Reaction - the basics

ox + red <=> red + ox Ch.14-16 1 2 1 2 Oxidizing Reducing Agent Agent

Redox reactions: involve transfer of from one species to another. (oxidant): takes electrons (reductant): gives electrons

Redox Reaction - the basics Balance Redox Reactions (Half Reactions)

Reduced Oxidized 1. Write down the (two half) reactions. ox1 + red2 <=> red1 + ox2 2. Balance the (half) reactions ( and Charge): a. Start with elements other than H and O. Oxidizing Reducing Agent Agent b. Balance O by adding . c. balance H by adding H+. Redox reactions: involve transfer of electrons from one d. Balancing charge by adding electrons. species to another. (3. Multiply each half reaction to make the number of Oxidizing agent (oxidant): takes electrons electrons equal. Reducing agent (reductant): gives electrons 4. Add the reactions and simplify.) VFe V Fe3+ V 2+ 2 +→+ VFe 3 ++

Example: Balance the two half reactions and redox Important Redox Titrants and the Reactions reaction equation of the of an acidic of Na2C2O4 ( oxalate, colorless) with KMnO4 (deep purple). Oxidizing (Oxidants) - 2- 2+ MnO4 (qa ) + C2O4 q(a ) → Mn q(a ) + CO2(g) (1) Permanganate +qa -qa 2-qa 16H ( ) + 2MnO4 ( ) + 5C2O4 ( ) → − + − 2+ 2+ MnO 4 8 5 +→++ 4 2OHMneH 2Mn q(a ) + 8H2O(l) + 10CO2 (g)

MnO − 34 −+ →++ + 2)( OHsMnOeH Example: Balance 4 2 2 Sn2+ + Fe3+ <=> Sn4+ + Fe2+ − − 2− 4 + → MnOeMnO 4 2+ - 3+ 2+ Fe + MnO4 <=> Fe + Mn

1 Important Redox Titrants and the Reactions Important Redox Titrants and the Reactions

Oxidizing Reagents (Oxidants) Oxidizing Reagents (Oxidants) (2) (3) Potassium Iodate

2− −+ 3+ 72 +→++ 72614 2OHCreHOCr 1 − 56 −+ +→++ 3 OHIeHIO 3 2 22 2− 4+ + 3+ 2+ 2 7 2 ++→++ 2OHUO3Cr2H2U3OCr

Important Redox Titrants and the Reactions Important Redox Titrants and the Reactions Reducing ( Reductants ) Reducing Reagent ( Reductants ) (1) (2)

2- 2- ¯ 1 2S2O3 S4O6 +2e − +→ II e− 2 2

Galvanic Cells - Components Galvanic Cells - Line Notation

A galvanic (voltaic) cell uses a spontaneous to generate .

– ( and ). – : cations move from anode to cathode, anions move from cathode to anode. Line notation

Cd(s) | Cd(NO3)2 (aq) || AgNO3 (aq) | Ag(s) Phase boundary Salt bridge Phase boundary

2 Standard Potentials Standard Reduction (Half-Cell) Potentials Standard electrode (S.H.E.) • The S.H.E. is the cathode. It consists of a Pt electrode in a Standard (Eo ) is the associated tube placed in 1 M H+ solution. H is bubbled through the with a reduction reaction at an electrode when all solutes are 2 1 M and all gases are at 1 atm. Cathode tube. • For the S.H.E., we assign + - 2H q(a , 1M) + 2e → H2 ( g, 1 atm)

Reduction Reaction • E°red of zero. • The potential of a cell can be calculated from standard 2H+ (1 M ) +2e- H (1 atm) 2 reduction potentials: Eo = 0 V E°cell = E°red(cathode)− E°red (anode)

Standard Reduction (Half-Cell) Potentials Standard Potentials We use hydrogen (S.H.E.)

• Consider Zn(s) → Zn2+q(a ) + 2e-. We measure E + − ° cell (aq)+ ↔ 2 g)(1/2HeH E ≡ 000.0 relative to the S.H.E. (cathode): We can measure Eº for other half-reactions, relative to E°cell = E°red(cathode) - E°red(anode) the hydrogen reaction, e.g. for : 0.76 V = 0 V - E°red(anode). +(aq) − ↔+ AgeAg ( s)E° = 799.0 • Therefore, E°red(anode) = -0.76 V. • Standard reduction potentials must be written as Standard reduction potentials are listed in Ap. H in º reduction reactions: your book. E for the H2 reaction is for the reaction º 2+ - at 25 C Zn q(a ) + 2e → Zn(s), E°red = -0.76 V.

Nernst Equation for a Half-Reaction • The relates the potential of the half reaction to reagent • For the half-reactions: b ° RT [B] - ° EE −= ln ↔+ bna BeA E nF []A a

- ° dc eBA +↔++ EdDcCnba ° RT [ ] [DC ] EE −= ln nF [][]BA ba RT EE −°= lnQ nF # of moles of electrons

3 Nernst Equation for a Half-Reaction Nernst Equation for a Complete Reaction

At 298K (25oC)

ox1 + red2 <=> red1 + ox2 V 05916.0 V EE ° −= logQ n

EE −= E−+

Example (Nernst) Example (Net Reaction) • Write the Nernst equation for the reduction of • Find the voltage for the Ag-Cd cell and state if the reaction is spontaneous if the right cell contained 0.50 M AgNO3(aq) and if phosphoric to solid white phosphorous: the left contained 0.010 M Cd(NO3)2(aq) + - • 1) + − ° 3 4 POH ++ 5H ↔ 4 s + 2 OH4)(1/4P5e E =° -.402 ↔+ s)(Ag2e2Ag2 E+ = 799.0 V + − ° 05916.0 1 ↔+ s)(Cde2Cd E− −= 402.0 V E 402.0 −−= log 5 5 []POH H + 43 [] • 2) 05916.0 1 E+ = 799.0 − log 2 = 781.0 V • Note that multiplying the reaction by any factor 2 []0.50 05916.0 1 does not affect Eº or the calculated E: • 3) E 402.0 −−= log −= V461.0 − 2 []0.010 10e 10H PO2H 10H + ++ 10e - ↔ s + OH8)(1/2P E =° -.402 3 4 4 2 • 4) = − EEE −+ = − − = + 242.1)461.0(781.0 V 05916.0 1 • 5) E 402.0 −−= log −+ 2 + 10 + ↔ s)(Ag2e2Ag2 10 []POH H + 2+ 43 [] −+ s s)(Ag2CdAg2)(Cd + ↔ s)(Cde2Cd ↔+ +

Determination of the Equivalence Point Curve

Aox + Bred <=> Ared + Box EXAMPLE: Derive the titration curve for 50.00 mL At equivalence point, Ecell=0: of 0.0500 M Fe2+ with 0.00, 0592.0 A ][ 0592.0 B ][ 15.00, 25.00, 26.00 mL ° red ° red 4+ EA − log = EB − log 0.1000 M Ce in a medium nA Aox ][ nB Box ][ that is 1.0 M in HClO4. Potential of saturated = and = ABBA at the eequivalenc point red ox r oxed electrode is 0.241 o o V. AA + nEn EBB Valid for simple E = Redox expressions + nn BA

4 EXAMPLE: Derive the titration curve for 50.00 mL of EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 2+ 0.0500 M Fe with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M M Ce4+ in a medium that is 1.0 M in HClO .Potential of 4+ 4 Ce in a medium that is 1.0 M in HClO4. Potential of saturated calomel electrode is 0.241 V. saturated calomel electrode is 0.241 V. Titration reaction: Fe2+ + Ce4+ <=> Ce3+ + Fe3+ Fe2+ + Ce4+ <=> Ce3+ + Fe3+ 4+ Fe3+ + e- <=> Fe2+ Eo = 0.767 V At 15.00 mL of Ce added, VFeMFe > VCeMCe Ce4++ e- <=> Ce3+ Eo = 1.70 V “Buffer region” At 0.00 mL of Ce4+ added, initial point no Ce4+ present; MV minimal, unknown [Fe3+]; thus, insufficient information to Fe3+ ][ = CeCe +VV calculate E CeFe mL M )1000.0)(00.15( 0592.0 Fe2+ ][ = ×= 10308.2 −2 M EE °= − log − 241.0 + )00.1500.50( mL n Fe3+ ][

EXAMPLE: Derive the titration curve for 50.00 mL of EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 4+ 4+ Ce in a medium that is 1.0 M in HClO4. Potential of Ce in a medium that is 1.0 M in HClO4. Potential of saturated calomel electrode is 0.241 V. saturated calomel electrode is 0.241 V. Fe2+ + Ce4+ <=> Ce3+ + Fe3+ Fe2+ + Ce4+ <=> Ce3+ + Fe3+ At 15.00 mL of Ce4+ added, V M > V M 4+ Fe Fe Ce Ce At 15.00 mL of Ce added, VFeMFe > VCeMCe “Buffer region” “Buffer region” Fe3+ ×= 10308.2][ −2 M Half-reactions: Fe3+ + e- <=> Fe2+ Eo = 0.767 V Fe3+ ×= 10308.2][ −2 M Fe2+ ×= 10538.1][ −2 M 2+ − MVMV Fe ][ = CeCeFeFe 0592.0 Fe2+ ][ +VV CeFe EE −°= log − 241.0 n Fe3+ ][ mL M − mL M )1000.0)(00.15()0500.0)(00.50( −2 = ×= 1054.1 M 0592.0 ×1054.1 −2 + )00.1500.50( mL 767.0 −= log =− 533.0241.0 V 1 2.038×10−2

EXAMPLE: Derive the titration curve for 50.00 mL of EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 4+ 4+ Ce in a medium that is 1.0 M in HClO4. Potential of Ce in a medium that is 1.0 M in HClO4. Potential of saturated calomel electrode is 0.241 V. saturated calomel electrode is 0.241 V.

Fe2+ + Ce4+ <=> Ce3+ + Fe3+ Fe2+ + Ce4+ <=> Ce3+ + Fe3+ 4+ 4+ At 25.00 mL of Ce added, VFeMFe = VCeMCe, At 26.00 mL of Ce added, VFeMFe < VCeMCe, Equivalence point After equivalence point Half-reactions: Fe3+ + e- <=> Fe2+ Eo = 0.767 V 3+ MV FeFe Ce4++ e- <=> Ce3+ Eo = 1.70 V Ce ][ = +VV CeFe + nEn E E = CeCeFeFe − 241.0 mL M )0500.0)(00.50( + nn = ×= 1029.3 −2 M CeFe + 00.2600.50( )mL + 70.1767.0 = =−=− 99.0241.023.1241.0 V 1+1

5 EXAMPLE: Derive the titration curve for 50.00 mL of EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 0.0500 M Fe2+ with 0.00, 15.00, 25.00, 26.00 mL 0.1000 M 4+ 4+ Ce in a medium that is 1.0 M in HClO4. Potential of Ce in a medium that is 1.0 M in HClO4. Potential of saturated calomel electrode is 0.241 V. saturated calomel electrode is 0.241 V. Fe2+ + Ce4+ <=> Ce3+ + Fe3+ 2+ 4+ 3+ 3+ 4+ Fe + Ce <=> Ce + Fe At 26.00 mL of Ce added, VFeMFe > VCeMCe 4+ After equivalence point At 26.00 mL of Ce added, VFeMFe < VCeMCe, 3+ - 2+ o After equivalence point Half-reactions: Fe + e <=> Fe E = 0.767 V Ce4++ e- <=> Ce3+ Eo = 1.70 V Ce3+ ×= 1029.3][ −2 M Ce3+ ×= 1029.3][ −2 M Ce4+ ×= 1032.1][ −3 M − MVMV Ce4+ ][ = FeFeCeCe 0592.0 Ce3+ ][ +VV EE −°= log − 241.0 CeFe n Ce4+ ][ mL M − mL M )0500.0)(00.50()1000.0)(00.26( −3 = ×= 1032.1 M 0592.0 ×1029.3 −2 + 00.2600.50( )mL 70.1 −= log =− 63.0241.0 V 1 1.32×10−3

Galvanic Cells A galvanic (voltaic) cell uses a spontaneous Theoretical curve for chemical reaction to generate electricity. titration of 100.0 mL 2+ – Electrodes (cathode and of 0.50 0 M Fe with anode). 4+ 0.100 M Ce in 1 M – Salt bridge: cations HClO4. (p.330-331) move from anode to cathode, anions move 2+ from cathode to anode. 0592.0 Fe ][ EE −°= log 3+ − 241.0 – produces electricity n Fe ][ when the cell reaction 0592.0 767.0 −= =− 526.0241.01log V is not at equilibrium. 1

Free Energy and Electrochemical Reaction E and Equilibrium Constants • A produces electricity because the cell • The free energy change, ∆G, for a chemical reaction at reaction is not at T, P equals the maximum possible electrical work • At equilibrium, E for the net reaction can thus be that can be done by the reaction on its surroundings. - ° related to K + ↔ CeA Ecna + - ↔+ DeD Ednd ° EG q - Work = −∆ = × c ≈ b ’ ° 05916. [C] ∆ ° 05916. [B] ÷ EEE −+ E+ −=−= log a ∆ E− −− log d ÷ −=∆ nFEG n []A « n []D ◊

dc ° 05916.0 [ ] [DC ] ° 05916.0 • A spontaneous reaction (∆G<0)‰ E>0 EE −= log ba E −= logQ °C25at n [][]BA n

• Think: how the equilibrium constant can be related to free ≈ ° ’ ° 05916.0 ∆ nE ÷ 0 EE =→= log K °C25at ∆ ÷ energy? n K =10« 05916.0 ◊ ° C25at When the cell is at equilibrium, E=0 and Q=K

6 Cells as Chemical Probes Example: Find the equilibrium constant Two Equilibria: (1) Equilibrium between two half-cells

3+ 2+ 2+ (2) Equilibrium within each half-cell s +↔+ Fe2Cu2Fe)(Cu − - ° sAgCl +↔+ )10.0,()(e)( EMaqClsAg + = 222.0 V This can be divided into two half reactions: + - o aqH M + ↔ 2 bargH )00.1,(2e)?,(2 - = 0E V 3 −+ ↔+ 2+ )(Fe2e22Fe Es ° = 771.0 V + •The half-reaction that you 2+ + - ↔ s)(Cu2eCu o = 339.0E V write must involve species that - appear in two oxidation states in the cell. ° ° EEE ° =−=−= 432.0339.0771.0 V + − •The reaction is in equilibrium in the right cell is not the net

≈ ° ’ cell reaction: ∆ nE ÷ (2)(0.432) ∆ 05916.0 ÷ K =10« ◊ =10 )05916.0( ×= 104 14 s ↔ + aq + − aq)(Cl)(Ag)(AgCl

Important Biochemical Reactions Survival Tips

Electrochemistry, , , complex formation, and acid- • Formal potential, Eº’, meant to define potentials (1) Write half-reactions and their standard potentials under conditions (2)Write Nernst equation for the net reaction and put in of all the known quantities. (3) Solve for the unknown and use that concentration in the chemical equilibrium equation to solve the problems.

Ex. p.286-287

S.H.E. Potentiometry (again) • Potentiometry: the use of electrodes to measure from chemical reactions. • Electroactive species: can donate or accept electrons at an electrode; can be measured as the part of a galvanic cell (analyte) • : we then connect the analyte half- The standard reduction potential, Eo, for each half- reaction to a second cell with a fixed composition (known cell is measured when potential), the 2nd half-cell is called reference electrode. different half-cells are • Indicator electrode: responds to analyte connected to S.H.E. - electrodes: inert , e.g., Pt, Au S.H.E. || Ag+ (aq. =1) | Ag(s) - -selective electrodes: respond to specific analytes Standard means that the activities of all species are unity. Not practical for regular use due to the hydrogen gas

7 Reference Electrodes Reference Electrodes

Detect Fe2+ /Fe3+ in solution: a Pt (indicator electrode) in the half-cell and connect this half cell to a 2nd half-cell at a constant potential.

Reference Electrode

+ -3 2+ ° Indicator Electrode ↔+ FeeFe E+ = 771.0 V The entire left half-cell containing s - s +↔+ Cl)(Age)(AgCl − E° = 222.0 V - appropriate and a salt bridge. ≈ 2+ ’ ∆ [Fe ]÷ E+ −= log059.771.0 ∆ 3+ ÷ + -3 2+ ° « []Fe ◊ ↔+ FeeFe E+ = 771.0 V + − EEE =- - - − ° E- = 222.0 []Cl.059log- s s +↔+ Cl)(Age)(AgCl E- = 222.0 V

Silver-Silver Electrode Saturated Calomel Electrode (S.C.E.)

Ag | AgCl Electrode : - − ° s s +↔+ Cl)(Age)(AgCl E- = + 222.0 V KCl saturatedw/ saturatedw/ KCl E += 197.0 V

Hg | Hg2Cl2 Electrode: • The difference in E is due to activity - − ° 22 s l +↔+ Cl)(Hge)(Cl1/2Hg E += 268.0 V coefficients. KCl saturatedw/ saturatedw/ KCl E += 241.0 V • A double junction electrode: avoid to mix Cl- with the analyte

Voltage Conversions between Different Indicator Electrodes - Metal Electrodes Reference Scales

• Metal electrodes: develop potential in response to a redox reaction on their surface • Pt is mostly inert, not participating in reactions – It simply allows transfer to/from solution • is the most common metal indicator electrode is also an inert metal indicator electrode An indicator electrode has a potential of -0.351 V with respect to an S.C.E., what’s its potential with • electrodes are often used because many redox respect to an S.H.E.? reactions are very fast on a carbon surface V241.0 0.351V- + V241.0 = -0.110V

8 Indicator Electrodes-Metal electrodes Example: 100.0 mL solution containing 0.100 M NaCl • Metal electrodes: develop potential in response to a redox reaction on was titrated with 0.100 M AgNO3 and monitored with a their surface S.C.E. What voltage reading would be observed after 65.0

• Inert metal indicator electrode: allows to/from the mL? −+ ClAg →+ s)AgCl( solution but not participating in reactions, e.g., Pt (the most common one), Au. Ve = 0.100 mL • Silver Indicator Electrode E += [Aglog05916.0558.0 + ] ≈100.0 ’ + - ° [ − ]= 100.0)(350.0(Cl M) ∆ ÷ = 021.0 M sg ↔+ )(Age)(A Es + = 799.0 V «165.0 ◊

- − ° 22 s l +↔+ Cl)(Hge)(Cl1/2Hg E− = 241.0 V −10 + K sp ×108.1 −9 ≈ ’ []Ag = − = ×= 105.8 ∆ 1 ÷ []Cl 021.0 EEE −+ −=−= log059.0799.0 ∆ ÷ − 241.0 « []Ag+ ◊ E = + 05916.0558.0 log -9 =× 081.0)108.5( V E += [Aglog059.0558.0 + ]

Ion Mobilities and Liquid Junction Potentials Junction Potential

• Junction potential: A voltage difference develops whenever dissimilar solutions are in contact. • Happens at the salt bridge/solution interface since different have different mobilities in water. • A major (fundamental) source of error in a potential measurement.

Ion-Selective (pH Combination Electrode) Electrodes (ISE) • The most common ISE • Responds selectively to one • A glass membrane ion using an ion- selectively binds H+ selective membrane • Two Ag|AgCl reference • Do not involve electrodes measure the redox reactions potential difference across • The electric the glass across the • An ion-exchange membrane depends equilibrium is on the on [analyte] surface of the glass • E is measured by membrane two reference electrodes 0.05916 + - + + − E = constant + log []C Out C25at n =° charge of analyte n aqaq(s)(s) out aqin ssaq )(Ag|)(AgCl|)(Cl),(H|)(H||)(Cl|AgCl|Ag

9 The Glass Membrane of a pH Electrode Errors in pH Measurement •Cross section of the glass membrane of a pH electrode. 1. Calibration standards (±0.01 pH) 2. Junction potential (~0.01 pH) •H+ can diffuse into the membrane to replace the 3. Junction potential drift (recalibrate every 2 hrs) metal ions through binding 4. Sodium error (when [H+] is low and [Na+] is high) to in glass (ion- 5. Acid error (strong acid, the glass surface is saturated exchange equilibrium). with H+) Response of the electrode: 6. Equilibration time (~30s with adequate stirring)

+ E constant += β (0.0592) []Hlog Out 7. Hydration of glass (A dry electrode requires several O of soaking) E constant −= β (0.0592) pH Out C25at 8. (calibrate at same T as measurement) β: ~1.00, electromotive 9. Cleaning (remove hydrophobic liquid) efficiency measured during calibration.

Specifications for Electrochemical Selectivity Coefficient Techniques

• The selectivity Advantages X toresponse X coefficient: the k XA, = Linear response to analyte over wide dynamic range A toresponse A relative response of Nondestructive the electrode to −5 Short response times k ,K Na ++ =1×10 different species Unaffected by color/turbidity (limited matrix effects)

k ++ = 44.0 Cs ,K Cs Cheap • The smaller k is, the k ++ = 8.2 Rb ,K Rb less interference there Disadvantages is due to ion X Sensitivity (High detection limits) Response of ion -selective electrode : Not universal 0.0592 O E ƒ )a(ka = constant ± β ( ) log [ A + xA,X ] C25at n x

Voltammetry

• A collection of techniques in which the relation between current and voltage is observed during electrochemical process.

• Can be used to (1) Study electroactivity of ions and at the electrode/solution interface (2) Probe coupled chemical reactions and measure electron p transfer rates i (3) Examine electrode surfaces

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

10 Stripping Analysis

Anodic stripping (ASV): 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

Cathodic stripping voltammetry (CSV): typically anions are oxidized and deposited onto an electrode with subsequent stripping via a negative potential scan.

- e.g. 2I + 2Hg -2e = Hg2I2 at a Hg electrode Deposition step - Hg2I2 + 2e = 2 Hg + 2I cathodic stripping, reduction Trace analysis (enhanced sensitivity) can be realized since sample analytes are preconcentrated from a large-volume dilute solution into (onto) a small-volume electrode under forced convection.

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