1. 03 REDOX EQM Half Cells with Inert Electrode C. Y. Yeung (CHW, 2009)p.01 Inert Electrode : In the metal-metal ion system, metal is the electrode, and metal ion is the electrolyte. M n+ M However, if the system does not involve metal, and “inert electrode” is required. (e.g. platinium / graphite)

2. p.02Pt Fe 2+ Fe 3+ - e - + e - (1) Metal Ion – Metal Ion Sytem FeSO 4 (aq) + Fe 2 (SO 4 ) 3 (aq) E Fe 3+,Fe 2+ | Pt = +0.77 V as cathode:Fe 3+ (aq), Fe 2+ (aq) | Pt(s) Cell Diagram as anode:Pt(s) | Fe 2+ (aq), Fe 3+ (aq) Half Cells using Inert Electrode : E MnO 4 -,Mn 2+ | Pt = +1.52 V MnO 4 - (aq) + 8H + + 5e - Mn 2+ (aq) + 4H 2 O(l) Fe 3+ (aq) + e - Fe 2+ (aq) Cell Diagram as cathode:[MnO 4 - (aq) + 8H + (aq)], [Mn 2+ (aq) + 4H 2 O(l)] | Pt(s) as anode:Pt(s) | [Mn 2+ (aq) + 4H 2 O(l)], [MnO 4 - (aq) + 8H + (aq)]

3. p.03 Ex ercise Calculate the cell e.m.f. and write the cell diagram of the electrochemical cell shown below in accordance with the IUPAC convention. (ref.: p. 207) E Fe 3+,Fe 2+ | Pt = +0.77 V E MnO 4 -,Mn 2+ | Pt = +1.52 V cathodeanode E cell = +1.52 – (+0.77) = +0.75V Cell Diagram C(graphite) | Fe 2+ (aq), Fe 3+ (aq) [MnO 4 - (aq) + 8H + (aq)], [Mn 2+ (aq) + 4H 2 O(l)] | C(graphite) [Ref.: p. 210 Example 20-5C] Oxidation Reduction

4. p.04Pt I-I-I-I- I2I2I2I2 - e - + e - (2) Non-metal Ion – Non-metal Ion Sytem I 2 (aq) + I - (aq) E I 2, I - | Pt = +0.54 V as cathode:I 2 (aq), 2I - (aq) | Pt(s) Cell Diagram as anode:Pt(s) | 2I - (aq), I 2 (aq) I 2 (aq) + 2e - 2 I - (aq) <Example> Ex ercise If iron(III) / iron(II) system acts as the cathode, while iodine / iodide acts as the anode, with using platinium as electrodes. Calculate the cell e.m.f. and write the cell diagram. (ref.: p.207) E cell = +0.77 – (+0.54) = +0.23V Cell Diagram: Pt(s) | 2I - (aq), I 2 (aq) Fe 3+ (aq), Fe 2+ (aq) | Pt(s)

5. p.05Pt (3) Metal – Sparingly Soluble Salt Sytem HCl(aq) Platinium Black coated with Ag and AgCl E AgCl | Ag = +0.54 V AgCl(s) + e - Ag(s) + Cl - (aq) + e - - e - AgAgCl as cathode:Cl - (aq) | AgCl(s) | Ag(s) Cell Diagram as anode:Ag(s) | AgCl(s) | Cl - (aq) electrolyte! E PbSO 4 | Pb = -0.35 V PbSO 4 (s) + 2e - Pb(s) + SO 4 2- (aq) as cathode:SO 4 2- (aq) | PbSO 4 (s) | Pb(s) Cell Diagram as anode:Pb(s) | PbSO 4 (s) | SO 4 2- (aq) i.e. H 2 SO 4 is used as electrolyte.

6. p.06Assignment p.202 Check Point 20.3 [due date: 11/5 (Mon)] Next …. Standard Hydrogen Electrode (S.H.E.), Measurement of Standard Electrode Potentials (p. 197-212) p.20.5(b) [due date: 11/5 (Mon)] p.202 Table 20.1…?