Balancing Oxidation-Reduction g Reactions - neutral and acid ...

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Balancing an oxidation-reduction reaction is a little simpler ... The steps to balancing oxidation-reduction (redox) reactions are: 1. ..... Practice Problems. Cu( s) + ...
Balancingg Oxidation-Reduction Reactions - neutral and acid solutions Balancing an oxidation-reduction reaction is a little simpler if the reaction is either neutral or acidic, 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.

Balancingg Oxidation-Reduction Reactions - neutral and acid solutions The steps to balancing oxidation-reduction (redox) reactions are: 1. Write separate equations for reduction and oxidation h lf reactions. half ti 2. For each half reaction a. First balance all elements except O and H. b Add H2O to balance O b. O. c. Add H+ to balance H. d. Add electrons to balance net charge. 3 If necessary, 3. 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. q together g and cancel any y 4. Add the two ½ reaction equations common terms. 5. Double check that all species and charges balance.

E ample 1. Example 1 Let’ss apply these steps to balance the reaction: Let MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(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. d t The first obvious ½ reaction is: Fe2+ = Fe3+ Here it is clear that the Fe is going from a +2 to a +3 oxidation state.

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(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 : M O4- = Mn MnO M 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.

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(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 d on the th right i ht an left l ft side id off the th equation. ti Thus, Th I would just focus on the two main elements, Fe and Mn and make ½ reactions that connect them as products and reactants: Fe2+ = Fe3+

MnO4- = Mn2+

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 2. For each half reaction: a. First Balance all elements except O and H b Add H2O to balance O b. c. Add H+ to balance H d. Add electrons to balance net charge Fe2+ = Fe3+ 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) D.

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 2. For each half reaction: Part d. Add electrons to balance net charge Fe2+ = Fe3+ Right now you have +2 charge on the left hand side of the equation ti andd a +3 on the th right. i ht To T gett these th in i balance b l 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: Fe2+ = Fe3+ +1e-

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Fe2 = Fe3+ +1eAt this point you now know that this is your oxidation ½ reaction. reaction How? Well, one way to tell is to look at the oxidation states. F iincreases it Fe its oxidation id ti state t t from f +2 to t +3, +3 andd oxidation id ti is i 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. reaction This is the easy way because you don don’tt have to figure out oxidation states!

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 2 for other half reaction: a. First Balance all elements except O and H b. Add H2O to balance O c Add H+ to balance H c. d. Add electrons to balance net charge MnO4-6Mn2+ A. Mn balanced no problem B. 4 O O’ss on the left, so you need to add 4 H2O O’ss on the right to balance the O’s MnO4-6Mn2+ + 4H2O C (Next Page) C.

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 2 for other half reaction: a. First Balance all elements except O and H b. Add H2O to balance O c Add H+ to balance H c. d. Add electrons to balance net charge MnO4-6Mn2+ + 4H2O C. Now you have 8 H’s on the right, so you have to add 8 H+’ss to the left to make the H H’ss balance: 8H+ + MnO4-6Mn2+ + 4H2O D. (Next Page)

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 2 for other half reaction: a. First Balance all elements except O and H b. Add H2O to balance O c Add H+ to balance H c. d. Add electrons to balance net charge 8H+ + MnO4-6Mn2+ + 4H2O D. Now you have +8-1 = +7 charge on the left hand side of the equation and a +22 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+ + MnO4-6Mn2+ + 4H2O

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) 55e- + 8H+ + MnO M O4-6Mn 6M 2+ + 4H2O At this point you now know that this is your reduction ½ reaction. How? Again Again, if you spent the time do determine the oxidation state of the Mn, you would find that it is +7 in MnO4- and +2 in Mn2+ so the oxidation date is reduced, h hence the th name reduction d ti reaction. ti But the far easier way is to look where the electrons are. Electrons on the left, reduction.

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 3 St 3. If necessary multiply lti l one or b both th ½ reaction ti 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: Fe2+ = Fe3+ +1e& 5 - + 8H+ + MnO 5e M O4- = Mn M 2+ + 4H2O There is only 1 electron in the first reaction, but there are 5 electrons in the second reaction.

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 3 St 3. If necessary multiply lti l one or b both th ½ reaction ti 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. Fe2+ = Fe3+ +1e- & 5e- + 8H+ + MnO4- = Mn2+ + 4H2O Since the total number of electrons in both ½ reactions h have t be to b equal, l we have h to t multiply lti l the th first fi t reaction ti by b 5 to turn it into the equation (Fe2+ = Fe3+ +1e-)×5 = 5 Fe2+ = 5Fe3+ +5eSo the number of electrons in BOTH reactions is the same.

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 4 4. Add the two ½ reaction equations together and cancel any common terms. 5 Fe F 2+ = 5Fe 5F 3+ +5e +5 5e- + 8H+ + MnO4- = Mn2+ + 4H2O = 5e- + 8H+ + MnO4- + 5 Fe2+ = Mn2+ + 4H2O + 5Fe3+ +5eThe only common terms are the 5 e- so we have 8H+ + MnO4- + 5 Fe2+6Mn2+ + 4H2O + 5Fe3+ Now fill in N i the th physical h i l forms f so I don’t d ’t take t k points i t off. ff 8H+(aq) + MnO4-(aq) + 5 Fe2+(aq) = Mn2+ (aq) + 4H2O(l) + 5Fe3+(aq)

Balance the equation: q MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq) Step 5 5. Double check that all species and charges balance balance. 8H+(aq)+MnO4-(aq)+5 Fe2+(aq) = Mn2+(aq)+4H2O(l)+5Fe3+(aq) H8 = 4(2) Mn 1 =1 O 4 = 4 Fe 5 = 5 Charge +8 -11 5(+2) = +2 0 5(+3) 8-1+10 = +17 = 2 + 15 = +17 It all ll balances b l so it mustt be b good. d

Example p 2: Cr2O7-2 + C2H5OH = Cr+3 + CO2 Step 1. Half reactions: Cr2O7-2 = Cr+3 & C2H5OH = CO2

Step p 2: Balancing g ½ reactions Cr2O7-2 = Cr+3 & C2H5OH = CO2 Atoms EXCEPT ‘O’ O and ‘H’ H Cr2O7-2 = 2 Cr+3 & Balancing B l i O Cr2O7-2 = 2 Cr+3 + 7 H2O &

C2H5OH = 2 CO2

3 H2O + C2H5OH = 2 CO2

Balancing H 14 H++Cr2O7-2 = 2 Cr+3+7 H2O &3 H2O+C2H5OH=2 CO2+12 H+ Balancing Charge 6 e- + 14 H+ + Cr2O7-2 = 2 Cr+3 + 7 H2O 3 H2O + C2H5OH = 2 CO2 + 12 H+ + 12 e-

Step p 3: Gettingg the same number of electrons in both equations. EQN 1: 6 e- + 14 H+ + Cr2O7-2 = 2 Cr+3 + 7 H2O EQN 2: 3 H2O + C2H5OH = 2 CO2 + 12 H+ + 12 eThe first equation has 6 electrons, the second has 12, so we need to multiply the first equation by 2 EQN 1: 2×(6 e- + 14 H+ + Cr2O7-2 = 2 Cr+3 + 7 H2O) EQN 1: 12 e- + 28 H+ + 2 Cr2O7-2 = 4 Cr+3 + 14 H2O EQN 2: 3 H2O + C2H5OH = 2 CO2 + 12 H+ + 12 e-

St 4: Step 4 Adding Addi equations ti ttogether…. th EQN 1:12 e- + 28 H+ + 2 Cr2O7-2 = 4 Cr+3 + 14 H2O + EQN 22: 3 H2O + C2H5OH = 2 CO2 + 12 H+ + 12 eNet: 12 e- + 28 H+ + 2 Cr2O7-2 + 3 H2O + C2H5OH = 4 Cr+3 + 14 H2O + 2 CO2 + 12 H+ + 12 e-

St 4: Step 4 …and d removing i common terms t Net: 12 e- + 28 H+ + 2 Cr2O7-2 + 3 H2O + C2H5OH = 4C Cr+3 + 14 H2O + 2 CO2 + 12 H+ + 12 e-

Removing terms: 16 H+ + 2 Cr2O7-2 + C2H5OH =

4 Cr+3 + 11 H2O + 2 CO2

St 4: Step 4 Double D bl checking h ki the th balance b l 16 H+ + 2 Cr2O7-2 + C2H5OH = 4 Cr+3 + 11 H2O + 2 CO2 H 16 6 = 11(2) C 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!

Final Ans Answer er 16 H+(aq) + 2 Cr2O7-2(aq) + C2H5OH(aq) = q) + 11 H2O ((l)) + 2 CO2(g) 4 Cr+3((aq) (I almost forgot to put in the physical forms)

Practice Problems Cu(s) + NO3-(aq) = Cu2+(aq) + NO(g) I-(aq) ( ) + ClO-(aq) ( ) I3-(aq) ( ) + Cl-(aq) ( ) H3AsO4(aq) + Zn(s) = AsH3(g) + Zn2+(aq) (Answers are on the next few pages)

Cu(s) ( ) + NO3-((aq) q) = Cu2+((aq) q) + NO(g) (g) ½ rxns: Cu(s) = Cu2+(aq) & NO3-(aq) = NO(g) Balance atoms: Cu(s) = Cu2+(aq) & NO3-(aq) = NO(g) + 2H2O 4H+(aq)+ NO3-(aq) = NO(g) + 2H2O Balance charge: Cu(s) = Cu2+(aq) + 2e3e- + 4H+(aq)+ NO3-(aq) = NO(g) + 2H2O Get # of electrons the same: (Cu(s) = Cu2+(aq) + 2e-)x3 2x( 3e- + 4H+(aq) (aq)+ NO3-(aq) = NO(g) + 2H2O) 3Cu(s) = 3Cu2+(aq) + 6e6e- + 8H+(aq)+ 2NO3-(aq) = 2NO(g) + 4H2O Adding together: 3Cu(s) + 6e- + 8H+(aq)+ 2NO3-(aq) = 3Cu2+(aq) + 6e- + 2NO(g) + 4H2O

Cu(s) ( ) + NO3- ((aq) q) = Cu2+((aq) q) + NO(g) (g) Removing common terms: 3Cu(s) + 8H+(aq)+ 2NO3-(aq) = 3Cu2+(aq) + 2NO(g) + 4H2O Checking Ch ki balance: b l Cu 3 H 8 N 2 O 2(3) Charge +8 2( 2(-1) 1)

=3 = = = = 3(2+)

4(2) 2 2(1)

4(1)

Finall Answer: Fi A 3Cu(s) + 8H+(aq)+ 2NO3-(aq) = 3Cu2+(aq) + 2NO(g) + 4H2O(l)

I-(aq) ( ) + ClO-(aq) ( ) = I3-(aq) ( ) + Cl-(aq) ( ) ½ Rxns: I-(aq) = I3-(aq) & ClO-(aq) = Cl-(aq) Balancing atoms: 3 I-(aq) = I3-(aq) & ClO-(aq) = Cl-(aq) + H2O 2H+ + ClO-(aq) ( ) = Cl-(aq) ( ) + H2 O Balancing Charge: 3 I-(aq) = I3-(aq) + 2e- & 2e- + 2H+ + ClO-(aq) = Cl-(aq) + H2O No need to multiply equations because they are the same. Adding equations: 3 I-(aq) + 2e- + 2H+ + ClO-(aq) = I3-(aq) + 2e- + Cl-(aq) + H2O Removing common terms: 3 I-(aq) + 2H+ + ClO-(aq) = I3-(aq) + Cl-(aq) + H2O

I-(aq) ( ) + ClO-(aq) ( ) = I3-(aq) ( ) + Cl-(aq) ( ) 3 I-(aq) + 2H+ + ClO-(aq) = I3-(aq) + Cl-(aq) + H2O Double checking the balance: I 3 = 1(3) H 2(1) = Cl 1 = O 1 Charge 3(-1) 3( 1) 2(+1) 2( 1) -11 = -11 -3+2-1 = -2 = -2

2 1 1 -11

Putting P tti in i physical h i l forms: f 3 I-(aq) + 2H+(aq) + ClO-(aq) = I3-(aq) + Cl-(aq) + H2O(l)

H3AsO A O4(aq) ( ) + Zn(s) Z ( ) = AsH A H3(g) ( ) + Zn Z 2+(aq) ( ) ½ Reactions: H3AsO4(aq) = AsH3(g); Zn(s) = Zn2+(aq) B l i Balancing: H3AsO4(aq) = AsH3(g) + 4H2O 8H+ + H3AsO4(aq) = AsH3(g) + 4H2O 8 e- +8H+ + H3AsO4(aq) = AsH3(g) + 4H2O Zn(s) = Zn2+(aq) + 2 eMultiplying right equations by 4: 4 Zn(s) = 4 Zn2+(aq) + 8 eAdding equations together: 8 e- +8H+ + H3AsO A O4(aq) ( ) + 4 Zn(s) Z ( )= AsH3(g) + 4H2O + 4 Zn2+(aq) + 8 eRemoving common terms: 8H+ + H3AsO4(aq) + 4 Zn(s) = AsH3(g) + 4H2O + 4 Zn2+(aq)

H3AsO A O4(aq) ( ) + Zn(s) Z ( ) = AsH A H3(g) ( ) + Zn Z 2+(aq) ( ) 8H+ + H3AsO4((aq) q) + 4 Zn(s) ( ) = AsH3(g) + 4H2O + 4 Zn2+((aq) q) Double checking Balance: H8 3 As 1 O 4 Z Zn 4 Charge ( ) 8(+1)

= 3 = 1 =4(1)

4(2) 4 4

=

4(+2) ( )

Putting in physical form: 8H+(aq) +H3AsO4(aq) +4 Zn(s) =AsH AsH3(g) +4H2O(l) +4 Zn2+(aq)