Presentation on theme: "Electrochemistry. It deals with reactions involving a transfer of electrons: 1. Oxidation-reduction phenomena 2. Voltaic or galvanic cell Chemical reactions."— Presentation transcript:
It deals with reactions involving a transfer of electrons: 1. Oxidation-reduction phenomena 2. Voltaic or galvanic cell Chemical reactions can be used to produce eletrical energy: 3.Electrolytic cells Electrical energy can be used to bring about chemical transformations:electrolysis
Oxidation and reduction Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) Zn (s) Zn 2+ (aq) + 2e - Oxidation half-reaction Cu 2+ (aq) + 2e - Cu (s) Reduction half-reaction
Galvanic /voltaic/ cells Electrode: e.g. a metal strip which can funcion as a cathode or an anode Half-cell: an electrode immersed in a solution containing the metal ions. Half of a voltaic cell in which an oxidation or a reduction occurs.
Electrode conventions CathodeAnode Ions attractedCationsAnions Direction of electron movement Into cellOut of cell Half-reactionReductionOxidation Sign Galvanic cell Electrolytic cell Positive Negative Positive There is a difference in electrical potentials between the solution and the electrode
Types of electrodes 1.Electrodes of the first kind: (metal electrodes) Zn, Cu, Fe, Au, Pt (indifferent) 2.Electrodes of the second kind: (calomel electrode) - metal: Hg - slightly soluble salt: Hg 2 Cl 2 precipitate - another soluble salt of its anion: KCl
Hg 2 Cl 2(s) Hg 2(aq) Cl (aq) - K sp = [Hg 2 2+ ][Cl - ] 2 /= 1.3 x / 2Hg Hg e - 3.Gas electrodes: H 2 /S.H.E./, Cl 2
Membrane electrodes Glass electrode Ion-selective electrode: Na +, K +
It can be calculated by means of NERNST-equation E = E 0 + ln c RT nF E = E 0 + log c n R = J/K mol T = K (25°C) F = C/mol n = number of moles of electrons transferred
Zn loses electrons, enters the solution. Cu gains electrons, deposits as metal. Electromotive force (E.M.F.) is the difference in potentials between the two half cells. E.M.F. = E cat. (+) – E an. (-)
Cell diagrams Zn (s) I Zn (aq) 2+ II Cu (aq) 2+ I Cu (s) half-cell salt bridge At the left: - oxidation - anode - negative At the right: - reduction - cathode - positive
Standard electrode potentials E = E 0 + log c n E = E 0 if c = 1 mol We can measure the potential differences only. Standard hydrogen electrode: S.H.E. ( + a=1) + 2e - H 2(q) 1 atm. E 0 = volt
Standard reduction potentials (at 25°C) Half-reactionE°(volts) Li + (aq) + e - Li (s) Ca 2+ (aq) + 2e - Ca (s) H + (aq) +2e- H 2(g) 0.00 Cu 2+ (aq) + 2e - Cu (s) Ag + (aq) + e - Ag (s) Cl 2(g) + 2e - 2 Cl - (aq) F 2(g) + 2e - 2F - (aq)
Concentration cells Zn I 0.01M Zn 2+ II 1M Zn 2+ I Zn E.M.F. = log ; [Zn 2+ ] 1 > [Zn 2+ ] 2 [Zn 2+ ] [Zn 2+ ] 2 Measurement of pH Pt H 2(g,1atm) I H + (xM) II H + (1M) I H 2(q,1atm) Pt E cell = log = (-log [H + ])  [H + ] E cell = pH
Redox cells Fe 2+ /Fe 3+ T = 25°C Fe 2+ Fe 3+ + e - oxidizedreduced E = E 0 + ln [ox] nF RT [red] E = E 0 + log [ox] n [red] S.H.E.
Electrode potentials of some reduction oxidation systems Co 2+ /Co V Pb 2+ / Pb V MnO 4 - /Mn V Fe 2+ /Fe V I 2 /2I V Sn 2+ /Sn V Cytochrom a Fe 3+ /Fe V Cytochrom c Fe 3+ /Fe V Hemoglobin Fe 3+ /Fe V Cytochrom b Fe 3+ /Fe V Vitamin C ox/red0.06V oxidizer reducer
Electrolysis: the use of electricity to bring about chemical change -Primary process: redox reaction, overvoltage -Secondary process: …..
Stoichiometry of electrolysis Faraday’s law m = kIt m = mass …k = electrochem. constant I = electric current in amperest = time A = 1 C/s ampere coulomb sec. The quantity of charge equivalent to one mole of electrons is called the faraday (F) 1F = C1F 1 g-equivalent