1. Electrochemistry Physical Chemistry

2. Daniel Cell 1. Electrochemistry is the study of the interconversion of electrical and chemistry energy. 2. Voltaic cell, in which a spontaneous reaction generates electrical energy. 3. Electrolytic cell: electrical energy is used to bring about a non spontaneous reaction. Oxidation-reduction (redox) reactions occur when electrons are given up by the substance being oxidized (the reducing agent) and simultaneously gained by the substance being reduced (the oxidizing agent). 2 Ag + + Zn  Zn 2+ + 2 Ag At Anode : Zn  Zn 2+ + 2e At Cathode: Ag + + e  Ag

3. 1. The two half-cells are connected by a salt bridge, which prevents free mixing of both electrolyte solutions but permits the proper movement of ions to maintain electrical neutrality. 2. When Zn is oxidized some anions must enter (or cations must leave) the Zn half-cell to compensate for the added positive charge of the Zn 2+ produced. Also when Ag+ reduced cations must enter (or anions must leave) the Ag half-cell. 3. Let’s say the salt bridge used is: KNO 3, K + ions move from the salt bridge into cathode half bridge. At the same time NO 3 - ions move into the anode half cell. 4. The Daniel Cell Diagram: Zn(s)/ZnSO 4 (aq) AgSO 4 (aq)/Ag(s) Daniel Cell

4. 1. When chlorine gas is bubbled through a water solution of NaBr, a spontaneous redox reaction occurs: Cl 2 (g) + 2Br-(aq)  2Cl-(aq) + Br 2 (g) a) What is the cathode and anode reaction? b) Which way do electrons move in the external circuit? c) Which way do anions and cations move within the cell? Solution: a) Cathode: Cl 2 + 2e   2Cl- Anode: 2Br-  Br 2 + 2e b) From anode to cathode c) Anion to anode, cations to cathode Practice

5. Oxidation Number Determine the oxidation number of the underlined atom: a) V 2 O 5 b) SO 4 2-- c) SO 3 2- d) OCl - e) HCO 3 - f) MnO 4 - Writing Redox Equation

6. 1. xA + yB  Product x(+a) + y(-b) = 0 Practice: 1. What are the oxidation numbers of KIO 3, KIO, Na 2 SO 3 and Na 2 SO 4. Thus, balance the equation. xKIO 3 + y Na 2 SO 3  cKIO + d Na 2 SO 4 2. Use the oxidation number method to balance the equation: a) Sn +HNO 3  SnO 2 + NO 2 + H 2 O Changes in Oxidation Number

7. Writing Half Equation ClO 3 - + I -  Cl- + I 2 1. Determine change of oxidation number: ClO 3 -  Cl- : +5 to -1 = -6….. Reduction I-  I 2 : -1 to 0 = +1…. Oxidation 2. ClO 3 - + 6e  Cl- 2I-  I 2 + 2e 3. H 2 O, H+ and OH- can be added to balance up the half equatons. ClO 3 - + 6H + + 6e  Cl- + 3H 2 O 2I-  I 2 + 2e ) x 3 --------------------------------------------------------- ClO 3 - + 6H + + 6I-  Cl- + 3H 2 O + 3I 2

8. 1. Cr 2 O 7 2- + H + + NO 2 -  Cr 3+ + NO 3 - 2. Cl 2 + SO 2  Cl- + SO 4 2- 3. C + H 2 SO 4  CO 2 + SO 2 4. H 2 O 2 + Mn 2+  MnO 2 + H 2 O 5. Cu + HNO 3  Cu 2+ + NO Practice 2

9. The Electrode Potential of Metal 1. Suppose you have a piece of magnesium in a beaker of water. There will be some tendency for the magnesium atoms to shed electrons and go into solution as magnesium ions. The electrons will be left behind on the magnesium Electrode Potential In a very short time, there will be a build-up of electrons on the magnesium, and it will be surrounded in the solution by a layer of positive ions (Helmholtz double layer). This produces a potential defference between the metal and the solution.

10. The Electrode Potential of Metal 1. Dynamic equilibrium will be established when the rate at which ions are leaving the surface is exactly equal to the rate at which they are joining it again. At that point there will be a constant negative charge on the magnesium, and a constant number of magnesium ions present in the solution around it. The potential difference produced between the electrode and the solution in half cell is called electrode potential. The magnitude of the electrode potential depends on the tendency of the metal to lose electrons and to form hydrated metal ions

11. The Electrode Potential of Metal 1. Copper is less reactive and so forms its ions less readily. The position of the magnesium equilibrium...... lies further to the left than that of the copper equilibrium.

12. The Electrode Potential of Metal All we need to know is that the magnesium equilibrium lies further to the left than the copper one. We need to know that magnesium sheds electrons and forms ions more readily than copper does. That means that we don't need to be able to measure the absolute voltage between the metal and the solution. It is enough to compare the voltage with a standardized system called a reference electrode. The system used is called a standard hydrogen electrode.

13. The Electrode Potential of Non Metal The standard hydrogen electrode The standard hydrogen electrode looks like this: As the hydrogen gas flows over the porous platinum, an equilibrium is set up between hydrogen molecules and hydrogen ions in solution. The reaction is catalyzed by the platinum. The ions in the solution have a concentrated of 1.0 M All gases involved have a pressure of 1 atm. A temperature of 25 o C

14. Using the standard hydrogen electrode 1. The standard hydrogen electrode is attached to the electrode system you are investigating - for example, a piece of magnesium in a solution containing magnesium ions.

15. Using the standard hydrogen electrode 1. The emf for the cell is 2.37 volt. 2. According to IUPAC, the standard potential for the magnesium electrode is -2.37. The negative sign is used to show that the electrode is the negative pole. 3. The voltmeter shows that the electrons flow from the magnesium electrode to the hydrogen electrode in the external circuit. This means that at Mg electrode: Mg  Mg 2+ + 2e 3. At hydrogen electrode: 2H + + 2e  H 2 4. The overall cell reaction: Mg + 2H +  Mg 2+ + H 2 5. The cell diagram: Mg(s)/Mg 2+ (aq, 1M) H + (aq, 1M)/ H 2 (g) /Pt(s) E Ө = -2.37 V

16. Using the standard hydrogen electrode 1. Half equation is written as a reduction process: Oxidized species + ne  reduced species. 2. Thus, Ag + (aq)/Ag(s) E Ө = +0.80 V means that the following half cell reaction has a standard electrode potential of +0.80 Ag + + e  Ag E Ө = +0.80 V 3. The cell diagram is written with: a) The half cell undergoing oxidation ( negative electrode) on the left of the diagram. b) The half cell undergoing reduction ( positive electrode) on the right of the diagram metal / metal ion combinationE° (volts) Mg 2+ / Mg-2.37 Zn 2+ / Zn-0.76 Cu 2+ / Cu+0.34 Ag + / Ag+0.80

17. Practice 1. Consider the following equilibrium involving I 2 and Fe 3+ : I 2 + 2e  2I - E Ө = +0.54 V Fe 3+ + e  Fe 2+ E Ө = +0.77 V a) Write the ionic equation for the overall cell reaction. b) State the observation after the half cell have been connected for a short while. c) Determine the polarity of the electrode in the electrochemical cell. Solutions a) 2Fe 3+ + 2I -  2Fe 2+ + I 2 b) The beaker on the right (Fe 3+ ) : yellow to green The beaker on the left (I-) : yellow to brown c) I 2 + 2e  2I - is anode cause oxidation happen. Fe 3+ + e  Fe 2+ is cathode cause reduction occur.

18. We could therefore write down the emf of the above cell (under standard conditions) as: emf of cell = potential of Cu - potential of Zn or conventionally For example, one could use the reaction to construct the cell

19. The electrochemical Series 1. The electrochemical series shows the ease of a species accept or loses electrons. 2. The higher a species the easier it loses electron (oxidized) 3. The E Ө value shows the possibility of a reaction but not the quantity of substances involved. 4. If E Ө for I 2 + 2e  2I - +0.54 V, then ½ I 2 + e  I - not (1/2 x 0.54V) but is +0.54V. 5. The reactivity of metals increases as the electrode potential become more negative while the reactivity if non metal increases as the electrode potential become more positive. 6. The strength of oxidizing agents increases on descending the series while the strength of reducing agents increases on ascending the series.

20. Practice 1. Arrange the following substances in the ascending order of their oxidizing strength: a) Na, Cl 2, Cu b) Zn, Ag +, Ag c) Cl 2, Br 2, Br - 2. The E Ө value for three elements, X, Y and Z are given below: X  X + + e E Ө = -0.34V Y  Y + + e E Ө = -0.80V Z  Z + + e E Ө = +2.46V

21. Predicting the Possibility of Redox Reaction. 1. The possibility of a spontaneous reaction occurring can be predicted by using: a) Electrochemical series b) E Ө value for the overall reaction 2. An oxidizing agent on the left of the electrochemical series will oxidize a reducing agent on the right if the reducing agent is above the oxidizing agent in the series. I 2 + 2e  2I- E Ө = +0.54 V Br 2 + 2e  2Br- E Ө = +1.07 V Cl 2 + 2e  2Cl- E Ө = +1.33 V Br 2 is able to oxidize I- into I 2 but cannot oxidize Cl- into Cl 2

22. Predicting the Possibility of Redox Reaction. 1. The possibility of a spontaneous reaction occurring can be predicted by using: a) E Ө value for the overall reaction 2. If E Ө for a combined reaction is positive, then the reaction will occur spontaneously. Sn 2+ + 2Fe 3+  Sn 4+ + 2Fe 2+ Sn 2+  Sn 4+ + 2e E Ө = -0.15 V 2Fe 3+ + 2e  2Fe 2+ E Ө = +0.77V --------------------------------------------------------------- E Ө = +0.62V ---------------------------------------------------------------

23. Practice 1. Predict whether the following reactions can take place under standard condition. a) Reduction of Sn 2+ to Sn by Mg b) Oxidation of Cl - to Cl 2 acidified Cr 2 O 7 2- Hints: Mg 2+ + 2e  Mg E Ө = -2.38 V Sn 2+ + 2e  Sn E Ө = -0.14 V -------------------------------------------------------------------- Cr 2 O 7 2- + 14H + + 6e  2Cr 3+ + 7H 2 O E Ө = +1.33 V Cl 2 + 2e  Cl- E Ө = +1.36 V

24. Predicting the Stability of Aqueous Ions 1. The relative stability of Cr and Cr 2+ can be explain as follow: Cr 2+ + 2e  Cr E Ө = -0.91 V 2H + + 2e  H 2 E Ө = 0.00 V ------------------------------------------------------- Cr + 2H +  Cr 2+ + H 2 E Ө = + 0.91V ------------------------------------------------------- 2. This implies that Cr 2+ is more stable than Cr because the forward reaction can occur spontaneously. 3. The Co 3+ ion can be stabilized in aqueous solution by complexing it with other ligands like ammonia.

25. The Nernst Equations 1. The value of the electrode potential depends on: a) The concentration of the ion and the temperature of the solution b) The pH value (acidity) of the solution c) The effect of complex formation. For the electrode system: oxidized species + ne  reduced species E = E Ө + RT/nF {ln [oxidized species]/[reduced species]} or E = E Ө + 0.059/n {lg [oxidized species]/[reduced species]}

26. The Nernst Equations a A + b B = c C + d D 1. Q = [C] c [D] d / [A] a [B] b 2. E = E Ө - 0.059/n {lg Q} Sn 2+ (aq) + Br 2 (l) Sn 4+ (aq) + 2 Br - (aq) Q = [Sn 4+ ][Br - ] 2 / [Sn 2+ ] The Nernst equation for this reaction is: E = 0.092 – 0.059/2 lg[ (0.0001)(0.0001) 2 / 0.05] E = 0.92 V - (-0.316 V) E = 1.24 V

27. Effect of the Concentration of an ion on the emf of a cell 1. Cu(s) / Cu 2+ (aq) // Ag + (aq) / Ag (s) Cu + 2Ag +  2Ag + Cu 2+ E = E Ө - 0.059/n {lg Q} If the concentration of Cu 2+ ions in the copper half cell is reduced, the E Cu is less than E Ө and the emf of the electrochemical cell increases. If the concentration of Cu 2+ ions in the copper half cell is reduced, the equilibrium will shift to the right.

28. The Nernst Equation and Solubility Product 1. When a silver electrode is immersed into a saturated silver chloride solution, the electrode potential at 298 K is 0.51 V. Calculate the solubility product of silver chloride at 298 K. E Ө Ag+/Ag = +0.80 V. E = E Ө + 0.059/1 lg[Ag+] 0.051 = 0.80 + 0.059 lg[Ag+] [Ag+] = 1.22 x 10 -5 moldm -3 Ksp = [Ag + ][Cl-] = (1.22 x 10 -5 ) 2 = 1.49 x 10 -10 mol 2 dm -6

29. The Nernst Equation and Solubility Product 1. Consider the cell: Pt(s) /H 2 (g, 1.0 atm)/HCl(0.01M)// AgCl(aq)/Ag(s) The measured cell emf is 0.046 V at 25 o C and E Ө Ag+/Ag = +0.80 V a) Write equation for the reaction occurring at the electrodes. b) Write an equation for the overall cell reaction. c) Calculate the emf of the cell under standard conditions. d) Calculate the [Ag + ] in equilibrium with AgCl and Cl-. e) What is the concentration of Cl-? f) Find the Ksp of AgCl.

30. The Nernst Equation and Equilibrium Constant 1. When the redox reaction in a Daniel cell has achieved equilibrium, E cell = 0.0 V. This is because at equilibrium, there is no overall change taking place in a chemical reaction. 0 = E Ө - 0.059/n lg Kc E Ө = 0.059/n [lg Kc] 2. Consider the following reaction: 2Fe 3+ + Cu  2Fe 2+ + Cu 2+ a) Calculate the E Ө the system. b) Calculate the equilibrium constant Kc for the reaction. [ 0.43V, 3.8 x10 14 ]

31. The Nernst Equation and pH 1. The cell reaction: Zn + 2H +  Zn 2+ + H 2 E Ө cell = 0.00-(-0.76) = + 0.76 V Using Nernst Equation: E cell = E Ө cell + 0.059/n lg [H + ] 2 /[Zn 2+ ]P H2 E cell = 0.76 + 0.059/2 lg [H + ] 2 / (1)(1) pH = (0.76 – E cell ) / 0.059 If the emf for the electrochemical cell is + 0.54 V, the pH would be pH = (0.76 – 0.54)/0.059 = 3.7