1. We’re given the materials used to construct an electrochemical cell and we are asked various questions about it, including the initial cell voltage. Here, we’ll work through this and answer all the questions.

2. An electrochemical cell is constructed using a silver electrode in 1 M AgNO 3 and a nickel electrode in 1 M Ni(NO 3 ) 2. A salt bridge with 1 M KNO 3 is placed between the beakers. a)Which electrode is the cathode? Write its half-reaction. b)Which electrode is the anode? Write its half-reaction. c)Write the overall redox equation and determine the initial voltage of this cell. d)When the cell is operating, electrons are flowing toward which electrode? e)When the cell is operating K + ions are moving toward which solution?

3. We’ll add a diagram to represent this cell. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 An electrochemical cell is constructed using a silver electrode in 1 M AgNO 3 and a nickel electrode in 1 M Ni(NO 3 ) 2. A salt bridge with 1 M KNO 3 is placed between the beakers. 1 M KNO 3 a)Which electrode is the cathode? Write its half-reaction. b)Which electrode is the anode? Write its half-reaction. c)Write the overall redox equation and determine the initial voltage of this cell. d)When the cell is operating, electrons are flowing toward which electrode? e)When the cell is operating K + ions are moving toward which solution?

4. We’re asked which electrode is the cathode and told to write its half-reaction. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 An electrochemical cell is constructed using a silver electrode in 1 M AgNO 3 and a nickel electrode in 1 M Ni(NO 3 ) 2. A salt bridge with 1 M KNO 3 is placed between the beakers. 1 M KNO 3 a)Which electrode is the cathode? Write its half-reaction. b)Which electrode is the anode? Write its half-reaction. c)Write the overall redox equation and determine the initial voltage of this cell. d)When the cell is operating, electrons are flowing toward which electrode? e)When the cell is operating K + ions are moving toward which solution?

5. We’re also asked to identify the anode and write its half-reaction V 1 M AgNO 3 1 M Ni(NO 3 ) 2 An electrochemical cell is constructed using a silver electrode in 1 M AgNO 3 and a nickel electrode in 1 M Ni(NO 3 ) 2. A salt bridge with 1 M KNO 3 is placed between the beakers. 1 M KNO 3 a)Which electrode is the cathode? Write its half-reaction. b)Which electrode is the anode? Write its half-reaction. c)Write the overall redox equation and determine the initial voltage of this cell. d)When the cell is operating, electrons are flowing toward which electrode? e)When the cell is operating K + ions are moving toward which solution?

6. Next, we are asked to write the overall redox equation and find the initial voltage of this cell. Unless otherwise stated, we always assume we are at standard conditions. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 An electrochemical cell is constructed using a silver electrode in 1 M AgNO 3 and a nickel electrode in 1 M Ni(NO 3 ) 2. A salt bridge with 1 M KNO 3 is placed between the beakers. 1 M KNO 3 a)Which electrode is the cathode? Write its half-reaction. b)Which electrode is the anode? Write its half-reaction. c)Write the overall redox equation and determine the initial voltage of this cell. d)When the cell is operating, electrons are flowing toward which electrode? e)When the cell is operating K + ions are moving toward which solution?

7. We are asked which electrode electrons are flowing toward while this cell is operating. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 An electrochemical cell is constructed using a silver electrode in 1 M AgNO 3 and a nickel electrode in 1 M Ni(NO 3 ) 2. A salt bridge with 1 M KNO 3 is placed between the beakers. 1 M KNO 3 a)Which electrode is the cathode? Write its half-reaction. b)Which electrode is the anode? Write its half-reaction. c)Write the overall redox equation and determine the initial voltage of this cell. d)When the cell is operating, electrons are flowing toward which electrode? e)When the cell is operating K + ions are moving toward which solution?

8. And finally, we are asked which way potassium cations are moving in the salt bridge? We’ll answer these questions one at a time. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 An electrochemical cell is constructed using a silver electrode in 1 M AgNO 3 and a nickel electrode in 1 M Ni(NO 3 ) 2. A salt bridge with 1 M KNO 3 is placed between the beakers. 1 M KNO 3 a)Which electrode is the cathode? Write its half-reaction. b)Which electrode is the anode? Write its half-reaction. c)Write the overall redox equation and determine the initial voltage of this cell. d)When the cell is operating, electrons are flowing toward which electrode? e)When the cell is operating K + ions are moving toward which solution?

9. In Part a, we’re asked to identify the cathode and write its half-reaction. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 1 M KNO 3 a) Which electrode is the cathode? Write its half-reaction.

10. We look at the reduction table in the data booklet and (click) locate the half- reactions for silver and nickel. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 1 M KNO 3 a) Which electrode is the cathode? Write its half-reaction.

11. The rule is, the higher half-reaction on this table is the cathode. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 1 M KNO 3 a) Which electrode is the cathode? Write its half-reaction. cathode anode The higher half-reaction on the table is the CATHODE

12. And the lower half-reaction on the table is the anode. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 1 M KNO 3 a) Which electrode is the cathode? Write its half-reaction. cathode anode The lower half-reaction on the table is the ANODE

13. So we know that the cathode is the silver electrode. a) Which electrode is the cathode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E 0 = +0.80 v Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v cathode anode

14. We’re asked to write the half-reaction at the cathode. a) Which electrode is the cathode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E 0 = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v cathode anode

15. Remember that reduction occurs at the cathode. a) Which electrode is the cathode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E 0 = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v cathode anode

16. The reduction half-reaction for Ag+ can be copied directly from the table. The half-reactions on this table are all written as reductions. So the half-reaction is (click) Ag + plus 1 e – gives Ag solid a) Which electrode is the cathode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E 0 = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v cathode anode

17. Because this half-reaction is not reversed, we note that its standard reduction potential, E naught, (click) is + 0.80 volts. a) Which electrode is the cathode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v cathode anode

18. The (b) part of this question asks us for the anode and its half-reaction b) Which electrode is the anode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v

19. The anode is the lower of the two metals on the table, so it is nickel in this case. b) Which electrode is the anode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v

20. So we’ll mark nickel as the anode. b) Which electrode is the anode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v anode

21. We’re asked to write the half-reaction occurring at the anode. b) Which electrode is the anode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v

22. We know that at the anode, oxidation occurs b) Which electrode is the anode? Write its half-reaction. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v

23. The half-reactions on the table are all written as reductions (click) so for oxidation (click), the half-reaction on the table must be reversed. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v For Oxidation, the reduction half-reaction must be REVERSED b) Which electrode is the anode? Write its half-reaction.

24. So reversing this half-reaction we get (click) Ni (solid) gives Ni (2+) plus 2 electrons. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v For Oxidation, the reduction half-reaction must be REVERSED b) Which electrode is the anode? Write its half-reaction.

25. Now when a half-reaction is reversed, (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v When a half-reaction is Reversed, the SIGN on the E 0 is switched. b) Which electrode is the anode? Write its half-reaction.

26. The sign on the E naught is switched. On the table it is –0.26 Volts, (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E 0 = +1.26 v Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +2.06 v When a half-reaction is Reversed, the SIGN on the E° is switched. b) Which electrode is the anode? Write its half-reaction.

27. So the E naught value for this half-reaction written as an oxidation is positive 0.26 volts. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v When a half-reaction is Reversed, the SIGN on the E° is switched. b) Which electrode is the anode? Write its half-reaction.

28. Because the half-reaction at the anode is oxidation, (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v When a half-reaction is Reversed, the SIGN on the E° is switched. b) Which electrode is the anode? Write its half-reaction. Oxidation

29. Positive 0.26 volts is called the Oxidation Potential of Nickel metal. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v b) Which electrode is the anode? Write its half-reaction. Called Oxidation Potential of Ni

30. Whereas the negative 0.26 volts on the table (click) is called the reduction potential of Ni (2 plus). (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v b) Which electrode is the anode? Write its half-reaction. Called Oxidation Potential of Ni Called Reduction Potential of Ni 2+

31. So now we have the half-reaction at the anode, along with its E naught value. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v b) Which electrode is the anode? Write its half-reaction.

32. The “c” part of this question asks us to write the overall redox equation and find the initial voltage of the cell. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

33. First we write the equation for the overall redox reaction (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

34. We get this by adding up the half-reactions at the cathode and anode. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

35. Notice Ag+ gains one electron, (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v Gains 1 e – c) Write the overall redox equation and determine the initial voltage of this cell.

36. While the nickel loses 2 electrons. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v Gains 1 e – Loses 2 e – c) Write the overall redox equation and determine the initial voltage of this cell.

37. Electrons need to be balanced, so that the number lost is equal to the number gained. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v Electrons need to be balanced c) Write the overall redox equation and determine the initial voltage of this cell.

38. So we multiply the silver half-reaction by 2. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v × 2 c) Write the overall redox equation and determine the initial voltage of this cell.

39. So now we have 2 electrons gained by the Ag+ ions. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

40. And 2 electrons lost by the nickel atoms, so electrons are now balanced. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

41. It is very important to know that when we multiply a half-reaction by a factor like 2, (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v When we multiply a half-reaction by a factor, we DO NOT multiply the E 0 value! It stays the same! c) Write the overall redox equation and determine the initial voltage of this cell.

42. The we do NOT multiply the E naught value by anything. It stays the same. Voltage is the energy per electron. Doubling the number of electrons does not change the energy possessed by each electron, so the voltage remains the same. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v When we multiply a half-reaction by a factor, we DO NOT multiply the E° value! It stays the same! c) Write the overall redox equation and determine the initial voltage of this cell.

43. Remember, the only way we alter an E naught value is, (click) we switch its sign when the half-reaction is reversed. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

44. So now we can add these two half-reactions to obtain the equation for the overall redox reaction. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

45. Starting out on the left side, we have 2 Ag + ions from the top left. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

46. And 1 Nickel atom from the bottom left, (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

47. And on the right side, we have 2 silver atoms from the top right. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

48. And one nickel (2+) ion from the bottom right. Remember the number of electrons gained is equal to the number lost, so we just cancelled out electrons. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

49. So this is the balanced equation for the overall redox reaction. If you check, you’ll see that atoms and charges are balanced. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E 0 = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

50. Now we can calculate the E naught value for the overall redox reaction. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

51. We calculate this by ADDing up the E naught values for the half-reactions as written here. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

52. 0.80 plus 0.26 adds up to (click) 1.06 volts (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

53. This is called the standard cell potential for this electrochemical cell. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell. This is called the standard cell potential for this electrochemical cell. It is also the initial voltage of this cell at standard conditions

54. It is also the initial voltage this cell would have if it was set up at standard conditions, which are 25°C, 1 Molar solutions, and 1 atmosphere of pressure. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell. This is called the standard cell potential for this electrochemical cell. It is also the initial voltage of this cell at standard conditions

55. So we’ve found the initial voltage of this cell at standard conditions, and its 1.06 V, just a little over 1 volt. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v c) Write the overall redox equation and determine the initial voltage of this cell.

56. The “d” part of this question asks toward which electrode electrons are flowing as the cell operates. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v d) When the cell is operating, electrons are flowing toward which electrode?

57. We learned in the previous video that electrons always flow from the anode toward the cathode in the wires. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v d) When the cell is operating, electrons are flowing toward which electrode? Electrons always flow: From the ANODE  the CATHODE

58. So if we replace the voltmeter with a light bulb, so that current can pass through, we see that electrons move from the nickel anode, toward the silver cathode in this cell. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v d) When the cell is operating, electrons are flowing toward which electrode? Electrons always flow: From the ANODE  the CATHODE e–e–

59. So we can say that electrons are flowing toward the silver electrode as this cell operates. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v d) When the cell is operating, electrons are flowing toward which electrode? Electrons always flow: From the ANODE  the CATHODE e–e– Electrons are flowing toward the Silver Electrode, which is the CATHODE

60. This makes sense when you look at the half-reactions. The half reaction for the nickel electrode (click) tells us electrons are LOST from the nickel electrode. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v d) When the cell is operating, electrons are flowing toward which electrode? Electrons are lost from the nickel electrode

61. And the half-reaction for silver tells us (click) electrons are gained at the silver electrode. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v d) When the cell is operating, electrons are flowing toward which electrode? Electrons are gained at the silver electrode

62. Therefore, electrons must be flowing from the nickel electrode where they’re lost, to the silver electrode where they’re gained. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v d) When the cell is operating, electrons are flowing toward which electrode? e–e– Electrons flow from the Ni electrode to the Ag electrode Electrons are gained at the silver electrode Electrons are lost from the nickel electrode

63. The “e” part of this question asks us which way the K+ ions are moving as the cell operates. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v e) When the cell is operating K + ions are moving toward which solution?

64. K+ ions are cations because they are positive. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v e) When the cell is operating K + ions are moving toward which solution? Cations

65. And from the previous video, we learned that cations in the salt bridge always move toward the solution surrounding the cathode. (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v e) When the cell is operating K + ions are moving toward which solution? Cations always flow Toward the CATHODE Cations

66. So potassium ions are moving toward the silver nitrate solution as the cell operates (Ag + + e –  Ag (s) )2 E° = +0.80 V Cathode (Reduction) Ni (s)  Ni 2+ + 2e – E° = +0.26 V Anode (Oxidation) Overall Redox Reaction 2Ag + + Ni (s)  2Ag (s) + Ni 2+ E° = +1.06 v e) When the cell is operating K + ions are moving toward which solution? K + ions are moving Toward the AgNO 3 solution K+K+

67. So we see that given the materials an electrochemical cell is constructed from, and the standard reduction table, we were able to answer many questions about the cell. V 1 M AgNO 3 1 M Ni(NO 3 ) 2 An electrochemical cell is constructed using a silver electrode in 1 M AgNO 3 and a nickel electrode in 1 M Ni(NO 3 ) 2. A salt bridge with 1 M KNO 3 is placed between the beakers. 1 M KNO 3 a)Which electrode is the cathode? Write its half-reaction. b)Which electrode is the anode? Write its half-reaction. c)Write the overall redox equation and determine the initial voltage of this cell. d)When the cell is operating, electrons are flowing toward which electrode? e)When the cell is operating K + ions are moving toward which solution?