Redox Equations under Basic Conditions

Transcription

1 Redox Equations under Basic Conditions Basic conditions means that t you have more OH - and very little H + in your solution. In fact, you have so little H + that it can t appear in the equation because it doesn t exist! The way you will balance redox reactions under basic conditions is exactly the same way you balanced equations under neutral and acidic solution, the only difference is that we will include one additional step to remove the H + in our equation and replace it with OH -.

2 Redox Equations under Basic Conditions The steps to balancing oxidation-reduction (redox) reactions under basic conditions are: 1. Write separate equations for reduction and oxidation half reactions. 2. For each half reaction a. First balance all elements except O and H. b. Add H 2 O to balance O. c. Add H + to balance H. d. Add electrons to balance net charge. 3. If necessary, multiply one or both ½ reaction equations by integers so the total number of electrons used in one reaction is equal to the total number of electrons furnished by the other reaction. 4. Add the two ½ reaction equations together and cancel any common terms. 5. Add OH - to BOTH sides of the equation to change any H + s into H 2 O. 6. Double check that all species and charges balance.

3 Example ample1. Let s try this example Balance the equation: NO 2- (aq) + Al(s) = NH 3 (g) + AlO 2- (aq) under basic conditions.

4 NO 2- (aq)+al( Al(s) = NH 3 ()+AlO( (g) 2- (aq) ½ Reactions: NO 2- (aq) = NH 3 (g) Al(s) = AlO 2- (aq) Balancing: NO 2- (aq) = NH 3 (g) + 2H 2 O 7H + + NO 2- (aq) = NH 3 (g) + 2H 2 O 6e - + 7H + + NO 2- (aq) = NH 3 3(g) + 2H 2O 2H 2 O + Al(s) = AlO 2- (aq) 2H O+Al(s)= - +4H + 2 AlO 2 (aq) 2H 2 O + Al(s) = AlO 2- (aq) + 4H + + 3e - l l h d b 2 h 6 l Multiplying the second equation by 2 so we have 6 electrons: 4H 2 O + 2 Al(s) = 2 AlO 2- (aq) + 8H + + 6e -

5 NO 2- (aq)+al( Al(s) = NH 3 ()+AlO( (g) 2- (aq) Adding equations together: 6e - + 7H + + NO 2- (aq) = NH 3 (g) + 2H 2 O + 4H 2 O + 2 Al(s) = 2 AlO 2- (aq) + 8H + + 6e - 6e - + 7H + + NO 2- (aq) + 4H 2 O + 2 Al(s) = NH 3 (g) + 2H 2 O + 2 AlO 2- (aq) + 8H + + 6e - Removing common terms: NO 2- (aq) +2 H 2 O +2Al(s) = NH 3 (g) +2AlO 2- (aq) +1H + Now to our added rule for basic solutions (next page)

6 NO 2- (aq)+al( Al(s) = NH 3 ()+AlO( (g) 2- (aq) NO 2- (aq) +2 H 2 O +2Al(s) () = NH 3 3(g) + 2 AlO 2- (aq) + 1 H + Now to our added rule for basic solutions: 5. Add OH - to BOTH sides of the equation to change any H + sintoh 2 O To apply this rule find the number of H + s in the equation. In this case there is 1 H + on the right hand side of the equation, so we add 1 OH - to BOTH sides of the equation so it remains balanced. The trick is that on the side with the H +, the H + and the OH - can be combined to make H 2 O, so the H + disappears from the equation! (Next page)

7 NO 2- (aq)+al( Al(s) = NH 3 ()+AlO( (g) 2- (aq) NO 2- (aq) +2 H 2 O +2 Al(s) () = NH 3 3(g) + 2 AlO 2- (aq) + 1 H + + OH - + OH - Rewriting: NO 2- (aq) +2 H 2 O +2Al(s) + OH - = NH 3 (g) +2AlO 2- (aq) +1 H + + OH - Combining H + and OH - : NO 2- (aq) +2 H 2 O +2Al(s) + OH - = NH 3 (g) +2AlO 2- (aq) +1H 2 O

8 NO 2- (aq)+al( Al(s) = NH 3 ()+AlO( (g) 2- (aq) Checking to remove common terms a second time: NO 2- (aq) +2H 2 O +2Al(s) + OH - = NH 3 (g) +2AlO 2- (aq) +1H 2 O Becomes: NO - O+2Al(s)+OH (aq) + H 2 + OH = NH 3 (g) +2AlO 2 (aq) Double checking our balance: NO 2- (aq) + H 2 O +2Al(s) +OH - = NH 3 (g) +2AlO 2- (aq) N 1 =1 O = 2(2) () H 2 1 = 3 Al 2 = 2 Charge -1-1 = 2(-1)

9 NO 2- (aq)+al( Al(s) = NH 3 ()+AlO( (g) 2- (aq) Adding physical forms we get our final answer: NO 2- (aq) + H 2 O(l) + 2Al(s) + OH - (aq) = NH 3 (g) +2 AlO 2- (aq)

10 Example ample2. Let s try this example Balance the equation: Balance the equation: Al(s) + MnO 4- (aq) = MnO 2 (s) + Al(OH) 4- (aq) under basic conditions.

11 Al(s) + MnO 4- (aq) = MnO 2 (s) + Al(OH) 4- (aq) ½ Reactions: Al(s) = Al(OH) 4- (aq) MnO 4- (aq) = MnO 2 (s) Balancing: 4 H 2 O + Al(s) = Al(OH) 4- (aq) 4 H 2 O + Al(s) = Al(OH) 4- (aq) + 4H + 4HO 2 + Al(s) = Al(OH) 4- (aq) + 4H + + 3e - M O ( ) M O ()+2HO MnO 4- (aq) = MnO 2 (s) + 2 H 2 O 4H + +MnO 4- (aq) = MnO 2 (s) + 2 H 2 O 3e - + 4H + +MnO 4- (aq) = MnO 2 (s) + 2 H 2 O 4 2 2

12 Al(s) + MnO 4- (aq) = MnO 2 (s) + Al(OH) 4- (aq) Both equations have the same # of electrons, so we don t have to multiply anything! Adding equations together: 4 H 2 O + Al(s) = Al(OH) 4- (aq) + 4H + + 3e - + 3e - + 4H + +MnO 4- (aq) = MnO 2 () (s) + 2HO 2 4 H 2 O + Al(s) () +3e - + 4H + +MnO 4- (aq) = Al(OH) 4- (aq) + 4H + + 3e - +MnO 2 (s) + 2 H 2 O Removing common terms: Removing common terms: 2 H 2 O + Al(s) +MnO 4- (aq) =Al(OH) 4- (aq) + MnO 2 (s)

13 Al(s) + MnO 4- (aq) = MnO 2 (s) + Al(OH) 4- (aq) 2 H 2 O + Al(s) () +MnO 4- (aq) = Al(OH) 4- (aq) +MnO 2 2( (s) How about that. There aren t any H + s so I don t have to do anything! This happens every once in a while. Double checking overall balance: 2HO 2 + Al(s) +MnO 4- (aq) = Al(OH) 4- (aq) + MnO 2 () (s) H 2(2) = 4 O 2 4 = 4 2 Al 1 = 1 Mn 1 1 Charge -1 = -1

14 Al(s) + MnO 4- (aq) = MnO 2 (s) + Al(OH) 4- (aq) Putting in physical forms to get our final answer: 2 H 2 O(l) + Al(s) +MnO 4- (aq) = Al(OH) 4- (aq) + MnO 2 (s)

15 Practice problems Here are a couple of problems for you to try. Balance the following two reactions under BASIC conditions: CN - (aq) + MnO 4- (aq) = CNO - (aq) + MnO 2 (s) Cl 2 (g) = Cl - (aq) + OCl - (aq)

16 Practice problems CN - (aq) +MnO - 4- (aq) = CNO - (aq) +MnO(s) 2 Answer: 3 CN - (aq) + 2 MnO 4- (aq) + 1 H 2 O(l) = 3 CNO - (aq) + 2 MnO 2 (s) + 2 OH - (aq) Cl = (g) Cl (aq) OCl (aq) Answer: 2 Cl 2 (g) + 4OH - (aq) = 2 Cl - (aq) + 2 OCl - (aq) + 2 H 2 O (l) If you didn t get these answers, look over my notes on the next few pages, otherwise you can exit from this tutorial now.

17 CN - (aq)+m MnO 4- (aq) = CNO - (aq)+m MnO() 2 (s) ½ Reactions: CN - (aq) = CNO - (aq) MnO 4- (aq) = MnO 2 (s) Balancing: H 2 O + CN - (aq) = CNO - (aq) H 2 O + CN - (aq) = CNO - (aq) + 2H + H 2 O + CN - (aq) = CNO - (aq) + 2H + +2e - MnO 4- (aq) = MnO 2 (s) + 2H 2 O 4H + + MnO 4- (aq) = MnO 2 (s) + 2H 2 O 3e - +4H + +MnO - 4 (aq) = MnO 2 (s) + 2H 2 O

18 CN - (aq)+m MnO 4- (aq) = CNO - (aq)+m MnO() 2 (s) ½ Reactions: H 2 O + CN - (aq) = CNO - (aq) + 2H + +2e - 3e - + 4H + + MnO 4- (aq) = MnO 2 (s) + 2H 2 O So we need to multiply the first reaction by 3 and the second by 2! 3x(H 2 O + CN - (aq) = CNO - (aq) + 2H + +2e - ) 2x( 3e - + 4H + + MnO 4- (aq) = MnO 2 (s) + 2H 2 O) So we have: 3 H 2 O + 3 CN - (aq) = 3 CNO - (aq) + 6H e - 6e H MnO 4- (aq) = 2 MnO 2 (s) + 4 H 2 O

19 CN - (aq)+m MnO 4- (aq) = CNO - (aq)+m MnO() 2 (s) Adding together: 3 H 2 O + 3 CN - (aq) = 3 CNO - (aq) + 6H e - + 6e H MnO 4- (aq) = 2 MnO 2 (s) + 4 H 2 O We get: 3 H 2 O + 3 CN - (aq) + 6e H MnO 4- (aq) = 3 CNO - (aq) + 6H e - +2 MnO 2 (s) + 4 H 2 O Removing common terms: Removing common terms: 3 CN - (aq) + 2 H MnO 4- (aq) = 3 CNO - (aq) +2 MnO 2 (s) + 1 H 2 O

20 CN - (aq)+m MnO 4- (aq) = CNO - (aq)+m MnO() 2 (s) Last equation: 3 CN - (aq) + 2 H MnO 4- (aq) = 3 CNO - (aq) +2 MnO 2 (s) + 1 H 2 O Adding 2OH - to both sides: 3 CN - (aq) + 2 H MnO 4- (aq) + 2OH - = 3 CNO - (aq) +2 MnO 2 (s) + 1 H 2 O + 2OH - Combining the H + and OH - on the left side to get H 2 O: Combining the H and OH on the left side to get H 2 O: 3 CN - (aq) + 2 MnO 4- (aq) + 2H 2 O = 3 CNO - (aq) +2 MnO 2 (s) + 1 H 2 O + 2OH -

21 CN - (aq)+m MnO 4- (aq) = CNO - (aq)+m MnO() 2 (s) Last equation: 3 CN - (aq) + 2 MnO 4- (aq) + 2H 2 O = 3CNO - (aq)+2mno(s) O+2OH 2 (s) + 1 H 2 - Removing common terms a second time: 3 CN - (aq) + 2 MnO 4- (aq) + 1 H 2 O = 3 CNO - (aq) +2 MnO 2 (s) + 2OH -

22 CN - (aq)+m MnO 4- (aq) = CNO - (aq)+m MnO() 2 (s) Checking the balance: 3 CN - (aq) + 2 MnO 4- (aq) + 1H 2 O = 3 CNO - (aq) +2 MnO 2 (s) + 2OH - C 3 = 3 N 3 = 3 Mn 2 = 2 O 2(4) 1 = 3 2(2) 2(1) H 2 = 2 Charge 3(-1) 2(-1) = 3(-1) 2(-1) Putting in the physical lforms to get our final equation: 3 CN - (aq) + 2 MnO 4- (aq) + 1 H 2 O(l) = 3 CNO - (aq) + 2 MnO 2 2( (s) + 2 OH - (aq)

23 Cl )+OCl 2 (g) = Cl - (aq) - (aq) ½ Reactions: Cl 2 2(g) = Cl - (aq) Cl 2 (g) = OCl - (aq) (Note: This is a tricky one, the same compound on the left is undergoing BOTH oxidation and reductions at the same time. This is called a disproportionation reaction, and you will see reactions like this every once in a while.)

24 Cl )+OCl 2 (g) = Cl - (aq) - (aq) Balancing: Cl 2 () (g) = 2 Cl - (aq) 2e - + Cl 2 (g) = 2 Cl - (aq) Cl 2 (g) = 2 OCl - (aq) 2 H 2 O + Cl 2 (g) = 2 OCl - (aq) 2HO+Cl(g) = - (aq)+4h OCl 2 H 2 O + Cl 2 (g) = 2 OCl - (aq) + 4 H e - Both equations have 2 electrons so we can combine without a multiplication: 2e - + Cl 2 (g) + 2 H 2 O + Cl 2 (g) = 2 Cl - (aq)+2ocl - (aq)+4h + +2 e -

25 Cl )+OCl 2 (g) =Cl - (aq) - (aq) Last equation: 2e - + Cl 2 () (g) + 2HO 2 + Cl 2 () (g) = 2 Cl - (aq) + 2 OCl - (aq) + 4H e - Removing common terms and combing Cl 2 : 2 Cl 2 (g) + 2 H 2 O = 2 Cl - (aq) + 2 OCl - (aq) + 4H + Adding 4OH- to both sides of the equation: 2Cl 2 (g) + 2H 2 O + 4OH - = 2 Cl - (aq) +2OCl - (aq) + 4H + +4OH - Combing the H+ and OH- on the right side of the equation: 2Cl 2 (g) + 2H 2 O + 4OH - = 2 Cl - (aq) +2OCl - (aq) + 4H 2 O Removing common terms: 2Cl 2 (g) + 4OH - = 2 Cl - (aq) + 2 OCl - (aq) + 2 H 2 O

26 Cl )+OCl 2 (g) = Cl - (aq) - (aq) Last equation: 2Cl( 2 (g) + 4OH - = 2 Cl - (aq) + 2 OCl - (aq) + 2 H 2 O Checking balance: 2 Cl 2 (g) + 4OH - = 2 Cl - (aq) + 2 OCl - (aq) + 2 H 2 O Cl 2(2) = 2 2 O 4(1) = 2(1) 2(1) H 4(1) = 2(2) Charge 4(-1) = 2(-1) 2(-1) Putting in physical forms for final answer: 2 Cl 2 (g) + 4OH - (aq) = 2 Cl - (aq) + 2 OCl - (aq) + 2 H 2 O (l) 2 g q q q 2