1. Here, we’ll go through an example of balancing a half-reaction in basic solution.

2. We’re asked to balance the half-reaction: P 4 gives HPO 2 minus, taking place in basic solution. Balance the half-reaction: (basic solution)

3. We start by adding P4 and HPO2 minus to the equation Balance the half-reaction: (basic solution)

4. Even though we are asked to balance this half-reaction in Basic solution, we start out by balancing it in acid solution. Later, we’ll change it to basic. Balance the half-reaction: (basic solution) Balance in Acid Solution First Balance in Acid Solution

5. The first step is to balance atoms other than oxygen or hydrogen. In this case it is phosphorus Balance the half-reaction: (basic solution) Balance phosphorus atoms

6. We have 4 P atoms on the left and only 1 P atom on the right Balance the half-reaction: (basic solution) 1 Balance phosphorus atoms 4

7. To balance P’s, we add a 4 here, in front of the HPO2 minus Balance the half-reaction: (basic solution) Balance phosphorus atoms 4

8. Now we have 4 P atoms on both sides, so phosphorus is balanced. Balance the half-reaction: (basic solution) 4 Balance phosphorus atoms 4

9. The next step is to balance Oxygen atoms Balance the half-reaction: (basic solution) Balance oxygen atoms

10. There are no oxygen atoms on the left side, and looking on the right, we have 4 times 2… Balance the half-reaction: (basic solution) Balance oxygen atoms

11. Which is equal to 8 oxygen atoms on the right Balance the half-reaction: (basic solution) Balance oxygen atoms 8

12. To balance O’s we add 8 water molecules to the left side. Balance the half-reaction: (basic solution) Balance oxygen atoms 8

13. So now we have 8 O’s on both sides, so oxygen atoms are balanced. Balance the half-reaction: (basic solution) Balance oxygen atoms 88

14. The next step is to balance hydrogen atoms Balance the half-reaction: (basic solution) Balance hydrogen atoms

15. At this point, we have 16 H’s on the left side, Balance the half-reaction: (basic solution) Balance hydrogen atoms 16

16. and 4 H’s on the right side. Balance the half-reaction: (basic solution) Balance hydrogen atoms 4 16

17. So we need 12 more H’s on the right side. Balance the half-reaction: (basic solution) Balance hydrogen atoms 4 16 Need 12 more H’s on the Right Side

18. To do this we add 12 H+ ions to the right side Balance the half-reaction: (basic solution) Balance hydrogen atoms 4 16 Add 12H + to the Right Side

19. So now we have a total of 4 + 12 (click) = 16 H atoms on the right side Balance the half-reaction: (basic solution) Balance hydrogen atoms 4 16 12 16

20. So hydrogen atoms are now balanced. Balance the half-reaction: (basic solution) Balance hydrogen atoms 4 16 12 16

21. Now we need to balance charge. Looking on the left side, P4 and 8H2O both have a charge of zero, so the total charge on the left side is zero Balance the half-reaction: (basic solution) Balance Charge 00 0

22. Looking on the right side, we see that 4 HPO2 minus ions have a total charge of negative 4, Balance the half-reaction: (basic solution) Balance Charges –4 00 0

23. And 12 H+ ions have a total charge of positive 12 Balance the half-reaction: (basic solution) Balance Charges –4 0 +12 0 0

24. Therefore, the total charge on the right side is negative 4 plus 12, Balance the half-reaction: (basic solution) Balance Charges –4 0 +12 0 0

25. Which is equal to positive 8 Balance the half-reaction: (basic solution) Balance Charges –4 0 +12 +8 0 0

26. To balance charge we must add electrons to the more positive side. Balance the half-reaction: (basic solution) Balance Charges +8 0 Add 8e – to the more positive side

27. Because the charge on the left side is zero, and the charge on the right side is positive 8, we must (click) add 8 electrons to the right side. Balance the half-reaction: (basic solution) Balance Charges +8 0 Add 8e – to the more positive side

28. So we’ll add 8 electrons to the right side here. Balance the half-reaction: (basic solution) Balance Charges +8 0 Add 8e – to the Right side

29. So the total charge on the right side is now (–4) + (+12) + (–8), Balance the half-reaction: (basic solution) Balance Charges 0 –4 +12 –8

30. Which add up to zero Balance the half-reaction: (basic solution) Balance Charges 0 –4 +12 –8 0

31. So charge is now balanced Balance the half-reaction: (basic solution) Balance Charges 0 –4 +12 –8 0

32. And this half-reaction is now balanced in acid solution. Balance the half-reaction: (basic solution) Balanced in Acid Solution

33. However, we’re asked to balance it in basic solution. Balance the half-reaction: (basic solution) Balance in Basic Solution

34. So we have to cancel out these 12 H+ ions on the right. Balance the half-reaction: (basic solution) Balance in Basic Solution Cancel out 12 H +

35. We do this by adding the equation H + plus OH minus gives H 2 O. Balance the half-reaction: (basic solution) Balance in Basic Solution

36. The H+ is on the left side, so it can cancel the (click) H+ on the right side in the top equation. Balance the half-reaction: (basic solution) Balance in Basic Solution

37. But because there are 12 H+ we need to cancel Balance the half-reaction: (basic solution) Balance in Basic Solution

38. We add the coefficient 12 to all three species in this equation. Balance the half-reaction: (basic solution) Balance in Basic Solution

39. Now we have 12H+ on the left side and on the right side, so these can be cancelled. Balance the half-reaction: (basic solution) Balance in Basic Solution

40. So cancelling these out, we’re left with this Balance the half-reaction: (basic solution) Balance in Basic Solution

41. Notice we now have 8 water molecules on the left side and (click) 12 water molecules on the right side. Balance the half-reaction: (basic solution) Balance in Basic Solution

42. So to simply, we remove 8 water molecules from both sides… Balance the half-reaction: (basic solution) Balance in Basic Solution Remove 8 H 2 O molecules from BOTH sides

43. Removing 8 water molecules from the left side leaves us with none, Balance the half-reaction: (basic solution) Balance in Basic Solution Remove 8 H 2 O molecules from BOTH sides

44. And Removing 8 water molecules from the right side leaves us with… Balance the half-reaction: (basic solution) Balance in Basic Solution Remove 8 H 2 O molecules from BOTH sides

45. 12 minus 8, which is equal to 4 water molecules Balance the half-reaction: (basic solution) Balance in Basic Solution 4 Remove 8 H 2 O molecules from BOTH sides

46. Now we add up what remains to get the overall half-reaction. We’ll start with the left side. On the left side, Balance the half-reaction: (basic solution) Balance in Basic Solution 4

47. We have P4 Balance the half-reaction: (basic solution) Balance in Basic Solution 4

48. We have P4 Balance the half-reaction: (basic solution) Balance in Basic Solution 4

49. One the right side, we have 4 HPO 2 minus Balance the half-reaction: (basic solution) Balance in Basic Solution 4

50. Plus 4 H2O, Balance the half-reaction: (basic solution) Balance in Basic Solution 4

51. Plus 8 electrons Balance the half-reaction: (basic solution) Balance in Basic Solution 4

52. So now we have the equation for this half-reaction balanced in basic solution. At this point, pause the video and check that all atoms and charges are balanced. Balance the half-reaction: (basic solution) Balanced in Basic Solution 4 Balanced Half-reaction in BASIC Solution