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Worksheet 1 - 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 +. Hydrogen –usually +1 , except when bonded to Group I or Group II, when it forms hydrides, -. Oxygen – usually -2 , except when it forms a O-O single bond, a peroxide, when it is -. Fluorine is always -1. Other halogens are usually -1, except when bonded to O or F.
- Assign oxidation numbers to each of the atoms in the following compounds:
Na 2 CrO 4 Na = O = Cr =
K 2 Cr 2 O 7 K = O = Cr =
CO 2 O = C =
CH 4 H = C =
HClO 4 O = H = Cl =
MnO 2 O = Mn =
SO 3 2-^ O = S =
SF 4 F = S =
a. What is the range of oxidation states for carbon?
b. Which compound has C in a +4 state?
c. Which compound has C in a -4 state?
2. 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 ) = N (in NH 3 ) =
H (in H 2 ) = H (in NH 3 ) =
When an oxidation number increases, that species has been oxidized.
b. Which reactant undergoes an increase in its oxidation number?
When an oxidation number decreases, that species has been reduced.
c. Which reactant undergoes a decrease in its oxidation number?
name formula oxidation state of N
NH 3
nitrogen
nitrite
NO 3 -
dinitrogen monoxide
NO (^2)
hydroxylamine NH 2 OH
nitrogen monoxide
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
___ Br -^ ___Br 2
b. Now balance charge by adding electrons ( e- )
___ Br -^ ___Br 2
This half-reaction is producing / consuming electrons. This is an oxidation / reduction half-reaction. Confirm this by assigning oxidation numbers to the bromine species.
- Next, balance the manganese half-reaction.
a. Balance the manganese atoms of the half-reaction
___ MnO 4 -^ ___ Mn2+
b. Next, balance oxygen by adding water molecules ( H 2 O )
___ MnO 4 -^ ___ Mn2+
c. Next, balance hydrogen by adding protons ( H+ )
___ MnO 4 -^ ___ Mn2+
d. Finally, balance charge by adding electrons ( e- ).
___ MnO 4 -^ ___ Mn2+
This half-reaction is producing / consuming electrons. This is a oxidation / reduction half-reaction. Confirm this by assigning oxidation numbers to the manganese atoms.
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 ____ multiply this half-reaction by ___
___Br -^ ___Br 2 + ___e-^ ___e-^ + ___H+^ + ___MnO 4 -^ ___Mn2+^ + ___H 2 O
Add the two half-reactions, canceling out species that appear on both sides (including electrons)
___Br -^ + ___H +^ + ___MnO 4 -^ ___Br 2 + ___Mn2+^ + ___H 2 O
Which compound is the oxidizing agent?
Which compound is the reducing agent?
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.
Balance the following redox-reaction which takes place in basic solutions.
Zn (s) + NO 2 -^ NH 3 + Zn(OH) 4 2-
