1. Reduction- Oxidation Reactions 2 nd lecture 1

2. Learning Objectives What are some of the key things we learned from this lecture? Types of electrochemical cells. √ How to measure Standard electrode potential. √ reference electrode. √ How to measure the electrode potential from the Nernst equation. √ Factors affecting the oxidation potential. √ 2

3. Electrochemical cell An electrochemical cell is a device used for generating an electromotive force (voltage) and current from chemical reactions, (galvanic, voltaic cell). electromotive forcevoltagechemical reactions or the reverse, inducing a chemical reaction by a flow of current, (electrolytic). The current is caused by the reactions releasing and accepting electrons at the different ends of a conductor. electrons

4. There are two types of electrochemical cells: Electrolytic cell, (electrical Energy chem. Energy) Galvanic cell (voltaic), (chemical energy) electrical energy)

5. In an electrochemical cell, an electric potential is created between two dissimilar metals. This potential is a measure of the energy per unit charge which is available from the oxidation/reduction reactions to drive the reaction. It is customary to visualize the cell reaction in terms of two half- reactions, an oxidation half-reaction and a reduction half- reaction.electrochemical cellelectric potential oxidation/reduction reactions Reduced species oxidized species + ne - OxidationOxidation at anode Oxidized species + ne - reduced species ReductionReduction at cathode

6. Electrolytic cell An electrolytic cell decomposes chemical compounds by means of electrical energy, in a process called electrolysis.electrolysis The result is that the chemical energy is increased. An electrolytic cell has three component parts: an electrolyte and two electrodes (a cathode and an anode). The electrolyte is usually a solution of water or other solvents in which ions are dissolved.electrolyte cathodeanodeelectrolytesolutionwatersolvents Molten salts such as sodium chloride are also electrolytes.

7. When driven by an external voltage applied to the electrodes, the electrolyte provides ions that flow to and from the electrodes, where charge-transferring, or redox, reactions can take place.voltage redox Only for an external electrical potential (i.e. voltage) of the correct polarity and large enough magnitude can an electrolytic cell decompose a normally stable, or inert chemical compound in the solution.electrical potential inert

8. At anode: 2Cl - Cl 2 +2e - (oxidation) At cathode: 2Na + + 2e - 2Na (reduction) Cell reaction: 2Na + + 2Cl - 2Na + Cl 2 (redox reaction)

9. 2- Galvanic Cells In redox reaction, electrons are transferred from the oxidized species to the reduced species. Imagine separating the two 1/2 cells physically, then providing a conduit through which the electrons travel from one cell to the other. 8H + + MnO 4 - + 5eMn 2+ + 4H2O Fe 2+ Fe 3+ + e- x 5

10. We need to “complete the circuit” by allowing positive ions to flow as well. We do this using a “salt bridge” which will allow charge neutrality in each cell to be maintained. Salt bridge/porous disk: allows for ion migration such that the solutions will remain neutral.

11. Galvanic Cell Electrochemical cell in which chemical reactions are used to create spontaneous current (electron) flow

12. Standard electrode potential (E 0 ) If the concentration of the components of the cell are unity (1M), measured at 25 0 c The potential is called standard potential. It is not possible to measure the absolute potential of an electrode, but it could be measured relative to a standard reference electrode.

13. How to measure E 0 standard electrode potential?  Suppose we have Cu rod immersed in its ions (half cells).  In order to measure potential of an electrode, it must be connected to another electrode (half cell) having known constant potential called reference electrode and we use “Normal (standard) Hydrogen Electrode“.  So an electric current will flow from that having a higher potential to the other one, this current is called e.m.f. Which is the algebric difference of the electrode potential of the two half cells.  If the reference electrode has a potential equal to zero, therefore, so the e.m.f. measured will be equal to E 0

14. Reference Electrodes

15. What is Normal Hydrogen Electrode (NHE)?  It consists of a piece of platinium foil coated electrolytically with platinum black and immersed in a solution of HCl (normal solu).  Hydrogen gas at pressure of 1 atmosphere is passed.  Platinum black layer absorbs a large amount of H 2 and can be considered as a bar of H 2.  Under fixed conditions: 1 atm pressure Normal solu of HCl  The potential is assumed to be zero.  By connecting the electrode of unknown potential to “NHE“, the E 0 can be measured by a potentiometer. (1 atm) (1 N) Pt wire Glass tube to contain H 2 (g) Bubbles of H 2 (g) H + + e - ½ H 2

16. Standard electrode potential E 0 : Is the e.m.f. produced when a half cell consisiting of an element immersed in one molar solution of its ions is coupled with NHE. By measuring E 0 for all elements and arrange into a series known as “Electrochemical Series “:Starting from E 0 = -ve value, to hydrogen where E 0 = zero, then to E 0 = +ve value From this series we can conclude that: 1- The sign of E 0 is similar to the charge on the metal electrode. 2- The greater the –ve value of E 0, the greater the tendency of the metal to pass into ionic state (oxidized) i.e. E 0 is a quantitative measure of the ability of the element to lose electrons giving its ions. 3- Metal with more negative potential will displace any other metal below it in the series from its solution. System E 0 Zn/Zn 2+ -0.76 Fe/Fe 2+ -0.44 Sn/Sn 2+ -0.13 H 2 (pt)/H + 0 Cu/Cu 2+ +0.34

17. Reduction Potentials Electrode Couple"E0, V" Na+ + e- --> Na-2.7144 Mg2+ + 2e- --> Mg-2.3568 Al3+ + 3e- --> Al-1.676 Zn 2+ + 2e- --> Zn-0.7621 Fe2+ + 2e- --> Fe-0.4089 Cd 2+ + 2e- --> Cd-0.4022 Tl+ + e- --> Tl-0.3358 Sn2+ + 2e- --> Sn-0.141 Pb2+ + 2e- --> Pb-0.1266 2H+ + 2e- --> H2(SHE)0 S4O62- + 2e- --> 2S2O32-0.0238 Sn4+ + 2e- --> Sn2+0.1539 SO42- + 4H+ + 2e- --> H2O + H2SO3(aq)0.1576 Cu 2+ + e- --> Cu+0.1607 S + 2H+ + 2e- --> H2S0.1739 AgCl + e- --> Ag + Cl-0.2221 Saturated Calomel (SCE)0.2412 UO22+ + 4H+ + 2e- --> U4+ + 4H2O0.2682

18. Reduction Potentials Hg2Cl2 + 2e- --> 2Cl- + 2Hg0.268 Bi3+ + 3e- --> Bi0.286 Cu2+ + 2e- --> Cu0.3394 Fe(CN)63- + e- --> Fe(CN)64-0.3557 Cu+ + e- --> Cu0.518 I2 + 2e- --> 2I-0.5345 I3- + 2e- --> 3I-0.5354 H3AsO4(aq) + 2H+ + 2e- -->H3AsO3(aq) + H2O0.5748 2HgCl2 + 4H+ + 2e- -->Hg2Cl2 + 2Cl-0.6011 Hg2SO4 + 2e- --> 2Hg + SO42-0.6152 I2(aq) + 2e- --> 2I-0.6195 O2 + 2H+ + 2e- --> H2O2(l)0.6237 O2 + 2H+ + 2e- --> H2O2(aq)0.6945 Fe3+ + e- --> Fe2+0.769 Hg22+ + 2e- --> Hg0.7955 Ag+ + e- --> Ag0.7991 Hg2+ + 2e- --> Hg0.8519 2Hg2+ + 2e- --> Hg22+0.9083 NO3- + 3H+ + 2e- -->HNO2(aq) + H2O0.9275

19.  If an inert electrode such as platinum is placed in a redox system e.g. Fe 3+ /Fe 2+ it will assume a definite potential (oxidation potential) indicative of the position of equilibrium.  Oxidation potential is the pot. difference arround an inert electrode immersed in solu. of oxidized and reduced form. If the oxidizing tendency predominate, the system will take electrons from platinum leaving it positively charged If the system has reducing properties, electrons will be given up to platinum and it will acquire a negative charge. The magnitude of the potential will be a measure of oxidizing or reducing character of the system. If +ve oxidizing and if –ve reducing.

20. To measure the oxidation potential on pt wire, the system should be connected to NHE. Standard Oxidation potential (SOP): Is the e.m.f. produced when a half cell consisting of an inert electrode (pt) dipped in equal concentration of both oxidized and reduced form of a system is connected with NHE. Depending on the SOP the systems were arranged in a series from this series we can conclude that: 1- The higher the SOP of the system, the stronger the oxidizing power of its oxidized form and the weaker the reducing power of its reduced form. 2- the most powerful oxidizing agents are those at the top of the series with higher +ve potential, and the most powerful reducing agents occupy the bottom with higher –ve potential. There should be adifference in oxidation potential between the two systems of about 0.4 v for oxidation to occur.

21. Nernst Equation FOR OXIDATION POTENTIAL Compensated for non unit activity (not 1 M) Relationship between cell potential and activities aA + bB +ne - cC + dD At 298K, 2.3RT/F = 0.0592 What is potential of an electrode of Zn(s) and 0.01 M Zn 2+ Zn 2+ +2e - Zn  °= -0.763 V activity of metal is 1

22. E ind.electrode = E 0 - 0.059 log [products] n [reactants] E ind.electrode = E 0 M/M+ - 0.059 log 1 n [M+]

24. Factors affecting the oxidation potential: 1- Common ion effect: If we determine a sample of Fe 2+ with KMnO 4 in presence of Cl - ions we have 3 systems: MnO 4 - /Mn 2+ (1.52) Cl 2 /Cl - (1.36) Fe 3+/ Fe 2+ (0.77) KMnO 4 will oxidize both Cl - and Fe 2+ so more KMnO 4 will be consumed. To overcome this problem the oxidation potential of MnO 4 - /Mn 2+ should be reduced below Cl 2 /Cl - HOW? By adding Zimmermann´s reagent It contains [MnSO 4, H 3 PO 4, H 2 SO 4 ] E 25 = E 0 - 0.059/n log [Mn 2+ ] / [MnO 4 - ]  Presence of MnSO 4 increases the conc of Mn 2+ ion.  So the oxidation potential will decrease below that Cl 2 /Cl -

25. 2-Effect of H + ion concentration: [H + ] has a deciding effect on the oxidation potential of oxidizing agents containing oxygen and so [H + ] should be included in Nernest equation. E increases with increasing [H + ] i.e.increasing acidity. MnO 4 - + 8H + + 5e ↔ Mn 2+ + 4H 2 O E MnO 4 - /Mn 2+ = E 0 - 0.059/5 log [Mn 2+ ] / [MnO 4 - ] [H + ] 8 At pH 5: KMnO 4 oxidizes I - but not Br - and Cl - At pH 3: KMnO 4 oxidizes I - and Br - but not Cl - At lower pH : KMnO 4 oxidizes I -, Br - and Cl -

26. AsO 4 3- + 2H + + 2e  AsO 3 3- + H 2 O E AsO 4 3 - / AsO 3 3- = E 0 - 0.059/2 log [ASO 3 3- ] / [AsO 4 3- ] [H + ] 2 If [H + ] increases: The oxidation potential of AsO 4 3- increases. If [H + ] decreases: The oxidation potential of AsO 4 3- decreases and the reducing power of AsO 3 3- increases. Therefore: AsO 4 3- oxidizes I - to I 2 in acid medium. While: I 2 oxidizes AsO 3 -3 in alkaline medium with NaHCO 3 AsO 4 -3 + 2 I - + 2H + AsO 3 -3 + I 2 + H 2 O acid alkaline

27. 3-Effect of complexing agents: E.g.1- Oxidation potential of I 2 /2I - system increases in presence of HgCl 2 WHY? Because HgCl 2 forms a complex with the I - ions [HgI 4 ] 2- I 2 + 2 e  2I - E I 2 /2I - = E 0 -  removing the I - ions from the share of the reaction.  minimizing its concentration.  decreasing the ratio of [I - ] 2 / I 2.  increasing the oxidation potential of I 2 /2I - system.

28. E.g.2- Oxidation potential of Fe 3+ / Fe 2+ system decreases on addition of F - or PO 4 3- WHY? Because F - and PO 4 3- form stable complexs with Fe 3+ which are [FeF 6 ] 3- or [Fe (PO 4 ) 2 ] -3 Fe 3+ + e  Fe 2+ E Fe 3+ /Fe 2+ = E 0 -  removing the Fe 3+ ions from the share of the reaction.  minimizing its concentration.  Increasing the ratio of Fe 2+ / Fe 3+.  lowering the oxidation potential of Fe 3+ / Fe 2+ system.

29. 4-Effect of precipitating agents: Oxidation potential of Ferricyanide/ Ferrocyanide system increases on addition of Zn 2+ WHY? Because Zn 2+ will precipitate Zn 2 [Fe(CN) 6 ] Fe (CN) 6 3- + e  Fe(CN) 6 4- E = E o -  removing the Ferrocyanide ions from the share of the reaction.  minimizing conc of ferrocyanide.  decreasing the ratio of [Fe(CN) 6] 4- / [Fe (CN) 6 ] 3-.  increasing the oxidation potential of [Fe(CN) 6] 3- / [Fe (CN) 6 ] 4-