1. Determine the oxidation number of the indicated atom in each of the following compounds or ions.
(a) each iron atom in Fe2O3
The oxidation number of each oxygen atom is – 2.
The total for all the oxygen atoms is – 6.
Thus, the total for both iron atoms is +6.
Therefore, the oxidation number of each iron atom is +3.
(b) sulfur in H2SO4
The oxidation number of each hydrogen atom is +1.
The total for both hydrogen atoms is +2.
The oxidation number of each oxygen atom is – 2.
The total for all the oxygen atoms is – 8.
Therefore, the oxidation number of sulfur is +6.
(c) nitrogen in NO2+
The oxidation number of each oxygen atom is – 2.
The total for both oxygen atoms is – 4.
Since this is an ion with a charge of 1+ the oxidation number of nitrogen is +5.
(d) each hydrogen atom in CaH2
Since calcium in a compound always has an oxidation number of +2, the oxidation number of each hydrogen atom is – 1.
2. Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that
undergo changes in oxidation number.
(a) PBr3(l) + 3 H2O(l) ® H3PO3(aq) + 3 HBr(aq)
Reactants Products
P Þ +3 P Þ +3
Br Þ – 1 H Þ +1
H Þ +1 O Þ – 2
O Þ – 2 Br Þ – 1
Since the oxidation numbers are unchanged in the products compared to the reactants this reaction is not an oxidation-reduction reaction.
(b) 3 SO2(g) + 2 HNO3(aq) + 2 H2O(l) ® 3 H2SO4(aq) + 2 NO(g)
Reactants Products
S Þ +4 H Þ +1
O Þ – 2 S Þ +6
H Þ +1 O Þ – 2
N Þ +5 N Þ +2
This is an oxidation-reduction reaction with S being oxidized (+4 to +6) and N being reduced (+5 to +2).
3. Determine oxidation numbers for each type of atom found in the reactants and products in the following oxidation-
reduction reactions. For each reaction indicate which atom is oxidized and which atom is reduced.
(a) Zn(s) + 2 NO3 – (aq) + 4 H+ (aq) ® Zn2+(aq) + 2 NO2 (g) + 2 H2O(l)
0 N Þ +5 +1 +2 N Þ +4 H Þ +1
O Þ – 2 O Þ – 2 O Þ – 2
Zinc is oxidized since its oxidation number increases from 0 to +2.
Nitrogen is reduced since its oxidation number decreases from +5 to +4.
(b) Ca(s) + 2 H2O(l) ® Ca(OH)2(s) + H2(g)
0 H Þ +1 Ca Þ +2 0
O Þ – 2 O Þ – 2
H Þ +1
Calcium is oxidized since its oxidation number increases from 0 to +2.
Hydrogen is reduced since its oxidation number decreases from +1 to 0 (in H2).
4. Complete and balance the following equations (in acidic solution), and identify the oxidizing and reducing agents.
(a) Cr2O72- (aq) + I – (aq) ® Cr3+(aq) + IO3 – (aq)
Reduction half-reaction:
Cr2O7 2- + 14 H+ + 6 e – ® 2 Cr 3+ + 7 H2O
Oxidation half-reaction:
I – + 3 H2O ® IO3 – + 6 H+ + 6 e –
Add the two half-reactions together to obtain the balanced chemical equation.
Cr2O7 2- + 8 H+ + I – ® 2 Cr 3+ + IO3 – + 4 H2O
Oxidizing agent Þ Cr2O7 2- Reducing agent Þ I –
(b) I2(s) + OCl – (aq) ® IO3 – (aq) + Cl – (aq)
Reduction half-reaction:
OCl – + 2 H+ + 2 e – ® Cl – + H2O
Oxidation half-reaction:
I2 + 6 H2O ® 2 IO3 – + 12 H+ + 10 e –
The reduction half-reaction has to be multiplied through by 5 in order to have the same number of electrons in each half-reaction.
5 OCl – + 10 H+ + 10 e – ® 5 Cl – + 5 H2O
Now we can add the two half-reactions together to obtain the balanced chemical equation.
5 OCl – + I2 + H2O ® 2 IO3 – + 5 Cl – + 2 H+
Oxidizing agent Þ OCl – Reducing agent Þ I2
5. Complete and balance the following equations (in basic solution), and identify the oxidizing and reducing agents.
Recall that oxygen atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.
(a) MnO4 – (aq) + Br – (aq) ® MnO2(s) + BrO3 – (aq)
Reduction half-reaction:
MnO4 – + 4 H+ + 3 e – ® MnO2 + 2 H2O
Oxidation half-reaction:
Br – + 3 H2O ® BrO3 – + 6 H+ + 6 e –
The reduction half-reaction has to be multiplied through by 2 in order to have the same number of electrons in each half-reaction.
2 MnO4 – + 8 H+ + 6 e – ® 2 MnO2 + 4 H2O
Now we can add the two half-reactions together to obtain the balanced chemical equation.
2 MnO4 – + Br – + 2 H+ ® 2 MnO2 + BrO3 – + H2O
To balance this equation in basic solution we need to add 2 OH – (aq) ions to both sides of the chemical equation.
2 MnO4 – + Br – + 2 H+ + 2 OH – ® 2 MnO2 + BrO3 – + H2O + 2 OH –
2 MnO4 – + Br – + 2 H2O ® 2 MnO2 + BrO3 – + H2O + 2 OH –
After the removal of excess water molecules we obtain the balanced
chemical equation.
2 MnO4 – + Br – + H2O ® 2 MnO2 + BrO3 – + 2 OH –
Oxidizing agent Þ MnO4 – Reducing agent Þ Br –
(b) H2O2(aq) + ClO2(aq) ® ClO2 – (aq) + O2(g)
Reduction half-reaction
ClO2 + 1 e – ® ClO2 –
Oxidation half-reaction:
H2O2 ® O2 + 2 H+ + 2 e –
The reduction half-reaction has to be multiplied through by 2 in order to have the same number of electrons in each half-reaction.
2 ClO2 + 2 e – ® 2 ClO2 –
Now we can add the two half-reactions together to obtain the balanced chemical equation.
H2O2 + 2 ClO2 ® 2 ClO2 – + O2 + 2 H+
To balance this equation in basic solution we need to add 2 OH – (aq) ions to both sides of the chemical equation.
H2O2 + 2 ClO2 + 2 OH – ® 2 ClO2 – + O2 + 2 H+ + 2 OH –
The balanced chemical equation is given below.
H2O2 + 2 ClO2 + 2 OH – ® 2 ClO2 – + O2 + 2 H2O
Oxidizing agent Þ ClO2 Reducing agent Þ H2O2
6. For each of the following redox reactions predict the direction (toward the products side or toward the reactants side)
that it proceeds under standard conditions.
Remember that redox reactions occur in the direction of converting the stronger of a pair of oxidizing agents and the stronger of a pair of reducing agents into a weaker oxidizing agent and a weaker reducing agent. You need to first identify the oxidizing agent and reducing agents on both sides of the chemical equation and then determine the stronger oxidizing agent and stronger reducing agent.
(a) Ni(s) + Zn 2+(aq) ® Ni 2+(aq) + Zn(s)
Ni(s) Þ reducing agent Zn(s) Þ reducing agent
Zn 2+(aq) Þ oxidizing agent Ni 2+(aq) Þ oxidizing agent
Ni 2+(aq) is a stronger oxidizing agent than Zn 2+(aq).
Zn(s) is a stronger reducing agent than Ni(s).
Therefore, this reaction will proceed toward the reactants side.
(b) Al(s) + 3 Ag +(aq) ® Al 3+(aq) + 3 Ag(s)
Al(s) Þ reducing agent Ag(s) Þ reducing agent
Ag +(aq) Þ oxidizing agent Al 3+(aq) Þ oxidizing agent
Ag +(aq) is a stronger oxidizing agent than Al 3+(aq).
Al(s) is a stronger reducing agent than Ag(s).
Therefore, this reaction will proceed toward the products side.
7. Consider the following voltaic cell:
(a) Determine the direction of electron flow and label the anode and cathode.
See the diagram below.
(b) Write a balanced chemical equation for the overall chemical reaction.
Ni 2+ (aq) + Mg(s) ® Ni(s) + Mg 2+ (aq)
(c) Label each electrode as negative or positive.
See the diagram below.
(d) Indicate the direction of anion and cation flow in the salt bridge.
See the diagram below.
