color that disappears when the last I3

1- is titrated (i.e., an endpoint).

– The blue color of the I3-(aq)

.starch indicator is visible to approximately

2 x 10-7 M concentration.

Thiocyanate.Iron(III) Complex

• Thiocyanate reacts with iron(III) to produce a deep,

red color:

• Fe3+ + SCN- FeSCN2+ (red)

• The red color can be used to detect the presence of

Fe3+ in titrations.

Example of redox indicators

1,10-phenanthrolines

Applications

Oxidizing Agents

• KMnO4 - titrant for Fe, Sn, and Oxalate

• Ce4+ - Very stable in strong sulfuric

acid, Titration of Fe and organics

• K2Cr2O7 - little used now due to hazards

and waste disposal of Cr

• I2 - add KI to increase solubility - Vit C,

wide range of applications

• Karl Fischer Detn of H2O

Reducing Agents

• Not too many of these. Fe2+ not very stable

Fe(NH3)2(SO4)2 Iodide

Thiosulfate - S2O32- ===> S4O6

2- + 2e-

• Many common REDOX procedures use a

reducing agent to reduce the sample and

then use an oxidizing agent titrant for

analysis rather than a reducing agent titrant.

eg. reduce Fe3+ to Fe2+ then titrate with Ce.