Redox Lecture Notes PDF

REDOXIMETRIC TITRATIONS Redox (Oxidation - Reduction) titrations are a type of volumetric analysis involves electron-transfer reactions. Oxidation: It is the process in which an atom, ion or molecule losses one or more electrons (loss of electrons) Reduction: It is the process in which an atom,...

REDOXIMETRIC TITRATIONS Redox (Oxidation - Reduction) titrations are a type of volumetric analysis involves electron-transfer reactions. Oxidation: It is the process in which an atom, ion or molecule losses one or more electrons (loss of electrons) Reduction: It is the process in which an atom, ion or molecule gains or accept one or more electrons (gain of electrons). 2Fe+2 + Cl2 2Fe+3 + 2 Cl- Both processes occur simultaneously.  Oxidants and Reductant Oxidant (oxidizing agent): It is an electron acceptor e.g KMnO4, K2Cr2O7, Ce(SO4)2, I2, KIO3, KBrO3, …………etc. Reductant (reducing agent): It is an electron donor e.g Fe2+, SO3-2, S2O32-, NO2-, C2O42-, ……….etc. During redox reactions ,electrons transferred from reductant to oxidant. i.e. from donor to acceptor Every process is represented by a half-reaction; showing ions involved and electrons transferred.  Redox Half-reactions 1- Oxidation of Fe2+ to Fe3+ by Ce4+ ion e.g Ce(SO4)2 Fe2+ = Fe3+ + e- ………………...… oxidation half reaction (i.e. takes place for reducing agent) Ce4+ + e- = Ce3+ ………………..…… reduction half reaction (i.e. takes place for oxidizing agent) __________________________________________________ Fe2+ + Ce4+ = Fe3+ + Ce3+ …….…... overall reaction. 2- Oxidation of Fe2+ to Fe3+ by MnO4- in the presence of H2SO4. MnO4- + 8H+ + 5e- = Mn++ + 4H2O Fe2+ = Fe3+ + 1e- Multiply by 5 where the first half reaction needs 5 electrons 5Fe2+ = 5Fe3+ + 5e- The over all reaction will be: MnO4- + 5Fe2+ + 8H+ = Mn2+ + 5Fe3+ + 4H2O Half-reactions of some Oxidants and Reductants Half-reactions of some Oxidants: MnO4- + 8H+ + 5e- = Mn2+ + 4H2O Ce4+ + e-= Ce3+ Cr2O72- + 14H+ + 6e- = 2Cr3+ + 7H2O I2 + 2e- = 2I- Half-reactions of some Reductants: C2O42- = 2CO2 + 2e- 2I- = I2 + 2e- 2S2O32- = S4O62- + 2e- NO2- + H2O = NO3- + 2H+ + 2e-  Electrode Potential If we immerse a metallic rod in its corresponding cation solution i.e., we have Mo / Mnn+ system. One of either possibilities may spontaneously takes place A- Oxidation process : Mo (s) = Mn+ + ne- There is a tendency for the metal to dissolve and going to the solution giving metal cation this tendency is termed Solution Pressure (S.P.) B- Reduction process: Mn+ + ne- = Mo There is a tendency for passage of metal cation from the solution to be deposited on the metal rod this tendency is termed Ionic Pressure (I.P.)  Which process will predominate? This will depend on the nature of metal (the ionic pressure is bigger or the electrolytic solution pressure) and on the concentration of the solute present. The metal rod will acquire either positive or negative charge; which will be surrounded by an equal number of the opposite charge i.e. Double Electric Layer is formed; this has a potential difference called Single Electrode Potential, which is the potential difference between metal rod (electrode) and its solution. This half-reaction forms what we call Half-Cell.  The potential difference between a metal and its ions solution is actually a measure of the tendency of the metal to be oxidized to metal ions or the tendency of the ions to be reduced to metal atoms.  Example: Zno / Zn2+ system; where SP > IP. i.e. Zno will goes in solution Zno (s) = Zn2+ + 2e- (oxidation) Other metals that behave similarly are Cd, Ni and Co.  Cuo / Cu2+ system; where IP > SP i.e. Cu2+ deposit on the metallic rod Cu2+ + 2e- = Cuo (s) (reduction) Other metals that behave similarly are Ag, Pt.  If these two half-cells are connected together, electrons will pass through from Zno/Zn+2 system to Cuo/Cu+2 system; i.e. this chemical change creates electric current in the cell which is called Galvanic cell or Voltaic cell. This particular cell is called Daniel cell (Zno / Zn2+ system - Cuo / Cu2+ system). The potential difference between the two half-cells is called electro- motive force (emf) of the cell. The overall reaction would be : Zno = Zn2+ + 2e- half-cell (oxidation half reaction) Cu2+ + 2e- = Cuo half-cell (reduction half reaction) Zno + Cu2+ = Zn2+ + Cuo.  Standard Electrode Potential “Eo” It is the electromotive force (emf) produced when a half cell consisting of the element immersed in a molar solution of its ions is coupled with a standard electrode such as standard hydrogen electrode (SHE, Eo = zero). The potential of a half-cell cannot be practically determined; however, the emf of a whole cell can be determined. Standard Electrode Potential (Eo) for some Systems System Eo System Eo Li / Li+ - 3.05 Co/Co2+ - 0.28 K / K+ - 2.92 Sn/Sn2+ - 0.13 Na/Na+ - 2.72 H2/2H+ 0.00 Mg/Mg2+ - 1.55 Cu/Cu2+ + 0.34 Al/Al3+ - 1.33 2I- / I2 + 0.58 Mn/Mn2+ - 1.10 Hg/Hg2+ + 0.79 Zn/Zn2+ - 0.76 Ag/Ag+ + 0.79 Cr/Cr3+ - 0.56 2Cl-/Cl2 + 1.36 Fe/Fe2+ - 0.44 Au/Au3+ + 1.68 Notes: The sign of the potential is similar to the charge on the metal electrode. The standard electrode potential is a quantitative measure of the readiness of the element to lose electrons and get oxidized, The standard electrode potential is a measure of the strength of the element as a reducing agent. The greater the negative value of the Eo of a metal, the greater is its tendency to pass to the ionic state and vice versa.  Normal (Standard) Hydrogen Electrode (NHE) It consists of a piece of platinum foil coated with platinum black and immersed in a solution of 1M HCl, H2 gas at 1 atm. pressure is pumped platinum black absorbs large amount of H2 and can be considered as a bar of hydrogen , it also catalyses the half reaction 2H+ + 2e- = H2 Under this conditions hydrogen electrode potential Eo equal Zero. This primary reference electrode is used to determine the standard potential of any system.  Assignment No. 1 Methods of calculation of equivalent weight of oxidizing agent or reducing agent.