1. In the previous videos, we looked at electrochemical cells with reactive metal electrodes and solutions containing their cations. However, some electrochemical cells have half-reactions involving gases. We’ll look at one of those here.

2. We’re given the diagram for an electrochemical cell V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum?

3. The left beaker has a platinum electrode V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) platinum electrode

4. Which is surrounded by hydrogen gas V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) hydrogen gas

5. And dipped into a solution of 1 molar hydrochloric acid V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) 1M HCl

6. Hydrochloric acid contains aqueous hydrogen ions and chloride ions. V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) H + (aq) and Cl – (aq) H+H+ Cl –

7. The beaker on the right has a tin electrode V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) tin electrode H+H+ Cl –

8. Immersed in a solution of 1 molar tin(II) nitrate. V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) H+H+ Cl – tin(II) nitrate

9. Which contains aqueous Sn (2 plus) ions, and nitrate ions. V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) H+H+ Cl – Sn 2+ (aq) NO 3 – (aq) Sn 2+ NO 3 –

10. Looking back at the half-cell on the left, it’s important to know that (click) platinum is an inert metal. The platinum electrode does not react in this cell. It only provides a surface upon which the half-reaction can take place. V 1 M HCl1 M Sn(NO 3 ) 2 1 M KNO 3 H 2(g) H+H+ Cl – The Platinum Electrode is Inert

11. Inert metals like platinum are often used in half-cells where gases are involved. Gases and aqueous ions need a solid surface to react with each other on. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – Used in half-cells where gases are involved

12. This particular half-cell is very important. (click) It’s called the standard half-cell. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell

13. The reactive species in the standard half-cell are H 2 gas and aqueous H + ions. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell Reactive species are H 2(g) and H + ions.

14. The chloride ions, Cl minus, are spectators in this particular half-cell. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell Reactive species are H 2(g) and H + ions. The chloride ions, Cl –, are spectators in this half-cell

15. The half-reaction for the standard half-cell is the shaded half-reaction right in the middle of the reduction table V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell

16. Here it is enlarged. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell The Half-reaction for the Standard Half-Cell

17. This half-cell is assigned a reduction potential of 0.00 Volts under standard conditions. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell E° = 0.00 V Temperature = 25°C Pressure = 1 atm (101.3 kPa) [H + ] = 1 M Standard Conditions

18. The little “naught” on the E naught means standard conditions. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell E° = 0.00 V Temperature = 25°C Pressure = 1 atm (101.3 kPa) [H + ] = 1 M Standard Conditions

19. At standard conditions (click) the temperature is 25°C. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell E° = 0.00 V Temperature = 25°C Pressure = 1 atm (101.3 kPa) [H + ] = 1 M Standard Conditions

20. The pressure is 1 atmosphere, or 101.3 kilopascals. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell E° = 0.00 V Temperature = 25°C Pressure = 1 atm (101.3 kPa) [H + ] = 1 M Standard Conditions

21. And the concentration of H + is 1 Molar V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell E° = 0.00 V Temperature = 25°C Pressure = 1 atm (101.3 kPa) [H + ] = 1 M Standard Conditions

22. The double arrow here reminds us that this half-reaction can occur either as a reduction and proceed to the right or as an oxidation and proceed to the left. The direction it goes depends on what other half-cell it is connected to. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – The Standard Half-Cell Can occur either as a Reduction or an Oxidation

23. Now that we know the half-cell on the left is the standard half-cell, we’ll look at the overall cell and go through a series of questions. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum? H+H+ Cl – 1 M KNO 3

24. The “a” part of the question asks us to write the half-reaction at the anode, along with its E naught value. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum? H+H+ Cl – 1 M KNO 3

25. We find the half-reactions on the reduction table. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 a) Write the half-reaction at the anode along with its E°.

26. The higher half-reaction on the table, the hydrogen half-cell, will act as the cathode V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 a) Write the half-reaction at the anode along with its E°. cathode

27. And the lower one, the tin half-cell will act as the anode. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 a) Write the half-reaction at the anode along with its E°. cathode anode

28. Because the tin half-reaction is the anode, oxidation is taking place, so the equation must be reversed (click) so its written as Sn solid gives Sn (2+) plus 2 electrons. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 a) Write the half-reaction at the anode along with its E°. Anode Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V anode

29. Also, because it the equation was reversed, the sign on the E naught must be switched, so (click) the E naught for this half-reaction is positive 0.14 Volts. This is the oxidation potential of tin metal V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 a) Write the half-reaction at the anode along with its E°. Anode Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V anode

30. The “b” part of the question asks us to write the half-reaction at the cathode, along with its E naught value. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum? H+H+ Cl – 1 M KNO 3

31. The cathode half-reaction is the hydrogen half-cell. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 a) Write the half-reaction at the cathode along with its E°. cathode Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V anode

32. And The cathode half-reaction is not reversed, so its (click) 2H + plus 2 electrons gives H 2 (gas) V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 a) Write the half-reaction at the cathode along with its E°. Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode anode cathode

33. And Its E naught value is equal to zero Volts V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 a) Write the half-reaction at the cathode along with its E°. Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode anode cathode

34. The “c” part of the question asks us to write the equation for the overall redox reaction, along with its E naught value. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum? H+H+ Cl – 1 M KNO 3 Cathode

35. To write the equation for the overall redox reaction, we add up the half-reactions the way they are written. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 c) Write the equation for the overall redox reaction and its E°. Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode anode cathode

36. Electrons gained are equal to electrons lost, so we don’t need to multiply any of the half-reactions and we can (click) cancel out the electrons. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 c) Write the equation for the overall redox reaction and its E°. Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Electrons Gained = Electrons Lost Cathode anode cathode

37. On the left side we have Sn (solid) and 2H (plus) V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 c) Write the equation for the overall redox reaction and its E°. Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode

38. And on the right side, we have Sn (2 plus) and H2 (gas) V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 c) Write the equation for the overall redox reaction and its E°. Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode anode cathode

39. To find the overall E naught, we add up positive 0.14 and zero, giving us (click) positive 0.14 volts V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 c) Write the equation for the overall redox reaction and its E°. Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode anode cathode

40. Remember the E naught value for an overall redox equation at standard conditions, is called (click) the standard cell potential V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 c) Write the equation for the overall redox reaction and its E°. Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Standard Cell Potential Cathode anode cathode

41. The “d” part of the question asks us to state the initial voltage of this cell. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum? H+H+ Cl – 1 M KNO 3 Cathode

42. Remember that standard cell potential is the same as the initial voltage of a cell. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) What is the initial voltage of this cell? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Standard Cell Potential Initial Voltage Cathode anode cathode

43. Initial Voltage So the initial voltage of this cell is positive 0.14 volts. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) What is the initial voltage of this cell? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode anode cathode

44. We’ll make a note of this up here by the voltmeter. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) What is the initial voltage of this cell? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V V = +0.14 V Cathode anode cathode

45. The “e” part of the question asks which way electrons are travelling as this cell operates. The cell will operate if we replace the voltmeter with a light bulb. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum? H+H+ Cl – 1 M KNO 3 Cathode

46. Remember that electrons always travel from the anode to the cathode through the wires. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates electrons will travel toward which electrode? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode Electrons travel from the Anode toward the Cathode in the Wires anode cathode

47. So they are travelling from the tin electrode toward the platinum electrode. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates electrons will travel toward which electrode? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode Electrons travel toward the Platinum Electrode e–e– anode cathode

48. The “f” part of the question asks what will happen to the pH around the platinum electrode as this cell operates. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum? H+H+ Cl – 1 M KNO 3 Cathode

49. To answer this question, we focus on the half-reaction taking place on the platinum electrode. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates what will happen to the pH of the solution near the Pt electrode? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode

50. We see that H + is on the left side of this half-reaction, so H + is (click) being consumed as it is reduced to hydrogen gas. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates what will happen to the pH of the solution near the Pt electrode? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode H + is being consumed as it is reduced to H 2(g) anode cathode

51. Remember from the Acid-base unit, that (click) pH is the NEGATIVE log of the hydrogen ion concentration. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates what will happen to the pH of the solution near the Pt electrode? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode pH = –log[H + ] So as [H + ] decreases, the pH increases anode cathode

52. So as H + is used up and the concentration of H +, or acidity decreases, (click) the pH will gradually increase near the platinum electrode. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates what will happen to the pH of the solution near the Pt electrode? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode pH = –log[H + ] So as [H + ] decreases, the pH increases anode cathode

53. The “g” part of the question asks what will happen to the mass of the platinum electrode as this cell operates. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) Given the electrochemical cell shown in the diagram: a)Write the half-reaction at the anode along with its E°. b)Write the half-reaction at the cathode along with its E°. c)Write the equation for the overall redox reaction and its E°. d)What is the initial voltage of this cell? e)As the cell operates electrons will travel toward which electrode? f)As the cell operates, what will happen to the pH of the solution near the Pt electrode? g)As the cell operates, what will happen to the mass of Platinum? H+H+ Cl – 1 M KNO 3 Cathode

54. Remember the half-reaction taking place on the platinum electrode is the reduction of H+ ions to hydrogen gas. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates, what will happen to the mass of Platinum? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode

55. The platinum electrode itself is inert. It doesn’t undergo any reaction. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates, what will happen to the mass of Platinum? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode The Platinum Electrode is Inert

56. It simply provides a solid surface for this reaction to take place on. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates, what will happen to the mass of Platinum? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode The platinum electrode provides a surface for this half-reaction to take place on

57. So as the cell operates (click), the mass of the Platinum electrode will NOT change in any way. 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 d) As the cell operates, what will happen to the mass of Platinum? Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode The mass of the platinum electrode will NOT change The platinum electrode provides a surface for this half-reaction to take place on anode cathode

58. To review, by using a cell diagram, the table of reduction potentials, and the principals we learned, we were able to answer a number of questions about this electrochemical cell. V 1 M HCl1 M Sn(NO 3 ) 2 H 2(g) H+H+ Cl – 1 M KNO 3 Sn (s)  Sn 2+ + 2e – E°= +0.14 V 2H + + 2e –  H 2(g) E°= +0.00 V. Sn (s) + 2H +  Sn 2+ + H 2(g) E°= +0.14 V Cathode V = +0.14 V anode cathode