Reduction - Oxidation Titrations Theory and Application Ashraf M. Mahmoud, Associate Prof. 1

Oxidation-Reduction (Redox) Definitions & Terms 2 Fe 2+ + Cl 2 2Fe Cl  2 Fe e 2 Fe 3+ ( Loss of electrons : Oxidation ) Cl 2 + 2e 2Cl - ( Gain of electrons : Reduction )  In every redox reaction, both reduction and oxidation must occur.  Substance that gives electrons is the reducing agent or reductant.  Substance that accepts electrons is the oxidizing agent or oxidant. Reaction of ferrous ion with chlorine gas Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half equation. 4

Oxidation Number (O.N)  The O.N of a monatomic ion = its electrical charge.  The O.N of free uncombined elements = zero  The O.N of metals in combination is usually positive  The O.N of an element in a compound may be calculated by assigning O.N to the remaining elements of the compound using the above rules and the following additional rules:  The O.N. oxygen = -2 (except in peroxides O= -1).  The O.N. of hydrogen = +1 (except in metallic hydride H=-1)  The algebraic sum of the positive and negative O.N. of the atoms of the formula (O.N of a compound) = zero.  The algebraic sum of the positive and negative O.N. of the atoms in a polyatomic ion (radical )= the charge of the ion. 5

Oxidation Numbers of Some Substances SubstanceOxidation Numbers NaCl H 2 NH 3 H 2 O 2 LiH K 2 CrO 4 SO 4 2- KClO 3 Na = +1, Cl = -1 H = 0 N = -3, H = +1 H = +1, O = -1 Li = +1, H = -1 K = +1, Cr = +6, O= -2 O = -2, S = +6 K = +1, Cl = +5, O= -2 6

For manganese SpeciesMnMn 2+ Mn 3+ MnO 2 MnO 4 2  MnO 4  O.N For nitrogen SpeciesNH 3 N2H4N2H4 NH 2 OHN2N2 N2ON2ONOHNO 2 HNO 3 O.N.–3–2– Oxidation states of manganese and nitrogen in different species 7

O.N in covalent compounds as NH 3, CH 4 and H 2 O are as follows: 8

Eqivalent weights of oxidants & reductant a.Method of limited application (old definition of eq. wts). b.Ion-electron method (By the number of electrons lost or gained by one mole of the substance in a half reaction). c.Oxidation number method Is the weight of the substance that reacts with or contains 1 g of available hydrogen or 8 g of available oxygen. 2KMnO 4 + acid  K 2 O + 2MnO + 5O 2KMnO 4 → 5O, i.e Eq. W = 2x KMnO 4 /10 or KMnO 4 / 5. 2KMnO 4 + alkaline  K 2 O + 2MnO 2 +3O Eq. W = MW/3 K 2 Cr 2 O 7  K 2 O + Cr 2 O 3 + 3O Eq. W = MW/6 KMnO 4 = MW/5 9

Al + HCl = AlCl 3 + H Oxidation: ( Al = Al e - )  2 = 6 e - Reduction: ( 2 H + + 2e = H 2 )  3 = 6 e - Redox: 2 Al + 6 H + = 2 Al H 2 Balancing Redox Reactions Using the Half-Reaction Method 10

 Balance each half reaction: MnO 4  + 5é Mn 2+ C 2 O 4 2  2 CO 2 + 2é 2 MnO 4  + 5 C 2 O 4 2  2 Mn CO H 2 O  Balance oxygen atoms by adding water 2 MnO 4  + 5 C 2 O 4 2  + 16 H + 2 Mn CO H 2 O  Balance hydrogen atoms by adding H +  Make number of electrons gained in each reaction equal 2 MnO 4  + 5 C 2 O 4 2  2 Mn CO 2 2 MnO 4  + 5 C 2 O 4 2  + 16 H + 2 Mn CO H 2 O Balancing Redox Reactions Using the Half-Reaction Method 11

Balancing Redox Reactions Using the Half-Reaction Method Assign oxidation numbers Identify substance oxidized Identify substance reduced Write 1/2 reactions (eliminate spectator ions, if necessary, adding appropriate number of electrons to balance charge Balance electrons gained & lost Add 1/2 reactions to get overall reaction Add spectator ions (if necessary) and finish the balancing 12

Cl 2 + 2Fe 2+  2Fe Cl - Ce 4+ + Fe 2+ → Ce 3+ + Fe 3+ Cr 2 O 7 2  + Fe 2+ +  Cr 3+ + Fe 3+ + KClO 3 + Fe 2+ + → Fe 3+ + KCl + MnO Fe 2+ +  Mn 2+ + Fe 3+ Cl 2 + Fe 2+  Cl 2 + Fe 2+  Fe Cl - KClO 3 + 6Fe 2+ + → 6Fe 3+ + KCl + KClO 3 + 6Fe H + → 6Fe 3+ + KCl + 3H 2 O Cr 2 O 7 2  + 6Fe H +  2Cr Fe H 2 O MnO Fe H +  Mn Fe H 2 O Examples 13

BrO 3  + 5Br  + 6H + → 3Br 2 + 3H 2 O SnCl 2 + 2HgCl 2 → Hg 2 Cl 2  + SnCl 4 3As 2 O 3 + 2BrO 3  → As 2 O 5 + 2Br  K 2 S 2 O 8 + 2Fe H + → 2Fe KHSO 4 I 2 + 2Na 2 S 2 O 3  2NaI + Na 2 S 4 O 6 2KMnO HCl → 2 KCl + 2MnCl 2 + 5Cl H 2 O 2MnO 2 + 2AsO H +  2Mn AsO H 2 O BrO 3  + Br  + → 3Br 2 + Examples 14

Nernest Equation for Electrode Potential (E) E t = Electrode potential at temperature t. E  = Standard electrode potential (constant depend on the system) R = Gas constant T = Absolute Temp. (t°C + 273) F = Faraday constant (96500 Coulombs) Log e = ln (natural logarithm = log) n = Valency of the ion [M n+ ]=Molar concentration of metal ions in solution E t = E o + log [M n+ ] RT nF E 25 °C = E o + log [M n+ ] n 17

Standard Oxidation Potential (E) Standard Oxidation Potential (E o ) E 25 °C = E o + log [Oxidized] / [Reduced] n The higher E°, the stronger the oxidizing power of its oxidant and the weaker the reducing power of its reduced form. The most powerful oxidizing agents are those at the top (higher +ve) and the most powerful reducing agents are at the bottom (higher – ve). 22

If any two redox systems are combined, the stronger oxidizing agent gains electrons from the stronger reducing agent with the formation of weaker reducing and oxidizing agents. Cl 2 /Cl - (E° = V) & Fe 3+ /Fe 2+ (E°= V) Cl 2 + 2Fe 2+  2Fe Cl - strong oxid. + strong red. → weak oxid. + weak red. agent agent agent agent 23 Standard Oxidation Potential (E) Standard Oxidation Potential (E o )

Oxidation Potentials: Electrochemical Series 25

 The potential of MnO 4  /Mn 2+ varies with the ratio [MnO 4  ]/[Mn 2+ ].  If ferrous is titrated with MnO 4  in presence of Cl , chloride will interfere by reaction with MnO 4  and gives higher results. Factors Affecting Oxidation Potential 1. Common Ion E 25 °C = E o + log [Oxid] /[Red] n Zimmermann’s Reagent (MnSO 4, H 3 PO 4 and H 2 SO 4 )  MnSO 4 has a common ion (Mn 2+ ) with the reductant that lowers the potential of MnO 4  /Mn 2+ system:  Phosphoric acid lowers the potential of Fe 3+ /Fe 2+ system by complexation with Fe 3+ as [Fe(PO 4 ) 2 ] 3 .  Sulphuric acid is used for acidification. 26

2. Effect of pH The oxidation potential of an oxidizing agent containing oxygen increases by increasing acidity and vice versa.  Potassium permenganate: MnO 4  + 8H + + 5e   Mn H 2 O E = E o + log MnO 4  /Mn 2+ [MnO 4  ][H + ] 8 [Mn 2+ ]  Potassium dichromate: Cr 2 O 7 2  + 14H + +6e   2Cr H 2 O E = E o + log Cr 2 O 7 2  /Cr 3+ [Cr 2 O 7 2  ][H + ] 14 [Cr 3+ ] 27 Factors Affecting Oxidation Potential

As 5+ or Sb 5+ can oxidize I   I 2 or I 2 oxidizes As 3+ and Sb 3+  As 5+ or Sb 5+ depending on the pH of the medium:  In presence of much acid, The Eº of As 5+ /As 3+ or Sb 5+ /Sb 3+  (it oxidizes I   I 2 ) As 2 O 5 + 4I  + 4H + → As 2 O 3 + 2I 2 + 2H 2 O Sb 2 O 5 + 4I  + 4H + → Sb 2 O 3 + 2I 2 + 2H 2 O  if E º is reduced by decreasing [H + ] (in presence of sodium bicarbonate), arsenite is oxidized with iodine. 28 Arsenate/ arsenite system:  Arsenate /arsenite & antimonate/antimonite  In slightly acid or neutral medium, The E º of As 5+ /As 3+ or Sb 5+ /Sb 3+  (I 2 oxidizes As 2 O 3 and Sb 2 O 3  As 2 O 5 and Sb 2 O 5 ) Factors Affecting Oxidation Potential

E (I 2 /2I  ) system increases by the addition of HgCl 2 since it complexes with iodide ions. Hg I   [HgI 4 ] 2  (low dissociation complex) 3. Effect of Complexing Agents E (Fe 3+ /Fe 2+ ) is reduced by the addition of F  or PO 4 3  due to the formation of the stable complexes [FeF 6 ] 3  and [Fe(PO 4 ) 2 ] 3  respectively. Thus, ferric ions, in presence of F  or PO 4 3  cannot oxidize iodide although E o (Fe 3+ /Fe 2+ ) = 0.77 while E o (I 2 /2I  ) = Iodine: I 2 + 2e   2I  E =E o + log I 2 /I  [I 2 ] [I  ] 2 Ferric: Fe 3+ +e   Fe 2+ E =E o + log Fe 3+ /Fe 2+ [Fe 3+ ] [Fe 2+ ] 29 Factors Affecting Oxidation Potential

Addition of Zn 2+ salts which precipitates ferrocyanide: [Fe(CN) 6 ] 4- + Zn 2+  Zn 2 [Fe(CN) 6 ]  The oxidation potential of ferri/ferrocyanide system in presence of Zn 2+ ions can oxidize iodide to iodine, although the oxidation potential of I 2 /2I - system is higher. 4. Effect of Precipitating Agents Ferricyanide: [Fe(CN) 6 ] 3  + e   [Fe(CN) 6 ] 4  E = E o + log Ferri/Ferro [[Fe(CN) 6 ] 3  ] [[Fe(CN) 6 ] 4  ] 30 Factors Affecting Oxidation Potential

In presence of tartarate or citrate I 2 oxidizes Cu + to cu +2 as these anions form stable complexes with Cu +2 Cu +2 /Cu + system E°= 0.15 I 2 /2I  system E°= 0.54 Expected Actually 2Cu I  → 2CuI 2 The instability of CuI 2 results in formation of Cu 2 I 2 ppt which  Cu + → ↑ E° of Cu +2 /Cu + system to become able to oxidize I  → I 2 unstable 31 I 2 oxidize Cu + → Cu +2 The reverse occurs due to instability of CuI 2 → Cu 2 I 2  + I 2  S 2 O 3 -2 We add KSCN… Why? Factors Affecting Oxidation Potential 4. Effect of Precipitating Agents

It is the plot of potential (E, volts) versus the volume (ml) of titrant. Redox Titration Curve Example: Titration of 100 ml 0.1 N Ferrous sulphate ≠ 0.1 N ceric sulphate. Ce 4+ + Fe 2+ (sample) → Ce 3+ + Fe 3+ The change in potential during titration can be either measured or calculated using Nernest equation as follows: 32  At the beginning (0.00 mL of Ce +4 added, inital point): E = E o /n log [Fe 3+ ]/[Fe 2+ ] No Ce +4 present; minimal, unknown [Fe +3 ]; thus, insufficient information to calculate E

36 Redox Titration Curve

 After adding 10 mL Ce 4+ : E= /1 log 10/90 =0.69 V  After adding 50 mL Ce 4+ : E = /1 log 50/50 = 0.77 V  After adding 90 mL Ce 4+ :E= /1 log 90/10 = 0.81 V  After adding 99 mL Ce 4+ : E = /1 log 99/1 = 0.87 V  After adding 99.9 mL Ce 4+ : E= log 99.9/0.1= 0.93 V  After adding 100 mL Ce 4+ : the two potentials are identical 33 Redox Titration Curve

E = E O /1 log [Fe 3+ ]/[Fe 2+ ] E = E O /1 log [Ce 4+ ]/[Ce 3+ ] Summation of two equations: 2 E = E o 1 + E o log [Fe 3+ ][Ce 4+ ] [Fe 2+ ][Ce 3+ ] At the end point: Fe 2+ = Ce 4+, and Fe 3+ = Ce 3+ 2 E = E o 1 + E o 2 E = ( E o 1 + E o 2 ) / 2 E = = 1.10 V  After adding 101 ml Ce 4+ : E= /1 log1/100= 1.33 V  After adding 110 mL Ce 4+ : E= log 10/100=1.39 V 34 Redox Titration Curve

Potential (V) Ce 4+ (mL) 35 Redox Titration Curve

Detection of End Point in Redox Titrations 1. Self Indicator (No Indicator) When the titrant solution is coloured (KMnO 4 ): KMnO 4 (violet) + Fe 2+ + H +  Mn 2+ (colourless) + Fe 2+. The disappearance of the violet colour of KMnO 4 is due to its reduction to the colourless Mn 2+. When all the reducing sample (Fe 2+ ) has been oxidized (equivalence point), the first drop excess of MnO 4  colours the solution a distinct pink. 37

In Titration of Fe 2+ by Cr 2 O 7 2  Cr 2 O 7 2  + 3Fe H +  2Cr Fe H 2 O The reaction proceeds until all Fe 2+ is converted into Fe 3+ Fe 2+ +Ferricyanide (indicator)  Fe 2+ ferricyanide (blue)]. Fe 2+ + [Fe(CN) 6 ] 3  → Fe 3 [Fe(CN) 6 ] 2. The end point is reached when the first drop fails to give a blue colouration with the indicator (on plate) 2. External Indicator 38 Detection of End Point in Redox Titrations

3. Internal Redox Indicator Redox indicators are compounds which have different colours in the oxidized and reduced forms. In ox + n e  In ox They change colour when the oxidation potential of the titrated solution reaches a definite value: E = E° /n log [In OX ]/[In red ] When [In ox ] = [In red ], E = E° Indicator colours may be detected when: [In ox ]/[In red ] = 1/10 or 10/1 hence, Indicator range: E = E° In ± /n 39 Detection of End Point in Redox Titrations

4. Irreversible Redox Indicators In acid solutions, methyl orange is red. Addition of strong oxidants (Br 2 ) would destroy the indicator and thus it changes irreversibly to pale yellow colour Some highly coloured organic compounds that undergo irreversible oxidation or reduction Methyl Orange 42 Detection of End Point in Redox Titrations

Importance of Redox Reactions in Biological systems and Pharmaceutical Applications  Oxidizing agents act on human tissues by one of the following mechanisms: 1.Germicides through liberation of oxygen in the tissues such as hydrogen peroxide 2.Bactericidal against anaerobes as KMnO 4 (for wounds) or Cl 2 (for dental therapy) through denaturing the proteins by direct oxidation.  Oxidation can have deleterious effect on the cells and tissues of human body because it produces reactive oxygen and nitrogen species, such as OH , NO , NO 2  and alkoxyl radicals RO  which damage cells and other body components.  Antioxidants (reducing agents) can counteract the effect of the reactive oxygen and nitrogen species and protect the other compounds in the body from oxidation. 44

Importance of Redox Reactions in Biological systems and Pharmaceutical Applications  Typical antioxidants include vitamin A, C and E; minerals such as selenium; herbs such as rosemary.  Antioxidants act by one or more of the following mechanisms: 1.Reaction with the generated free radicales can counteract the effect of the reactive oxygen and nitrogen species 2.Binding the metal ions needed for catalyzing the formation of reactive radicales 3.Repairing the oxidative damage to the biomolecules or induction of the enzymes that catalyze the repair mechanisms. 45

Redox reactions are included not only in inorganic or organic drug analysis but also in the metabolic pathways of drugs. Catabolic pathways: oxidation of C-atoms of carbohydrate or lipid or protein Anabolic pathways: reduction of C-atoms of carbohydrate or lipid or protein Phase-I Metabolic Pathways (Functionalization reactions) I- Oxidative reactions A.Oxidation of aromatic compounds → hydroxy aromatic compounds B.Oxidation of olefins → Epioxides C.Oxidative dealkylation Redox Reactions and Biological Action 46

RH (Parent drug) P-450-( Fe 3 + ) RH─P-450-(Fe 3+ ) RH─P-450-(Fe 2+ ) NADPH + flavoprotein reductase H2OH2O ROH (oxidized product) RH─P-450-(Fe 2+ ) O2-O2- Redox Reactions in Biological System 48