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
1 The half-equation method
b) Balancing redox equation using half-equation methods
Rules for balancing redox equation using half-equation method
Solution
Step I : Write down the oxidation number of every atom present
Step II: Divide the equation into two half-equations (the oxidation half and the reduction half)
Step III: Balance each half-equation thus
a) Balance other atoms apart from O and H, then balance O and finally balance the 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).
Step IV: Cross-multiply each half-equation by the electron co-efficient of the other half-equation (to balance electron gain and loss)
Step 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 +5 –2 +1 +2 +4 –2
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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 as shown below
2 Rules for balancing complex ionic equation occurring in an acidic medium
Step I : Write down the oxidation number of every atom present
Step II: Divide the equation into two half-equations (the oxidation half and the reduction half)
Step III: Balance each half-equation thus
a) To the reduction half-equation, add hydrogen ions (H+) to the LHS and water H2O to the RHS of the equation. Now balance the atoms using appropriate coefficients
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).
Step IV: Cross-multiply each half-equation by the electron co-efficient of the other half-equation (to balance electron gain and loss)
Step V: Add the two half-equations to get the net-balanced equation; then include the state of matter.
Example
1. Balance the following ionic equations occurring in and acidic medium
