Balancing Oxidation-Reduction Reactions - neutral and acid solutions
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1 Balancing Oxidation-Reduction Reactions - neutral and acid solutions Balancing an oxidation-reduction reaction is a little simpler if the reaction is either neutral or acidic, and a little more difficult if the solution is basic. If the problem doesn t specifically state that the reaction conditions are basic, you should always use the simpler procedure given in this tutorial for acid or neutral conditions.
2 Balancing Oxidation-Reduction Reactions - neutral and acid solutions The steps to balancing oxidation-reduction (redox) reactions 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. Double check that all species and charges balance.
3 Example ample1. Let s apply these steps to balance the reaction: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq)
4 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 1. Write separate equations for reduction and oxidation half reactions. In making ½ reactions you are looking for elements that are in one oxidation state as a reactant, and a different oxidation state as a product. The first obvious ½ reaction is: Fe 2+ =Fe 3+ Here it is clear that the Fe is going from a +2 to a +3 oxidation state.
5 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 1. Write separate equations for reduction and oxidation half reactions. By process of elimination, your guess for the other ½ reaction would then be : MnO 4- = Mn 2+ If you go through the oxidation state calculation you would find that the Mn here goes from a +7 state to a +2 state, but I don t usually do this. It takes too much time, and it doesn t really help with balancing the equation.
6 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 1. Write separate equations for reduction and oxidation half reactions. I usually don t even bother to think about elements changing oxidation states. I just look for the same element in different compounds on the right an left side of the equation. Thus, I would just focus on the two main elements, Fe and Mn and make ½ reactions that connect them as products and reactants: Fe 2+ = Fe 3+ MnO 4- = 2+ Mn
7 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 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 Fe 2+ = Fe 3+ A. Fe balanced no problem B. No O s to balance so don t worry- be happy C. No H s either - don t worry - be happy D. (next page)
8 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 2. For each half reaction: Part d. Add electrons to balance net charge Fe 2+ = Fe 3+ Right now you have +2 charge on the left hand side of the equation and a +3 on the right. To get these in balance you always add electrons to the side with the largest positive charge to lower the charge to match the other side of the equation. So here you add 1 e - to the right hand side so the net charge on both sides is +2: Fe 2+ = Fe 3+ +1e -
9 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Fe 2 = Fe 3+ +1e - At this point you now know that this is your oxidation ½ reaction. How? Well, one way to tell is to look at the oxidation states. Fe increases its oxidation state t from +2 to +3, and oxidation is defined as the increase in an oxidation state. The other way is from the electrons. Whenever the electrons are on the right had side of the equation, you have an oxidation ½ reaction. This is the easy way because you don t have to figure out oxidation states!
10 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 2 for other 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 MnO 4-6Mn 2+ A. Mn balanced no problem B. 4 O s on the left, so you need to add 4 H 2 O s Os on the right to balance the O s MnO 4-6Mn H 2 O C. (Next Page)
11 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 2 for other 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 MnO 4-6Mn H 2 O C. Now you have 8 H s on the right, so you have to add 8 H + s to the left to make the H s balance: 8H + + MnO 4-6Mn H 2 O D. (Next Page)
12 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 2 for other 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 8H + + MnO 4-6Mn H 2 O D. Now you have +8-1 = +7 charge on the left hand side of the equation and a +2 on the right. So here you add 5 e - to the left hand side so the net charge on both sides is +2: 5e - + 8H + + MnO 4-6Mn H 2 O
13 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) 5e - +8H + +M MnO 4-6Mn 2+ +4HO 2 At this point you now know that this is your reduction ½ reaction. How? Again, if you spent the time do determine the oxidation state of the Mn, you would find that it is +7 in MnO 4- and +2 in Mn 2+ so the oxidation date is reduced, hence the name reduction reaction. But the far easier way is to look where the electrons are. Electrons on the left, reduction.
14 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 3. If necessary multiply l 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. Here are your two ½ reactions: Fe 2+ = Fe 3+ +1e - & 5e - +8H + +M MnO 4- = Mn 2+ +4HO 2 There is only 1 electron in the first reaction, but there are 5 electrons in the second reaction.
15 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 3. If necessary multiply l 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. Fe 2+ = Fe 3+ +1e - & 5e - +8H + +MnO 4- = 2+ Mn +4H 2 O Since the total number of electrons in both ½ reactions have to be equal, we have to multiply l the first reaction by 5 to turn it into the equation (Fe 2+ = Fe 3+ +1e - ) 5 = 5 Fe 2+ = 5Fe 3+ +5e - So the number of electrons in BOTH reactions is the same.
16 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 4. Add the two ½ reaction equations together and cancel any common terms. 5F Fe 2+ = 5Fe 3+ +5e - 5e - + 8H + + MnO 4- = Mn H 2 O = 5e - + 8H + + MnO Fe 2+ = Mn H 2 O + 5Fe 3+ +5e - The only common terms are the 5 e - so we have 8H + + MnO Fe 2+ 6Mn H 2 O + 5Fe 3+ N fill i th h i l f I d t t k i t ff Now fill in the physical forms so I don t take points off. 8H + (aq) + MnO 4- (aq) + 5 Fe 2+ (aq) = Mn 2+ (aq) + 4H 2 O(l) + 5Fe 3+ (aq)
17 Balance the equation: MnO 4- (aq) + Fe 2+ (aq) = Fe 3+ (aq) + Mn 2+ (aq) Step 5. Double check that all species and charges balance. 8H + (aq)+mno 4- (aq)+5 Fe 2+ (aq) = Mn 2+ (aq)+4h 2 O(l)+5Fe 3+ (aq) H 8 = 4(2) Mn 1 = 1 O 4 = 4 Fe 5 = 5 Charge (+2) = (+3) = +17 = = +17 It all balances so it must be good.
18 Example 2: Cr 2 O C 2 H 5 OH = Cr +3 + CO 2 Step 1. Half reactions: Cr 2 O 7-2 = Cr +3 & C 2 H 5 OH = CO 2
19 Step 2: Balancing ½ reactions Cr 2 O -2 7 = Cr +3 & C 2 H 5 OH = CO 2 Atoms EXCEPT O and H Cr 2 O -2 7 = 2 Cr +3 & C 2 H 5 OH = 2 CO 2 Bl Balancing O Cr 2 O -2 7 = 2 Cr H 2 O & 3 H 2 O + C 2 H 5 OH = 2 CO 2 Balancing H 14 H + +Cr 2 O 7-2 = 2 Cr H 2 O &3 H 2 O+C 2 H 5 OH=2 CO H + Balancing Charge 6 e H + + Cr 2 O -2 7 = 2 Cr H 2 O 3HO+CH 2CO +12H 2 2 H 5 OH = e -
20 Step 3: Getting the same number of electrons in both equations. EQN 1: 6 e H + + Cr 2 O 7-2 =2 Cr H 2 O EQN 2: 3 H 2 O + C 2 H 5 OH = 2 CO H e - The first equation has 6 electrons, the second has 12, so we need to multiply the first equation by 2 EQN 1: 2 (6 e H + + Cr 2 O 7-2 = 2 Cr H 2 O) EQN 1: 12 e H Cr 2 O -2 7 = 4 Cr H 2 O EQN 2: 3 H 2 O + C 2 H 5 OH = 2 CO H e -
21 Step 4: Adding equations together. th EQN 1:12 e H Cr 2 O -2 7 = 4 Cr H 2 O + EQN 2: 3 H O+C +12H 2 2 H 5 OH = 2 CO e - Net: 12 e H Cr 2 O H 2 O + C 2 H 5 OH = 4 Cr H 2 O + 2 CO H e -
22 Step 4: and removing common terms Net: 12 e H Cr 2 O H 2 O + C 2 H 5 OH = 4C Cr HO+2CO +12H e - Removing terms: 16 H Cr 2 O C 2 H 5 OH = 4 Cr H 2 O + 2 CO 2
23 Step 4: Double checking the balance 16 H Cr 2 O C 2 H 5 OH = 4 Cr H 2 O + 2 CO 2 H 16 6 = 11(2) Cr 2(2) = 4 O 2(7) 1 = 11 2(2) C 2 = 2(1) Charge 16+ 2(-2) 0 = 4(+3) 0 0 It all seems to balance so it looks good!
24 Final Answer 16 H + (aq)+2cro (aq) +CH 2 5 OH(aq) = 4 Cr +3 (aq) + 11 H 2 O (l) () + 2 CO 2 2(g) (I almost forgot to put in the physical forms)
25 Practice Problems Cu(s) + NO 3- (aq) = Cu 2+ (aq) + NO(g) I - (aq)+clo - (aq)i 3- (aq)+cl - (aq) H 3 AsO 4 (aq) + Zn(s) = AsH 3 (g) + Zn 2+ (aq) (Answers are on the next few pages)
26 Cu(s) () + NO 3- (aq) = Cu 2+ (aq) + NO(g) ½ rxns: Cu(s) = Cu 2+ (aq) & NO 3- (aq) = NO(g) Balance atoms: Cu(s) = Cu 2+ (aq) & NO 3- (aq) = NO(g) + 2H 2 O 4H + (aq)+ NO 3- (aq) = NO(g) + 2H 2 O Balance charge: Cu(s) = Cu 2+ (aq) + 2e - 3e - + 4H + (aq)+ NO 3- (aq) = NO(g) + 2H 2 O Get # of electrons the same: (Cu(s) = Cu 2+ (aq) + 2e - )x3 2x( 3e - + 4H + (aq)+ NO 3- (aq) = NO(g) + 2H 2 O) 3Cu(s) = 3Cu 2+ (aq) + 6e - 6e - + 8H + (aq)+ 2NO 3- (aq) = 2NO(g) + 4H 2 O Adding together: 3Cu(s) + 6e - + 8H + (aq)+ 2NO 3- (aq) = 3Cu 2+ (aq) + 6e - + 2NO(g) + 4H 2 O
27 Cu(s) () + NO 3- (aq) = Cu 2+ (aq) + NO(g) Removing common terms: 3Cu(s) + 8H + (aq)+ 2NO 3- (aq) = 3Cu 2+ (aq) + 2NO(g) + 4H 2 O Checking balance: Cu 3 = 3 H 8 = 4(2) N 2 = 2 O 2(3) = 2(1) 4(1) Charge +8 2(-1) = 3(2+) Final Answer: 3Cu(s) + 8H + (aq)+ 2NO 3- (aq) = 3Cu 2+ (aq) + 2NO(g) + 4H 2 O(l)
28 I - (aq)+clo - (aq) = I - )+Cl 3- (aq) - (aq) ½ Rxns: I - (aq) = I 3- - = - (aq) & ClO (aq) Cl (aq) Balancing atoms: 3 I - (aq) = I 3- (aq) & ClO - (aq) = Cl - (aq) + H 2 O 2H + + ClO - (aq) = Cl - (aq)+ho 2 Balancing Charge: 3 I - (aq) = I 3- (aq) + 2e - & 2e - + 2H + + ClO - (aq) = Cl - (aq) + H 2 O No need to multiply equations because they are the same. Adding equations: 3I - (aq) + 2e - + 2H + + ClO - (aq) = I (aq) 2e Cl (aq) + H 2 O Removing common terms: 3 I - (aq) + 2H + + ClO - (aq) = I 3- (aq) + Cl - (aq) + H 2 O
29 I - (aq)+clo - (aq) = I - )+Cl 3- (aq) - (aq) 3I - (aq) +2H + + ClO - (aq) = I 3- (aq)+cl - (aq)+h 2 O Double checking the balance: I 3 = 1(3) H 2(1) = 2 Cl 1 = 1 O 1 1 Charge 3(-1) 2(+1) -1 = = -2 = -2 P tti i h i l f Putting in physical forms: 3 I - (aq) + 2H + (aq) + ClO - (aq) = I 3- (aq) + Cl - (aq) + H 2 O(l)
30 H AO( )+Z )+Z 3 AsO 4 (aq) Zn(s) = AsH 3 (g) Zn 2+ (aq) ½ Reactions: H 3 AsO 4 (aq) = AsH 3 (g); Zn(s) = Zn 2+ (aq) Bl Balancing: H 3 AsO 4 (aq) = AsH 3 (g) + 4H 2 O 8H + + H 3 AsO 4 (aq) = AsH 3 (g) + 4H 2 O 8 e - +8H + + H 3 AsO 4 (aq) = AsH 3 (g) + 4H 2 O Zn(s) = Zn 2+ (aq) + 2 e - Multiplying right equations by 4: 4 Zn(s) = 4 Zn 2+ (aq) + 8 e - Adding equations together: 8 e - +8H + +HAO( 3 AsO 4 (aq)+4z() Zn(s) = AsH 3 (g) + 4H 2 O + 4 Zn 2+ (aq) + 8 e - Removing common terms: 8H + + H 3 AsO 4 (aq) + 4 Zn(s) = AsH 3 (g) + 4H 2 O + 4 Zn 2+ (aq)
31 H AO( )+Z )+Z 3 AsO 4 (aq) Zn(s) = AsH 3 (g) Zn 2+ (aq) 8H + + H 3AsO 4 4( (aq) + 4 Zn(s) () = AsH 3 3(g) + 4H 2 O + 4 Zn 2+ (aq) Double checking Balance: H8 3 = 3 4(2) As 1 = 1 O 4 =4(1) 4 Zn 4 4 Charge 8(+1) = 4(+2) Putting in physical form: 8H + (aq)+haso (aq)+4zn(s) =AsH (g)+4ho(l) Zn (aq)
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