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Home , Electrochemistry, Nernst equation, Redox, Salt bridge, Titration

An Introduction to

Redox Reaction

ox1 + red2 <=> red1 + ox2

1 Outline

1. Review  Concepts: Reaction, Cell, Potential  Balance an Equation  2. Redox 3. Potentiometry 4. and Electrochemical Sensor

12.1 Balancing Oxidation- Reduction Reactions

• Balance: the amount of each element present at the beginning of the reaction must be present at the end. • Charge Balance: are not lost in a . Half Reactions • Half-reactions are a convenient way of separating oxidation and reduction reactions.

2 Half Reactions • The half-reactions for Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe3+(aq) are Sn2+(aq)  Sn4+(aq) +2e- 2Fe3+(aq) + 2e-  2Fe2+(aq) • Oxidation: electrons are products. • Reduction: electrons are .

Balancing Equations by the Method of Half Reactions

• Consider the titration of an acidic of Na2C2O4 ( , colorless) with KMnO4 (deep purple).

3 How?

1. Write down the two half reactions. 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding . c. Then balance H by adding H+. d. Finish by balancing charge by adding electrons. 3. Multiply each half reaction to make the number of electrons equal. 4. Add the reactions and simplify. 5. Check!

Balancing Equations by the Method of Half Reactions

+ - 2- 2+ 16H (aq) + 2MnO4 (aq) + 5C2O4 (aq)  2Mn (aq) + 8H2O(l) + 10CO2(g)

4 Exercise:

Try to Balance: Sn+2 + Fe+3 <=> Sn+4 + Fe+2 +2 - +3 +2 Fe + MnO4 <=> Fe + Mn

Sn+2 + Fe+3 <=> Sn+4 + Fe+2

5 +2 - +3 +2 Fe + MnO4 <=> Fe + Mn

Important Redox Titrants and the Reactions

Oxidizing Reagents (Oxidants) (1) Permanganate

 +  2+ MnO4 + 8H + 5e  Mn + 4H2O

 +  MnO 4 + 4H + 3e  MnO2 (s) + 2H2O  2 MnO4 + e  MnO4

6 Important Redox Titrants and the Reactions

Oxidizing Reagents (Oxidants) (2)

2 +  3+ Cr2O7 +14H + 6e  2Cr + 7H2O

2 4 + + 3+ 2+ Cr2O7 + 3U + 2H  2Cr + 3UO2 + H2O

Important Redox Titrants and the Reactions

Oxidizing Reagents (Oxidants) (2) Potassium Iodate

1 IO  + 6H + + 5e  I + 3H O 3 2 2 2

7 Important Redox Titrants and the Reactions

Reducing ( Reductants ) (2) Potassium 1 I  I + e 2 2

Important Redox Titrants and the Reactions

Reducing Reagent ( Reductants ) (2) Sodium

2- 2- 2S2O3 S4O6 +2e

8 12.2 Nernst Equation

• Voltaic cells consist of – : Zn(s)  Zn2+(aq) + 2e2 – : Cu2+(aq) + 2e-  Cu(s) – (used to complete the electrical circuit): cations move from anode to cathode, anions move from cathode to anode. • The two solid are the (cathode and anode).

9 The salt bridge allows charge transfer through the solution and prevents mixing.

Fig. 12.1. Voltaic cell.

10 Galvanic Cells

anode cathode oxidation reduction

spontaneous redox reaction

19.2

11 • This rearranges to give the Nernst equation: RT E = E°  lnQ nF • The Nernst equation can be simplified by collecting all the constants together using a of 298 K: 0.0592 E = E°  logQ n • (Note that change from natural to -10 log.) • Remember that n is number of moles of electrons.

Nernst Equation

aA + bB + ne- <=> cC + dD

0.0592 [C]c[D]d E = Eo ------log ------at 25oC n [A]a[B]b

0.0592 [Re d] E = E 0  log n [Ox]

12 Standard Potentials

Standard (E0) is the associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm.

Reduction Reaction

- + 2e + 2H (1 M) H2 (1 atm)

E0 = 0 V

Standard electrode (SHE) 19.3

Standard Reduction (Half-Cell) Potentials

•The SHE is the cathode. It consists of a Pt electrode in a + tube placed in 1 M H solution. H2 is bubbled through the tube. • For the SHE, we assign + - 2H (aq, 1M) + 2e  H2(g, 1 atm)

• E°red of zero. • The potential of a cell can be calculated from standard reduction potentials:

E°cell = E°red (cathode)  E°red (anode)

13 Standard Reduction (Half-Cell) Potentials

2+ - • Consider Zn(s)  Zn (aq) + 2e . We measure Ecell relative to the SHE (cathode):

E°cell = E°red(cathode) - E°Ox(anode)

0.76 V = 0 V - E°red(anode).

• Therefore, E°red(anode) = -0.76 V. • Standard reduction potentials must be written as reduction reactions: 2+ - Zn (aq) + 2e  Zn(s), E°red = -0.76 V.

14 12.3 Cell Diagram

anode//cathode

Zn/ZnSO4 (0.1 M)//CuSO4 (0.2 M)/Cu

If a cell,

Ecell = Ecathode - Eanode

if Ecell => positive, spontaneous as written

15 o - The more positive the E , the better is the oxidized form (e.g., MnO4 ). The more negative the Eo, the better is the reduced form (e.g., Zn).

©Gary Christian, Analytical , 6th Ed. (Wiley)

16 12.4 Curve

12.4 Redox Titration Curve

17 12.4 Redox Titration Curve

Derivation of a titration curve Fe+2 + Ce+4 <=> Ce+3 + Fe+3 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M +4 Ce in a medium that is 1.0 M in H2SO4.

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3 E = Eo - (0.0592/n) log([red]/[ox]) At 0.00 mL of Ce+4 added, initial point no Ce+4 present; minimal, unknown [Fe+3]; thus, insufficient information to calculate E

18 Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with +4 0.1000 M Ce in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe

V M (15.00 mL)(0.1000 M) [Fe+3] = ------Ce Ce = ------VFe + VCe (50.00 + 15.00)mL = 2.308 x 10-2 M

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe [Fe+3] = 2.308 x 10-2 M

VFe MFe - VCe MCe [Fe+2] = ------VFe + VCe

19 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe [Fe+3] = 2.308 x 10-2 M (50.00 mL)(0.0500 M) - (15.00 mL)(0.1000 M) [Fe+2] = ------(50.00 + 15.00)mL = 1.54 x 10-2 M

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe [Fe+3] = 2.308 x 10-2 M[Fe+2] = 1.54 x 10-2 M for Fe+2 -> Fe+3 Eo = 0.69 v 0.0592 [Fe+2] o Ecell = Ecell ------log ------n [Fe+3]

20 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 15.00 mL of Ce added, VFeMFe > VCeMCe [Fe+3] = 2.308 x 10-2 M[Fe+2] = 1.54 x 10-2 M for Fe+2 -> Fe+3 Eo = 0.69 v 0.0592 1.54 x 10-2 Ecell = 0.68 v ------log ------v = 0.69 v 1 2.31 x 10-2

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3 At 25.00 mL of Ce+4 added,

VFeMFe = VCeMCe, equivalence point

o o nFe EFe + nCe ECe 1(1.44) + 1(0.68) Ecell = ------= ------v nFe + nCe 1 + 1 = 1.06 v

21 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 26.00 mL of Ce added, VCeMCe > VFeMFe V M (50.00 mL)(0.0500 M) [Ce+3] = ------Fe Fe = ------VFe + VCe (50.00 + 26.00)mL = 3.29 x 10-2 M

22 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 26.00 mL of Ce added, VCeMCe > VFeMFe [Ce+3] = 3.29 x 10-2 M

VCe MCe - VFe MFe [Ce+4] = ------VFe + VCe

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4. Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 26.00 mL of Ce added, VCeMCe > VFeMFe [Ce+3] = 3.29 x 10-2 M

(26.00 mL)(0.1000 M) - (50.00 mL)(0.0500 M) [Ce+4] = ------(50.00 + 26.00)mL = 1.32 x 10-3 M

23 EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 26.00 mL of Ce added, VCeMCe > VFeMFe [Ce+3] = 3.29 x 10-2 M[Ce+4] = 1.32 x 10-3 M for Ce+4 -> Ce+3 Eo = 1.44 v 0.0592 [Ce+3] o Ecell = Ecell ------log ------n[Ce+4]

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that

is 1.0 M in H2SO4.

Fe+2 + Ce+4 <=> Ce+3 + Fe+3 +4 At 26.00 mL of Ce added, VCeMCe > VFeMFe [Ce+3] = 3.29 x 10-2 M[Ce+4] = 1.32 x 10-3 M for Ce+4 -> Ce+3 Eo = 1.44 v 0.0592 3.29 x 10-2 E = 1.44 v ------log ------v = 1.39 v cell 1 1.32 x 10-3

24 25 26 Electrochemical Methods: Potentiometry •Reference Electrodes •Indicator Electrodes • pH Electrodes • -Selective Electrodes

Reference Electrodes

• A half-cell with a constant potential that is used as a comparison for other measurements. • Made of some stable , is readily available and usually simple to use. • Should be non-toxic if you are going to use it in a biological system. • Should be rugged and portable if you are going to use it in the field! • Several Varieties: – S.H.E – Ag/AgCl –

27 S.H.E. (again) • Cell made with hydrogen gas and hydrogen ion using a Pt electrode • Not practical for regular use due to the hydrogen gas • Assigned a standard potential of zero by definition

Ag/AgCl • Constructed of a , coated with silver , in a solution containing – Sometimes AgCl is in solution saturated with KCl – Saturated KCl keeps the chloride activity constant and helps keep the electrode stable

  AgCl(s) + e  Ag(s) + Cl (aq)

E° = 222mV (Relative to S.H.E.)

E (sat. KCl) + 197 mV (relative to S.H.E.)

RT RT E = E°  ln[Cl ] = 0.05916@25°C nF nF

IF we replace ln with base ten log!

28 29 Calomel (S.C.E) • Calomel is (I) chloride • One of the most common reference electrodes

  Hg2Cl2(s) + 2e  Hg(1) + 2Cl (aq )

E° = 268 mV (relative to S.H.E.)

E (sat. KCl) = 241 mV (relative to S.H. E.)

RT E = E°  ln[Cl ]2 nF

30 Conversions among different scales (different electrodes) • The reduction potential for (III) under standard conditions is ? (look it up) – What is it if measured compared to the Ag/AgCl ? – A saturated calomel electrode?

Indicator Electrodes • Used for indicating • Electrodes the potential caused – by some chemical – species as compared – to the reference • ? electrode. • Solid carbon • Usually connected to • Carbon Paste the + (cathode) side – Analyte of Interest of the potentiometer • Ion Selective • Made of a variety of – Glass membrane often allegedly, but – Crystalline never actually, inert • Solid State materials. – Liquid Membrane

31 Typical Metal Electrode Arrangement

pH (Glass Membrane) Electrodes • One of the simpler ion-selective electrodes (ISE) • Hydrogen Ion imparts a charge across a hydrated glass membrane

• Generally include an internal reference electrode (Ag/AgCl) and a separate Ag/AgCl electrode for sensing the charge imparted by the hydrogen

• Not as simple to use as you think!

32 33 Basic Nernst Equation of a pH Electrode (only the pH sensing cell, not the reference electrode)….

  AH + outside E pH cell = constant + x0.05916 x log  A +  H inside 

The constant is cancelled out by calibration in similar is an efficiency factor that cancels out by calibration in similar Ionic strength solutions. So, what really matters are the activities of hydroben ions! Don’t forget, electrodes respond to ACTIVITIES, even if we pretend That they are to simplify our calculations.

34 Ionic Activity and Debye and Huckel in 1923

Ionic Activity ai = riC

2  0.51z I log ri = 1+ I

Ionic Strength

ISFET pH electrode

(Ion Selective Field Effect Transistor)

35 What does 0.05916 mean? • It is a constant, if there is a one- reaction • It can be considered as the of a constant of 59.16 mV (relative pH mV readings) • A pH meter is a high-impedance 100.00 2.00 potentiometer (measures voltage) 159.16 3.00 • A pH change of “1” imparts a 218.32 4.00 change in 59.16 mV to the 277.48 5.00 potential recorded by the pH 336.64 6.00 meter! 395.80 7.00 • 1 pH unit change= 59.16 mV

Errors in pH Measurement…. 1. Junction potential due to the salt bridge and differences in junction potentials over time due to contamination of the junction • Overcome by regular recalibration 2. Sodium Error will result in high concentration of sodium solutions. The sodium can also impart a charge across the glass membrane. 3. error (strong ) can saturate or contaminate the membrane with hydrogen ion! 4. Equilibration Error is overcome by letting the electrode equilibrate with the solution 5. Dried out glass membrane (ruins electrode) 6. Temperature. Since temperature affects activities, it is best to have all solutions at the same, constant temperature! 7. Strong bases. Strongly basic solutions (>pH 12) will dissolve the glass membrane! 8. Uncertainty in your buffer pH due to normal weighing, diluting errors.

36 Ion Selective Electrodes…… • Selectivity Coefficient – Defines how an ISE responds to the species of interest versus some interfering species • Interferences cause a signal (voltage) to be imparted on the electrode that is NOT the result of the ion or chemical species of interest – You want the selectivity coefficient to be as SMALL as possible electrode response to X k = (A,X ) electrode response to A Where X is the interfence and A is the analyte

0.05916 E = constant   log A + k  A n ()A ()(A,X ) X

Examples of Electrodes / Interferences

37 Solid State Electrodes

• Uses a small amount of a doped crystal to transport charge from the solution to an inner electrode • The inner (sensing) electrode can be a Ag/AgCl electrode, and a separate Ag/AgCl electrode can be present – Combination electrode • You can use a separate reference electrode also.

38 • Typical equation (Fluoride Electrode):

E = constant    0.05916  log()AF outside

•Use: – Prepare calibration and sample solutions to similar ionic strengths and – Connect ISE and reference electrode to potentiometer (pH meter in mV mode) – Record potentials of calibration solutions – Prepare – Measure potential of sample solutions and calculate fluoride activity!

39 Liquid Membrane ISE’s

• Replace the solid state crystal with a liquid ion-exchanger filled membrane • Ions impart a charge across the membrane • The membrane is designed to be SELECTIVE for the ion of interest……

40 Voltammetry

Voltammetry can be used to • Study electroactivity of ions and at the electrode/solution interface • Probe coupled chemical reactions and measure rates • Examine electrode surfaces

An consists of a working (analyzing) electrode, an auxiliary (counter) electrode, and a reference electrode. The control device is a .

Clarifications of disk electrodes:

In quiescent solutions,

Conventional-sized Microelectrode Ultramicroelectrode Nanode (>mm) (>25 mm) (0.1-25 μm) (nm)

In flowing or agitated solutions,

Rotating disk electrode, channel electrode, wall-jet electrode, and flow-through tubular electrode.

Disk Pt/Au wire or coil Imbedded In Teflon or Kel-F

41 Stripping Analysis

In anodic stripping voltammetry (analysis), analytes are reduced and deposited into (onto) an electrode. They are reoxidized during the stripping step. e.g. Cd2+(aq) + 2e = Cd(Hg) Deposition Step Cd(Hg) – 2e = Cd2+ (aq) Stripping

In the deposition step, solution is stirred or the electrode is rotated.

In cathodic stripping voltammetry, typically anions are oxidized and deposited onto an electrode with subsequent stripping via a negative potential scan.

- 2I + 2Hg -2e = Hg2I2 at a Hg electrode (deposition step)

- Hg2I2 + 2e = 2 Hg + 2I (cathodic stripping, reduction)

Because sample analytes are preconcentrated from a large-volume dilute solution into (onto) a small-volume electrode under forced convection, trace analysis (enhanced sensitivity) can be realized.

Tissue-based (voltammetric)

PPO:

42