1. Here, we’ll use the half-reactions we came up with and balanced in Part 1 to build an equation for the overall Redox Reaction taking place. We’ll balance it in acid solution here. Then in Part 3, we will convert it to Basic Solution

2. Here are the two balanced half-reactions we obtained in Part 1. Write a balanced redox equation in basic solution, given the skeleton equation: The 2 half-reactions we balanced in Part 1

3. In order to add these up and obtain the overall redox equation, we must look at the electrons. Write a balanced redox equation in basic solution, given the skeleton equation:

4. We see that 3 electrons and gained by the lower half-reaction, and (click) 2 electrons are lost by the top half-reaction. Write a balanced redox equation in basic solution, given the skeleton equation:

5. So the electrons gained are NOT equal to the electrons lost. We must multiply these half-reactions by factors which will make the electrons gained equal to those lost Write a balanced redox equation in basic solution, given the skeleton equation: e – ’s gained ≠ e – ’s gained

6. The lowest common multiple of 3 and 2, is 6. Therefore we multiply by factors which will make electrons gained and electrons lost, both equal to 6. Write a balanced redox equation in basic solution, given the skeleton equation: Lowest common multiple of 3 and 2 is 6

7. So we multiply the top half-reaction by 3, Write a balanced redox equation in basic solution, given the skeleton equation:

8. And the bottom half-reaction by 2 Write a balanced redox equation in basic solution, given the skeleton equation:

9. So 2 × 3 = 6 electrons are gained by the bottom half-reaction Write a balanced redox equation in basic solution, given the skeleton equation: 6 e – gained

10. And 3 × 2 = 6 electrons are lost by the top half-reaction. Write a balanced redox equation in basic solution, given the skeleton equation: 6 e – lost

11. So, because electrons gained are now equal to electrons lost, Write a balanced redox equation in basic solution, given the skeleton equation: 6 e – lost 6 e – gained

12. The electrons can be cancelled from the half-reactions Write a balanced redox equation in basic solution, given the skeleton equation:

13. So now we’re left with these. Write a balanced redox equation in basic solution, given the skeleton equation:

14. At this point, we add what we have on the left and right of the arrows to obtain the equation for the overall redox reaction. Write a balanced redox equation in basic solution, given the skeleton equation:

15. Starting on the top left, we have (click) 3 × 2 NH2OH Write a balanced redox equation in basic solution, given the skeleton equation:

16. Which gives us 6 NH2OH Write a balanced redox equation in basic solution, given the skeleton equation:

17. And on the bottom left, we have 2 × HPO3 (2 minus) Write a balanced redox equation in basic solution, given the skeleton equation:

18. Which we’ll add here, Write a balanced redox equation in basic solution, given the skeleton equation:

19. And 2 × 5 H +, Write a balanced redox equation in basic solution, given the skeleton equation:

20. Which equals 10 H +. Write a balanced redox equation in basic solution, given the skeleton equation:

21. Now for the right side. On the top right we have (click) 3 × 1 N2, Write a balanced redox equation in basic solution, given the skeleton equation:

22. Which is 3 N 2, Write a balanced redox equation in basic solution, given the skeleton equation:

23. 3 × 2 H2O, Write a balanced redox equation in basic solution, given the skeleton equation:

24. Which is equal to 6 H2O, Write a balanced redox equation in basic solution, given the skeleton equation:

25. And 3 times 2 H +, Write a balanced redox equation in basic solution, given the skeleton equation:

26. Which equals 6 H +. Write a balanced redox equation in basic solution, given the skeleton equation:

27. On the bottom right, we have 2 × P Write a balanced redox equation in basic solution, given the skeleton equation:

28. Which is equal to 2 P Write a balanced redox equation in basic solution, given the skeleton equation:

29. And 2 × 3 H 2 O, Write a balanced redox equation in basic solution, given the skeleton equation:

30. Which is equal to 6 H 2 O. Write a balanced redox equation in basic solution, given the skeleton equation:

31. Now we have the balanced redox equation. Write a balanced redox equation in basic solution, given the skeleton equation: Balanced Redox Equation

32. But it has H+ ions on both sides, Write a balanced redox equation in basic solution, given the skeleton equation:

33. And two sets of water molecules on the right side, Write a balanced redox equation in basic solution, given the skeleton equation:

34. So this equation must be simplified Write a balanced redox equation in basic solution, given the skeleton equation: Must Be Simplified

35. We start by removing 6 H+ ions from both sides, Write a balanced redox equation in basic solution, given the skeleton equation: Remove 6 H + ions from Both Sides

36. Removing 6 H+ ions from the right side, leaves us with none Write a balanced redox equation in basic solution, given the skeleton equation: Remove 6 H + ions from Both Sides

37. And Removing 6 H+ ions from the left side leaves us with 10 – 6 (click), or 4 H+ ion Write a balanced redox equation in basic solution, given the skeleton equation: Remove 6 H + ions from Both Sides 4

38. We simplify the water by adding the two sets of 6 H2O molecules on the right side Write a balanced redox equation in basic solution, given the skeleton equation: Add up the two sets of H 2 O molecules 4

39. To give us 12 water molecules altogether. Write a balanced redox equation in basic solution, given the skeleton equation: 6H 2 O + 6H 2 O = 12H 2 O +12 4

40. We’ll re-write this equation in more compact form here Write a balanced redox equation in basic solution, given the skeleton equation: +12 4

41. So now we have the redox equation balanced in acid solution. At this point, pause the video and check to see that all atoms are balanced and total ionic charge is balanced Write a balanced redox equation in basic solution, given the skeleton equation: +12 4 Balanced Redox Equation in Acid Solution

42. However, the original question wants us to balance this equation in BASIC solution Write a balanced redox equation in basic solution, given the skeleton equation: +12 4 Balanced Redox Equation in Acid Solution

43. So this equation needs to be converted to Basic Solution Write a balanced redox equation in basic solution, given the skeleton equation: +12 4 Convert to Basic Solution

44. In the next video, which is Part 3, we will convert this redox equation to basic solution. Write a balanced redox equation in basic solution, given the skeleton equation: +12 4 In PART 3 (the next video) we’ll Convert this to Basic Solution