Download Worksheet25 Redox Reduction and more Study Guides, Projects, Research Chemistry in PDF only on Docsity!
Worksheet 25 - Oxidation/Reduction Reactions
Oxidation number rules:
Elements have an oxidation number of 0 Group I and II – In addition to the elemental oxidation state of 0, Group I has an oxidation state of +1 and Group II has an oxidation state of +2. Hydrogen –usually +1, except when bonded to Group I or Group II, when it forms hydrides, -1. Oxygen – usually -2, except when it forms a O-O single bond, a peroxide, when it is -1. Fluorine is always -1. Other halogens are usually -1, except when bonded to O.
- Assign oxidation numbers to each of the atoms in the following compounds:
Na 2 CrO 4 Na = +1 O = -2 Cr = +
K 2 Cr 2 O 7 K = +1 O = -2 Cr = +
CO 2 O = -2 C = +
CH 4 H = +1 C = -
HClO 4 O = -2 H = +1 Cl = +
MnO 2 O = -2 Mn = +
SO 3 2-^ O = -2 S = +
SF 4 F = -1 S = +
a. What is the range of oxidation states for carbon? -4 to +
b. Which compound has C in a +4 state? CO (^2)
c. Which compound has C in a -4 state? CH 4
- Nitrogen has 5 valence electrons (Group V). It can gain up to 3 electrons (-3), or lose up to 5 (+5) electrons. Fill in the missing names or formulas and assign an oxidation state to each of the following nitrogen containing compounds:
- During chemical reactions, the oxidation state of atoms can change. This occurs when compounds gain or lose electrons, or when the bonds to an atom change. This is illustrated by the reaction between nitrogen and hydrogen to make ammonia: N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)
a. Assign oxidation numbers to each of the atoms in this reaction.
N (in N 2 ) = 0 (in NH 3 ) = -
H (in H 2 ) = 0 H (in NH 3 ) = +
When an oxidation number increases, that species has been oxidized.
b. Which reactant undergoes an increase in its oxidation number? H 2
When an oxidation number decreases, that species has been reduced.
c. Which reactant undergoes a decrease in its oxidation number? N 2
name formula oxidation state of N
ammonia NH 3 -
nitrogen N 2 0
nitrite NO 2 -^ +
nitrate NO 3 -^ +
dinitrogen monoxide N 2 O +
nitrogen dioxide NO 2 +
hydroxylamine NH 2 OH -
nitrogen monoxide (^) NO +
hydrazine N 2 H 4 -
Balancing Redox Reactions
Oxidation/Reduction (Redox) reactions can be balanced using the oxidation state changes, as seen in the previous example. However, there is an easier method, which involves breaking a redox reaction into two half- reactions. This is best shown by working an example.
Hydrobromic acid will react with permanganate to form elemental bromine and the manganese(II) ion. The unbalanced, net reaction is shown below,
Br -^ + MnO 4 -^ → Br 2 + Mn2+
- Break this into two half-reactions, one involving bromine and the other involving manganese.
Bromine half-reaction Manganese half-reaction
Br -^ → Br 2 MnO 4 -^ → Mn 2+
- First balance the bromine half-reaction first.
a. Balance the bromine atoms of the reaction
_ 2 __ Br -^ → _ 1 __Br (^2)
b. Now balance charge by adding electrons (e-)
_ 2 __ Br -^ → _ 1 __Br 2 + 2e-
This half-reaction is producing/consuming electrons. This is an
oxidation/reduction half-reaction. Confirm this by assigning oxidation
numbers to the bromine species. Br -^ = -1 Br 2 = 0 (oxidation state increases; oxidation)
- Next, balance the manganese half-reaction.
a. Balance the manganese atoms of the half-reaction
_ 1 __ MnO 4 -^ → _ 1 __ Mn2+
b. Next, balance oxygen by adding water molecules (H 2 O)
_ 1 __ MnO 4 -^ → __ 1 _ Mn2+^ + 4H 2 O
c. Next, balance hydrogen by adding protons (H+)
8H+^ + _ 1 __ MnO 4 -^ → _ 1 __ Mn2+^ + 4H 2 O
d. Finally, balance charge by adding electrons (e-).
5e-^ + 8H+^ + _ 1 __ MnO 4 -^ → _ 1 __ Mn2+^ + 4H 2 O
This half-reaction is producing/consuming electrons. This is a
oxidation/reduction half-reaction. Confirm this by assigning oxidation
numbers to the manganese atoms. MnO 4 -^ = +7 Mn2+^ = +2 +7 Æ +2 is a reduction Notice that the number of electrons equals the change in oxidation number.
- Now put the two half-reactions together. The number of electrons produced must equal the number of electrons consumed.
2 Br -^ → Br 2 + 2e-^ 5e-^ + 8H+^ + MnO 4 -^ → Mn 2+^ + 4H 2 O
multiply this half-reaction by _ 5 ___ multiply this half-reaction by _ 2 __
_ 10 _Br -^ → _ 5 _Br 2 + _ 10 e-^ _ 10 _e-^ + _ 16 _H+^ + _ 2 _MnO 4 -^ → _ 2 _Mn2+^ + _ 8 _H 2 O
Add the two half-reactions, canceling out species that appear on both sides (including electrons) 10 e-^ appear on both sides. Cancel those out to obtain
_ 10 __Br -^ + _ 16 __H+^ + _ 2 __MnO 4 -^ → _ 5 __Br 2 + _ 2 __Mn2+^ + _ 8 __H 2 O
Which compound is the oxidizing agent? MnO 4 -^ (gets reduced)
Which compound is the reducing agent? Br -^ (gets oxidized)
Notice that there are protons (H+) present in the reactants. This indicates that the reaction is carried out in an acidic solution. To carry this out in a basic solution, simply add enough hydroxide ions (OH-) to each side of the equation to neutralize the protons. The product of the neutralization reaction will be water. Add 16 OH-^ to each side. On the left side, 16 H+^ + 16 OH-^ = 16H 2 O The overall balanced reaction under basic conditions is:
_ 10 Br -^ + _ 2 _MnO 4 -^ + _ 16 _H 2 O → _ 5 _Br 2 + _ 2 _Mn2+^ + _ 8 _OH-^ + 8 H 2 O
_ 10 Br -^ + _ 2 _MnO 4 -^ + _ 8 _H 2 O → _ 5 _Br 2 + _ 2 _Mn2+^ + _ 8 _OH-
Balance the following redox-reaction which takes place in basic solutions. 0 +3 -2 -3 +1 +2 -2 + Zn (s) + NO 2 -^ → NH 3 + Zn(OH) 4 2-
4H 2 O + Zn Æ Zn(OH) 4 -2^ + 4H+^ + 2e-^ 7H+^ + 6e-^ + NO 2 -^ Æ NH 3 + 2H 2 O (multiply by 3) (multiply by 1)
Combine the two equations to get… 12H 2 O + 3Zn + 7H+^ + 6e-^ + NO 2 -^ Æ 3Zn(OH) 4 -2^ + 12H +^ + 6e-^ + NH 3 + 2H 2 O 10H 2 O + 3Zn + NO 2 -^ Æ 3Zn(OH) 4 -2^ + 5H+^ + NH 3
Add 5 OH-^ to each side (b/c there are currently 5H+^ on the right-hand side) to create basic conditions… 10H 2 O + 3Zn + NO 2 -^ + 5OH-^ Æ 3Zn(OH) 4 -2^ + 5H+^ + NH 3 + 5OH- 10H 2 O + 3Zn + NO 2 -^ + 5OH-^ Æ 3Zn(OH) 4 -2^ + 5H 2 O + NH 3
Simplify the expression by eliminating extra water: 5H 2 O + 3Zn + NO 2 -^ + 5OH-^ Æ 3Zn(OH) 4 -2^ + NH 3
