Ionic equations
Ionic
equations are equations that show only the oxidized and the reduced species.
For example, given the reaction below
1. Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
blue colourlesss
the
oxidation number of Zn changes from 0 to +2 (R.A) while the O.N of Cu
changes from +2 to 0 (oxidizing agent); but the O.N’s of the other
elements (S and O) remains unchanged and so will cancel out.
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
blue colourlesss
Ionic equations show only the oxidized and the reduced specie. So, the
ionic equation for the reaction above can be written as
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
2 The
reaction between potassium iodide and iron (III) tetraoxosulphate (VI) is
another example. It results in the formation of brown a mixture due to the
formation of iodine. Here, iodide ions are oxidized to iodine while iron (III)
ions are reduced to iron (II) ions.
Fe2(SO4)3(s) + KI(aq) → FeSO4(aq) + K2SO4(aq) + I2(l)
The
oxidation number of iodine changes from (–1) to 0 due to loss of electrons
while the Fe3+ gains electron to become Fe2+. There
is no change in the O.N of K
Ionically,
the above equation is written as
Fe3+(aq) + I-(aq)
→ Fe2+(aq) + I2(l)
Half
equations
Half-equations are equations
shows only the oxidation half or the reduction half equation. Using both
reactions discussed above that is,
Zn(s)
+ Cu2+(aq) →Zn2+(aq) + Cu(s)
R.
A O.A
Zn(s) → Zn2+ Oxidation half
Cu2+(aq) → Cu(s) Reduction half
Similarly
Fe3+(aq)
+ I-(aq) →
Fe2+(aq) + I2(l)
R.A O.A
We have
1 I-(aq) → I2(l) Oxidation half
2 Fe3+(aq) → Fe2+(aq) Reduction half
Balancing
redox equations
It is important to make sure that the number
of electrons lost by the reducing agent equals the number of electrons gained
by the oxidizing agent in redox reactions.
There are two main methods
that can be used to balance redox equations, they are
1.
The half-equation method and
2.
The oxidation number method and
1 The
half-equation method
b) Balancing
redox equation using half-equation methods
Rules
for balancing redox equation using half-equation method
Solution
i) Step
I
Write
down the oxidation number of every atom present
ii) Divide
the equation into two half-equations (the oxidation half and the reduction
half)
iii) Balance
each half-equation thus
a) Balance
the atoms other than O and H, then O … finally H
b) Balance
the charges by adding electrons to the left side for reduction half-equation (gain) and to the
right side for oxidation half-equation (loss).
iv) Cross-multiply
each half-equation by the electron co-efficient of the other half-equation (to
balance electron gain and loss)
v) Add
the two half-equations to get the net-balanced equation; then include the state
of matter.
Example:
Balance the following ionic equation using half-equation methods
i) Zn(s)
+ Cu2+(aq) →Zn2+(aq) + Cu(s)
ii) Br-(aq) + Cl2(g) → Cl–(aq) + Br2(l)
Solution
I Step
I
a) Write
out the O.N of every atom present
0 +2 +2 0
↑ ↑ ↑ ↑
Zn(s)
+ Cu2+(aq) →Zn2+(aq) + Cu(s)
b) Divide
the equation into 2 half-equations (oxidation-half and the reduction half)
Zn(s) → Zn2+ Oxidation half
Cu2+(aq) → Cu(s) Reduction half
c) Balance
the atoms and then the charges using appropriate coefficients
Zn → Zn2+ + 2e– [Add
electrons to the product side for oxidation]
Cu2+ +
2e– → Cu [ Add
electrons to the reactant side for reduction.]
d) Cross-multiply
the electron coefficient to balance the electron gain or loss.
Zn → Zn2+ + 2e–
Cu2+ +
2e– → Cu
e) Add
the two half-equations to get the net balanced equation.
Zn(s)
+ Cu2+(aq) →Zn2+(aq) + Cu(s)
II Example
2
Write
down the O.N of each atom or ion present
a) Br– + Cl2 → Cl– + Br2
b) Divide
the equation into two half-equations (oxidation half and the reduction half)
Br– → Br2 Oxidation
Cl2 + 2e- → Cl– Reduction
c) Balance
each half equation by suing appropriate number of atoms and number of
electrons.
Br– → Br2 + 2e
Cl2 + 2e → 2Cl–
d) Cross-multiply
the electron coefficient to balance the electron gain or loss.
2Br– → Br2 + 2e–
Cl2 + 2e– → 2Cl–
e) Add
the 2–half-equation to get the net balanced equation.
2Br-(aq) + Cl2(g) → 2Cl–(aq) + Br2(l)
The
oxidation number method
This
method involves or follows five steps to balance redox equation. These steps
are:
a) Assign
O.N to all the elements in the reaction
b) Identify
the oxidized species and the reduced specie from the O.N assigned.
c) Deduce
the number of electrons lost by the R.A in the oxidation and gained by the O.A
in the reduction (use lines to show the changes)
d) Multiply
these numbers by appropriate factors (numbers) to make the electron lost equal
the electron gained using these factors as balancing coefficients.
e) Inspect
the equation to complete the balancing assigning states be sure it is complete
hen add the states of matter.
b) ZnS(s) + O2(g) → ZnO + SO2(g)
Solution
i) Step
I
Assign
O.N for all the atoms present
+2 -2 0 +2 -2 +4 -2
↑ ↑ ↑ ↑
↑ ↑ ↑
ZnS(s) + O2(g) → ZnO + SO2(g)
ii) Step
II
Identify
the oxidized specie and the reduced specie
Loses
6 electrons
ZnS + O2 → ZnO + SO2
Gains
6e–
ZnS
is oxidized; the O.N of S increased from –2 in ZnS to +4 in SO2 (thereby
losing 6 electrons) and O.N of O decreased from 0(zero) in O2 to
–2 in ZnO and SO2.
iii) Step
III
Deduce
the electron gain and electron loss and indicate with connecting lines.
iv) Step
IV
Multiply
the electron gain and loss with co-efficient to make them equal. The sulphur
atom loses 6 electrons and each O atom in O2 gains 2 electrons,
as well as O in ZnO for a total of 6-electrons. If we put the coefficient in
front of O2 it will give 3-atoms of O on the right-hand side of
the equation; balancing the total number of electron gain and loss.
2ZnS + 3O2 → 2ZnO + SO2
v) Step
V
inspect
the balancing to see if it is complete. The atoms are balanced but our
coefficients must be whole numbers and so we multiply through by 2.
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
Check (Zn=2,
S = 2, O =
6) (Zn =
2, S = 2, O = 6)
Example
I
Balance
the following redox equations by oxidation number method
i) Cu(s) + HNO3 →Cu(NO3)2 + 2H2O + O2
ii) ZnS(s) + O2(g) → ZnO(s) + SO2(g)
Solution
a) Step
I
Assign
oxidation number to all elements in the reaction.
0 +1+5 –2
+2 +
–2
+1 +2 +4 –2
|
| | |
| | | | | | |
Cu + 2 H N O3 → Cu(NO3)2 + H2O + NO2
b) Step
II
Identify
the oxidized and the reduced species from the change in the oxidation numbers.
That is O.N of Cu increased from 0 to +2, so Cu was oxidized and the O.N of N
decreased from +5 (in HNO3) to +4 (in NO2); HNO3 was
reduced.
3 Step
III
Deduce
electron lost and electron gained using connecting lines between the atoms. 2
electrons were lost by Cu in the oxidation and 7 electron was gained by N in
the reduction.
loss
2e–
Cu + HNO3 →Cu(NO3)2 + H2O + NO2
gain
1e–
4 Step
IV
Multiply
each equation by electron coefficient to balance electron gain or loss. So we
will multiply the electron gain by the N by 2 and then put the coefficient of 2
in front of NO2 and HNO3.
Cu + 2HNO3 →Cu(NO3)2 + H2O + 2NO2
5 Step
V
Complete
the balancing (process) by inspection: The N has become 4 on the
right-hand side of the equation and so a 4 should be placed in front of HNO3.
Cu + 4HNO3 → Cu(NO3)2(q) + H2O(l) + 2NO2(g)
We
also place a 2 in front of H2O on the right-hand side of the
equation to balance the H; and then add the states of matter.
Cu(s) + 4HNO3 → Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)
Check
Reactants → PRODUCT
[Cu = 1, N = 4, O = 12, H = 4] [Cu = 1, N = 4, O = 12, H = 4]
Balancing complex ionic equations
To balance complex ionic equations; we take into consideration the reaction medium; where the reaction is occurring in an acidic medium or in an alkaline medium, and the half-equation method works best
1 To balance complex ionic equation occurring in acidic medium: The rules are the same as stated earlier for balancing redox reactions but for some few additions which will be made as we solve some exercises.
2
