1. Here we are given a diagram of an electrochemical cell which involves a gas and we will work though a series of questions about this cell.

2. We’re given the diagram for an electrochemical cell Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V e–e–

3. In the left beaker we have a platinum electrode. Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V e–e– platinum electrode (inert)

4. Remember platinum is inert and doesn’t react itself. Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V e–e– platinum electrode (inert)

5. Chlorine gas surrounds the platinum electrode V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V e–e– chlorine gas

6. And it is dipped into a 1 Molar solution of hydrochloric acid, which contains (click) H + and Cl minus ions V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V e–e– 1 M hydrochloric acid H+H+ Cl –

7. The electrode in the right beaker is an unknown metal, which we’ll call metal X. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V e–e– H+H+ Cl – Metal X

8. The solution in this beaker, X(NO3)3 has 3 nitrates for each X in the formula, which means (click) the charge on each X ion must be positive 3. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V e–e– H+H+ Cl – X(NO 3 ) 3 X3+X3+ NO 3 –

9. The diagram also shows that the initial voltage of this cell is 3.02 volts. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V e–e– H+H+ Cl – X3+X3+ NO 3 – initial voltage

10. And that electrons are travelling away from metal X and toward the platinum electrode. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – Electrons are travelling toward the platinum electrode e–e–

11. In this video, we will go through a series of 7 questions involving this electrochemical cell. You may want to pause the video, take a screen shot, print this page, and try these questions on your own first. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e–

12. Question (a) asks us to write the half-reaction occurring at the cathode and its E naught if it can be determined. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e–

13. Because electrons always move from the anode toward the cathode, (click) and they are moving toward the platinum electrode, (click) platinum must be the cathode. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– cathode V a) Write the half-reaction at the cathode along with its E°, if it can be determined.

14. And metal X must be the anode.. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– cathode V anode a) Write the half-reaction at the cathode along with its E°, if it can be determined.

15. Even though platinum is the cathode, 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– cathode V a) Write the half-reaction at the cathode along with its E°, if it can be determined.

16. Platinum is inert, so it doesn’t react itself. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– inert V a) Write the half-reaction at the cathode along with its E°, if it can be determined.

17. The reactive species in this half-cell are chlorine gas and chloride ions. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– V a) Write the half-reaction at the cathode along with its E°, if it can be determined. Reactive species are Cl 2(g) and Cl – ions.

18. This time the hydrogen ions are spectators instead of being the active ions. There is no H2 gas present, so the H+ ions do not take part in a half-reaction. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– V a) Write the half-reaction at the cathode along with its E°, if it can be determined. Reactive species are Cl 2(g) and Cl – ions. The hydrogen ions, H +, are spectators in this half-cell

19. You may recall, that in a previous example, hydrogen gas was present along with aqueous HCl. In that half- cell, H+ ions and H2 gas were both present, so they participated in the half-reaction which occurred, and Cl minus was the spectator. a) Write the half-reaction at the cathode along with its E°, if it can be determined. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – Reactive species are H 2(g) and H + ions. 1 M KNO 3

20. So when HCl is the solution in the beaker, which ion is active and which ion is the spectator depends on what else is present. 1 M HCl 1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– V a) Write the half-reaction at the cathode along with its E°, if it can be determined. Which ion is active depends on what else is present

21. We had previously determined that the reactive species in this half-cell are Cl2 gas and Cl minus ions, 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– V a) Write the half-reaction at the cathode along with its E°, if it can be determined. Reactive species are Cl 2(g) and Cl – ions.

22. We look for the half-reaction with these on the reduction table. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– V a) Write the half-reaction at the cathode along with its E°, if it can be determined.

23. Here is a closer view of it. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– V a) Write the half-reaction at the cathode along with its E°, if it can be determined. CATHODE

24. This half-reaction is occurring at the cathode and it is a reduction, so it does not need to be reversed. We can say the reaction at the cathode is (click) C l 2 + 2 e– gives 2 Cl (minus) 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– V a) Write the half-reaction at the cathode along with its E°, if it can be determined. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g)

25. Because the equation is not reversed, the sign on the E naught stays the same. So (click) the E naught value is just positive 1.36 volts. We ARE able to determine it because this half-reaction is on the reduction table. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e– V a) Write the half-reaction at the cathode along with its E°, if it can be determined. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g)

26. Question (b) asks us to write the half-reaction occurring at the anode and its E naught if it can be determined. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e–

27. Electrode X is the anode and oxidation always takes place at an anode. The cation of X is X3+, so the half-reaction at the anode (click), the oxidation of X to form X 3+ (click) is … 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V b) Write the half-reaction at the anode along with its E°, if it can be determined. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

28. X gives X 3+ plus 3 electrons 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V b) Write the half-reaction at the anode along with its E°, if it can be determined. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

29. X is not listed on the reduction table, so we are not able to determine (click) its E naught value from there. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V b) Write the half-reaction at the anode along with its E°, if it can be determined. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

30. So at this point, we call it “question mark” volts 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V b) Write the half-reaction at the anode along with its E°, if it can be determined. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

31. Question (c) asks us to write the equation for the overall redox reaction and its E naught value. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e–

32. To get the overall redox equation, we need to add up the half reactions, but we see that (click) Cl 2 gains 2 electrons and (click) X loses 3 electrons, so (click) electrons are not balanced yet. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ Electrons are not balanced

33. To balance electrons (click), we multiply the cathode half reaction by 3 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

34. And the anode half-reaction by 2 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

35. So now we see that 6 electrons are gained 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

36. And 6 electrons are lost. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

37. So electrons are balanced (click) and can be cancelled out. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

38. To get the equation for the overall redox reaction, we add up what’s left in the two half-reactions. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

39. Starting on the top left we have (click) 3 times Cl2 (click) gives 3 Cl2 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

40. On the bottom left we have (click) 2 times X (solid) (click) which gives 2 X (solid) 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

41. On the top right we have (click) 3 times 2 Cl minus (click) which gives 6 Cl minus 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

42. And on the bottom right we have (click) 2 times X 3+ (click) which gives 2 X 3+ 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

43. So now we have the balanced equation for the overall redox reaction. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

44. But we’re also asked to find the E naught value for the overall redox reaction. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+

45. Now, the voltage on the voltmeter, 3.02 volts, (click) is just the overall cell potential, which is the same as (click) the E naught for the overall redox reaction 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential

46. So now we know that the E naught value for the overall reaction, or overall cell potential is 3.02 volts. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V c) Write the equation for the overall redox reaction and its E°. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential

47. Question (d) asks us to find the E naught value for the half-reaction occurring at the anode. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e–

48. We know that adding up the E naught values for the two half-reactions, gives the E naught value for the overall reaction. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V d) Find the E° value for the half-reaction occurring at the anode. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential

49. So we can say that 1.36 volts plus question mark (click) is equal to 3.02 volts 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential d) Find the E° value for the half-reaction occurring at the anode.

50. Subtracting 1.36 volts from both sides gives us, question mark is equal to 3.02 volts minus 1.36 volts, 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential d) Find the E° value for the half-reaction occurring at the anode.

51. Which is 1.66 volts. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential d) Find the E° value for the half-reaction occurring at the anode.

52. So we’ll substitute 1.66 volts in here for question mark. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential d) Find the E° value for the half-reaction occurring at the anode.

53. We now have the E naught value for the half-reaction at the anode. Its equal to 1.66 volts. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential d) Find the E° value for the half-reaction occurring at the anode.

54. Question (e) asks us to find the oxidation potential of the metal element X V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e–

55. Remember that multiplying a half-reaction by a factor like 2 does NOT change the value of its E naught. So the half-reaction by itself (click) still has an E naught value of 1.66 volts. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential e) What is the oxidation potential of element X? X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s)

56. In this half-reaction, element X is losing 3 electrons, and losing electrons is called oxidation, 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential e) What is the oxidation potential of element X? X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) X is losing 3 electrons

57. So this half-reaction can be called the oxidation of X 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential e) What is the oxidation potential of element X? X (s)  X 3+ + 3e – E° = 1.66 V X is losing 3 electrons Oxidation of X (s) :

58. Therefore 1.66 V is called (click) the oxidation potential of X 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential e) What is the oxidation potential of element X? X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) : The Oxidation Potential of X

59. So now, we’ve answered question e (click), the oxidation potential of metal X is equal to positive 1.66 volts 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential e) What is the oxidation potential of element X? X (s)  X 3+ + 3e – E° = 1.66 V The Oxidation Potential of X Oxidation of X (s) :

60. Question (f) asks us to find the reduction potential of the cation X 3+. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what metal? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e–

61. We know that the oxidation potential of metal X is positive 1.66 volts, as shown by this oxidation half-reaction. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) : The Oxidation Potential of X f) What is the reduction potential of X 3+ ?

62. Now if we reverse this oxidation half reaction, we get (click) X 3+ plus 3 electrons gives X (solid) 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) : The Oxidation Potential of X Reverse this Half-reaction X 3+ + 3e –  X (s) E° = –1.66 V f) What is the reduction potential of X 3+ ?

63. This is the half-reaction for the reduction of X 3+. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) : The Oxidation Potential of X Reverse this Half-reaction X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ : f) What is the reduction potential of X 3+ ?

64. Remember, when we reverse a half-reaction, the sign on the E naught is switched, so the E naught value for the reduction half-reaction is (click) negative 1.66 volts 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential f) What is the reduction potential of X 3+ ? X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) : The Oxidation Potential of X Reverse this Half-reaction X 3+ + 3e –  X (s) E° = –1.66 V Switch the sign on E° Reduction of X 3+ :

65. So we have the half-reaction for the reduction of X3+ and its E naught value of –1.66 volts 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential f) What is the reduction potential of X 3+ ? X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) : The Oxidation Potential of X X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ :

66. So negative 1.66 volts is the Reduction Potential of X 3+. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential f) What is the reduction potential of X 3+ ? X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) : The Oxidation Potential of X X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ : The Reduction Potential of X 3+

67. So we’ve now answered this question, the reduction potential of X3+ is negative 1.66 volts. 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 V = 3.02 V H+H+ Cl – NO 3 – e–e– V k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (X (s)  X 3+ + 3e – )×2 E° = ? V. 3Cl 2(g) + 2X (s)  6Cl – + 2X 3+ E° = 3.02 V cathode anode CATHODE Cl 2(g) X3+X3+ cell potential f) What is the reduction potential of X 3+ ? X (s)  X 3+ + 3e – E° = 1.66 V Oxidation of X (s) : The Oxidation Potential of X X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ : The Reduction Potential of X 3+

68. Question (g) asks us what element metal X actually is. V 1 M HCl1 M X(NO 3 ) 3 1 M KNO 3 Cl 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the cathode along with its E°, if it can be determined. b)Write the half-reaction at the anode along with its E°, if it can be determined. c)Write the equation for the overall redox reaction and its E°. d)Find the E° value for the half- reaction occurring at the anode. e)What is the oxidation potential of element X? f)What is the reduction potential of X 3+ ? g)Metal X is actually what element? V = 3.02 V H+H+ Cl – X3+X3+ NO 3 – e–e–

69. We have the reduction half-reaction and the reduction potential for X 3+. 1 M HCl1 M X(NO 3 ) 3 H+H+ Cl – NO 3 – CATHODE X3+X3+ g) Metal X is actually what element? X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ : The Reduction Potential of X 3+

70. So we look on the standard reduction table to see if we can find a match, 1 M HCl1 M X(NO 3 ) 3 H+H+ Cl – NO 3 – CATHODE X3+X3+ g) Metal X is actually what element? X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ : The Reduction Potential of X 3+

71. We see this exactly matches the half-reaction for the reduction of aluminum 1 M HCl1 M X(NO 3 ) 3 H+H+ Cl – NO 3 – CATHODE X3+X3+ g) Metal X is actually what element? X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ : The Reduction Potential of X 3+

72. So all the X’s in this equation (click) can be replaced by the symbol for aluminum 1 M HCl1 M X(NO 3 ) 3 H+H+ Cl – NO 3 – CATHODE X3+X3+ g) Metal X is actually what element? X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ : The Reduction Potential of Al 3+ Al

73. So now we’ve answered question “g”. Metal X is actually aluminum 1 M HCl1 M X(NO 3 ) 3 H+H+ Cl – NO 3 – CATHODE X3+X3+ g) Metal X is actually what element? X 3+ + 3e –  X (s) E° = –1.66 V Reduction of X 3+ : The Reduction Potential of Al 3+ Al

74. This diagram and equations summarizes the electrochemical cell this example deals with. (click) Aluminum has replaced unknown metal “X” as the anode. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (Al (s)  Al 3+ + 3e – )×2 E° = 1.66 V. 3Cl 2(g) + 2Al (s)  6Cl – + 2Al 3+ E° = 3.02 V cathode anode Al replaces metal “X”

75. The chlorine half-reaction on the reduction table, written as a reduction, (click) is the half-reaction taking place at the cathode, and the E naught value is 1.36 volts k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (Al (s)  Al 3+ + 3e – )×2 E° = 1.66 V. 3Cl 2(g) + 2Al (s)  6Cl – + 2Al 3+ E° = 3.02 V cathode anode

76. And the aluminum half-reaction (click), reversed and written as an oxidation, is the half-reaction taking place at the anode, and because it was reversed, the E naught value is +1.66 volts k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (Al (s)  Al 3+ + 3e – )×2 E° = 1.66 V. 3Cl 2(g) + 2Al (s)  6Cl – + 2Al 3+ E° = 3.02 V cathode anode

77. Adding the two half-reactions, (click) gives us the overall redox equation k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (Al (s)  Al 3+ + 3e – )×2 E° = 1.66 V. 3Cl 2(g) + 2Al (s)  6Cl – + 2Al 3+ E° = 3.02 V cathode anode

78. with an E naught value of 3.02 volts, which (click) corresponds to the given initial voltage of 3.02 volts for this cell. k (Cl 2(g) + 2e –  2Cl – )×3 E°= 1.36 V. (Al (s)  Al 3+ + 3e – )×2 E° = 1.66 V. 3Cl 2(g) + 2Al (s)  6Cl – + 2Al 3+ E° = 3.02 V cathode anode