When balancing redox equations, we often find it convenient to omit the spectator
ions (Section 4-3) so that we can focus on the oxidation and reduction processes. We use
the methods presented in this chapter to balance the net ionic equation. If necessary we
add the spectator ions and combine species to write the balanced formula unit equation.
Examples 11-15 and 11-16 illustrate this approach.EXAMPLE 11-15 Net Ionic Equations
Permanganate ions oxidize iron(II) to iron(III) in sulfuric acid solution. Permanganate ions are
reduced to manganese(II) ions. Write the balanced net ionic equation for this reaction.
Plan
We use the given information to write as much of the equation as possible. Then we follow
steps 2 through 6 in Section 11-5. The reaction occurs in H 2 SO 4 solution; we can add Hand
H 2 O as needed to balance H and O in the half-reactions (Step 3).
SolutionFe^2 MnO 4 88nFe^3 Mn^2
Fe^2 88nFe^3 (ox. half-reaction)
Fe^2 88nFe^3 1 e (balanced ox. half-reaction)MnO 4 88nMn^2 (red. half-reaction)
MnO 4 8H 88nMn^2 4H 2 O
MnO 4 8H 5 e 88nMn^2 4H 2 O (balanced red. half-reaction)The oxidation half-reaction involves one electron, and the reduction half-reaction involves
five electrons. Now we balance the electron transfer and then add the two equations term
by term. This gives the balanced net ionic equation.5(Fe^2 88nFe^3 1 e)
1(MnO 4 8H 5 e88nMn^2 4H 2 O)
5Fe^2 (aq)MnO 4 (aq)8H(aq)88n5Fe^3 (aq)Mn^2 (aq)4H 2 O()EXAMPLE 11-16 Total Ionic and Formula Unit Equations
Write the balanced total ionic and the formula unit equations for the reaction in Example
11-15, given that the reactants were KMnO 4 , FeSO 4 , and H 2 SO 4.
Plan
The Kis the cationic spectator ion, and the anionic spectator ion is SO 42 . The Fe^3 ion
will need to occur twice in the product Fe 2 (SO 4 ) 3 , so there must be an even number of Fe
atoms. So the net ionic equation is multiplied by two. It now becomes:10Fe^2 (aq)2MnO 4 (aq)16H(aq)88n10Fe^3 (aq)2Mn^2 (aq)8H 2 O()Based on the 10Fe^2 and the 16H, we add 18SO 42 to the reactant side of the equation; we
must also add them to the product side to keep the equation balanced. Based on the 2MnO 4 ,
we add 2Kto each side of the equation.418 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
In Section 4-3, we first wrote the
formula unit equation.We separated
any ionized or dissociated species into
ions to obtain the total ionic equation.
Then we eliminated the spectator ions
to obtain the net ionic equation.In
Examples 11-15 and 11-16, we reverse
the procedure.