Oxidation-reduction reactions
(redox reactions)
Redox reactions-reactions in which one or more electrons are transferred
Oxidation states-a concept that provides a way to keep track of electrons in oxidation-reduction reactions according to certain rules
Rules for finding oxidation states:
Oxidation state of elements=0 Ex. Mg=0
Oxidation state of monatomic ion=same as charge Ex. Cl- = -1
Oxidation state of fluorine in its compounds is always –1
Oxidation state of oxygen in covalent compounds is always –2, except in peroxide, when each oxygen is –1
Oxidation state of hydrogen is always +1 when combined with a nonmetal;when combined with a metal it is –1
Sum of oxidation states is 0 for electrically neutral compounds.
Ions-sum of oxidation states must equal the charge of the ion.
Find the oxidation states of the following.
NO3- = -1 KMnO4 = 0
N + 3(-2) = -1 1 + Mn + 4(-2) =0
N + -6 = -1 1 + Mn + -8 = 0
N = 5 -7 + Mn = 0
Mn = 7
Oxidation-increase in oxidation state(a loss of electrons)
Reduction-decrease in oxidation state(gain of electrons)
Oxidizing agent-electron acceptor
Reducing agent – electron donor
2Na + Cl2 ® 2NaCl
Na is the reducing agent and is oxidized
Cl2 is the oxidizing agent and is reduced
| Soluble | |
| All Nitrates, Acetates, Ammonium and Group I salts | |
| All Chlorides, Bromides, and Iodides, except Silver, Lead, and Mercury (I) | |
| All Fluorides except Group II, Lead (II), and Iron (III) | |
| All Sulfates except Calcium, Strontium, Barium, Mercury, Lead (II), and Silver | |
| Insoluble | |
| All Carbonates and Phosphates except Group I | |
| All Hydroxides except Group I, Strontium and Barium | |
| All Sulfides except Group I, II, and Ammonium | |
| All Oxides except Group I | |
| INSOLUBLE means a precipitate forms when equal volumes of 0.10 M solutions or greater are mixed | |
Half-reactions- one involving oxidation and the other involving reduction.
There are acidic and basic reactions.
1.write half reactions
2.find oxidation states
3.balance charge using electrons
4.balance all elements except hydrogen and oxygen
5.balance oxygen using H2O
6.balance hydrogen using H+
7.if necessary, multiply half-reactions by common multiple to make the number of electrons equal in both reaction.
8.add the half-reactions and cancel identical species on opposite sides
9.add what’s left for the final reaction
10.write the states of matter for each element or compound
Ex. Cu(s) + NO3 -(aq)®
Cu2+(aq) + NO(g)
Cu ® Cu2+ Cu=0 Cu2+=2
NO3- ® NO N + 3(-2) =-1 N +-2 = 0
N + -6 = -1 N = 2
N = 5
Cu ® Cu2+ + 2e-
3e- + NO3- ® NO
3e- + NO3- ® NO + 2H2O
4H+ + 3e- + NO3- ® NO + 2H2O
3 (Cu ® Cu2+ + 2e- )
2 (4H+ + 3e- + NO3- ® NO + 2H2O)
3Cu ® 3Cu2+ + 6e-
+ 8H+
+ 6e- +
2NO3- ®
2NO + 4H2O
8H+
+ 3Cu + 2NO3-
+ 6e- ®
3Cu2+ +2NO + 4H2O + 6e-
8H+(aq) + 3Cu(s) + 2NO3- (aq)® 3Cu2+(aq) + 2NO(g) + 4H2O(l)
1.do the same as is done for acidic through step 8
2.then add the correct number of OH- to each side to cancel the H+ and make H2O
3.cancel waters on opposite sides if necessary
4.add together any like species on the same side for final reaction
Ex. IO3-(aq) + H2S(g) ® I2(g) + SO32-(aq)
IO3- ® I2 I + 3(-2)=-1 I = 0
I + -6 = -1
I = 5
H2S ® SO32- 2(1) + S = 0 S + 3(-2) = -2
2 + S = 0 S + -6 = -2
S = -2 S = 4
2IO3- + 10e- ® I2 ( add 10 electrons because for one molecule of Iodine it would take 5 electrons to balance the charge, so for 2 molecules it takes 10)
H2S ® SO32- + 6e-
2IO3- + 10e- ® I2 + 6H2O
H2S + 3H2O ® SO32- + 6e-
2IO3- + 10e- + 12H+ ® I2 + 6H2O
H2S + 3H2O ® SO32- + 6e- + 8H+
3 ( 2IO3- + 10e- + 12H+ ® I2 + 6H2O )
5 (H2S + 3H2O ® SO32- + 6e- + 8H+ )
6IO3- + 30e- + 36H+ ® 3I2 + 18H2O
+ 5H2S + 15H2O ® 5SO32- + 30e- + 40H+
3H2O 4H+
6IO3-
+ 30e- + 36H+ + 5H2S + 15H2O
®
3I2 + 18H2O + 5SO32- + 30e-
+ 40H+
4H2O
4OH-
+ 6IO3- + 5H2S
® 3I2 + 3H2O + 5SO32-
+ 4H+ + 4OH-
4OH-(aq) + 6IO3-(aq) + 5H2S(g) ® 3I2(g) + 7H2O(l) + 5SO32- (aq)