Unit 3
Chapter 1: Electrochemistry
Exercises page 85
11234-
d
c
b
a
1234-
False, gained
True
True
False, a more, cathodic
2-
31- a- SO3 + H2O → H2SO4
+6 -2
+1 -2
+1 +6 -2
It’s not a redox reaction since there’s no transfer of electrons.
b- I2O5 + CO → I2 + CO2
+5 +2 +2 -2
Reduction
0 +4 -2
oxidation
It’s a redox reaction since there’s transfer of electrons.
2- a- When electrons are exchanged between the metal (Zn) which is in direct contact with
the cation Cu2+ in aqueous solution CuCl2 , energy is released in the form of heat
(exothermic reaction).
b- deposit of Cu on Zn strip
color of CuCl2 solution will be fainter
c- Zn strip formed oxidation i.e. lost 2 electrons to become Zn2+ ion. Cu2+ ions found in
CuCl2 solution gain the 2 electrons and form reduction to become Cu deposit. Since the
concentration of Cu2+ ions in the solution is decreasing, the color of the solution will
become fainter.
3- Since Mg has more tendency to lose electrons than Cu, then Mg is the anode and Cu is
the cathode.
Anode half reaction: Mg → Mg2+ + 2eCathode half reaction: Cu2+ + 2e- → Cu
4- N2 : free element → o.n. = 0
NH3 : H is in group 1 → o.n = +1
o.n(N) + 3o.n(H) = 0
o.n(N) = -3
(summation of oxidation numbers in a compound is zero)
N2O4 : O is in group 2 → o.n = -2
2o.n(N) + 4o.n(-2) = 0
o.n(N) = 4
(summation of oxidation numbers in a compound is zero)
N2O : o.n(N) = +1
(same reason as N2O4)
N2H4 : o.n(N) = -2
(same as NH3)
NO3- : o.n(N) + 3o.n(O) = -1
o.n(N) = +5 (summation of o.n in a polyatomic ion is equal to its charge)
Decreasing order of oxidation number of nitrogen atom:
NO3- > N2O4 > N2O > N2 > N2H4 > NH3
5- a-
b- The metal to be plated should be the cathode since the ions of the plating metal
found in the solution are attracted to the cathode, where they pick up electrons and are
deposited on the object.
6- a-
H2 is the oxidized element since it performed oxidation.
N2 is the reduced element since it performed reduction.
7-
Oxidizing agent is the reactant that gained electrons (formed reduction) which is H 2O2.
Reducing agent is the reactant that lost electrons (formed oxidation) which is I-.
8- FeO : O is in group 6 → o.n. = -2
o.n(Fe) + o.n(O) = 0
o.n(Fe) = +2
Fe : 0
CO : o.n(O) = -2 and o.n(C) = +2
CO2 : o.n(C) + 2o.n(O) = 0 → o.n(C) = +4
The substance that is oxidized is CO since it performed oxidation.
The substance that is reduced is FeO since it performed reduction.
The oxidizing agent is FeO (ferrous oxide)
The reducing agent is CO (carbon monoxide)
9-
a- Anode half reaction : Mg → Mg2+ + 2eCathode half reaction: Cu2+ + 2e- → Cu
b- The electrons flow from the anode (Mg) to the cathode (Cu).
c- The anions in the salt bridge move toward the anode.
10- The spoon would be the cathode since it is the object to be plated.
iThe other electrode is silver.
iiThe electrolyte is solution containing silver ions.
iiiMaterials:
Electrolytic cell
D.C. power supply
Connecting wires
Anode half reaction: Ag → Ag+ + 1eCathode half reaction: Ag+ + 1e- → Ag
11-
a- Mg is oxidized since it performed oxidation and lost electrons
b- Co2+ is reduced since it performed reduction and gained electrons.
c- The reducing agent is Mg since it performed oxidation.
d- The oxidizing agent is Co2+ since it performed reduction.
12- Oxidation half reaction: Al → Al3+ + 3eReduction half reaction: 2H+ + 2e- → H2
The number of electrons lost should be equal to the number of electrons gained.
2Al → 2Al3+ + 3e6H+ + 6e- → 3H2
Redox reaction: 2Al + 6H+ → 2Al3+ + 3H2
13-
14-
a- Ag is oxidized since it performed oxidation.
b- O2 is the oxidizing agent since it performed reduction.
15-
a- Anode half reaction: Ni → Ni2+ + 2eCathode half reaction: Cu2+ + 2e- → Cu
b- Overall reaction: Ni + Cu2+ → Ni2+ + Cu
Electrons flow from the anode to the cathode.