# Balancing Redox Equations a. Use the following steps to balance the redox reaction below. Al(s) + Ni2+(aq)  Ni(s) + Al3+(aq)       i. Write the oxidation and reduction half-reactions. Make...

Balancing Redox Equations

a. Use the following steps to balance the redox reaction below.

Al(s) + Ni2+(aq)  Ni(s) + Al3+(aq)
i. Write the oxidation and reduction half-reactions. Make sure each half-reaction is balanced for number of atoms and charge.

ii. Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions.

iii. Add the equations and simplify to get a balanced equation.

b. The following reaction takes place in a basic solution.

MnO4–(aq) + NO2–(aq)  MnO2(s) + NO3–(aq)
The half-reactions (balanced only for atoms) are the following:

MnO4– + 2H2O  MnO2 + 4OH–

NO2– + 2OH–  NO3– + H2O
Use the following steps to finish balancing the equation.

i. Balance each half-reaction for charge.

ii. Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions.

iii. Add the equations and simplify to get a balanced equation.

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Karyth Cara | Certified Educator

Redox balancing is used between acids. It can be easy when the rules are followed exactly.

The first rule for redox balancing is to split the equation in two hal-equations. Balance these on half at a time.

The second is: do not balance for O or H until all others are balanced.

The third is to balance Os from the H2Os.

See the link below for the other rules.

llltkl | Student

(a) The overall reaction is Al(s) + Ni2+(aq) →  Ni(s) + Al3+(aq)

The oxidation half-reaction is: Al(s) → Al3+(aq) + 3e-

And the reduction half-reaction is: Ni2+(aq) + 2e- → Ni(s)

In order to balance the charges, the two half-reactions have to be cross multiplied by the number of electrons of each other. Thus, we get

The oxidation half-reaction is: Al(s) → Al3+(aq) + 3e- × 2

After multiplication, for balancing charge,

2Al(s) → 2Al3+(aq) + 6e-

And the reduction half-reaction is: Ni2+(aq) + 2e- → Ni(s) × 3

After multiplication, for balancing charge,

3Ni2+(aq) + 6e- → 3Ni(s)

2Al(s) +3Ni2+(aq) + 6e-  → 2Al3+(aq) + 3Ni(s)  + 6e-

Simplifying by cancelling equal terms from both sides, we get the balanced equation for the overall chemical (redox) reaction as:

2Al(s) +3Ni2+(aq) → 2Al3+(aq) + 3Ni(s)

(b) The overall reaction is:

MnO4–(aq) + NO2–(aq) → MnO2(s) + NO3–(aq)

The reaction occurs in a basic solution. Hence oxygen/hydrogen balance has to be done by addition of the species OH-/H2O, to a suitable side.

The oxidation half-reaction is: NO2–(aq) + 2OH–  → NO3–(aq) + H2O

And the reduction half-reaction is: MnO4–(aq) + 2H2O +3e- → MnO2(s) + 4OH–(aq)

In order to balance the charges, the two half-reactions have to be cross multiplied by the number of electrons of each other. Thus, we get

The oxidation half-reaction, after atom balance, is:

NO2–(aq) + 2OH–  → NO3–(aq) + H2O + 2e- × 3

After multiplication, for balancing charge,

3NO2–(aq) + 6OH–(aq)  → 3NO3–(aq) + 3H2O + 6e-

And the reduction half-reaction is:

MnO4–(aq) + 2H2O +3e- → MnO2(s) + 4OH–(aq) × 2

After multiplication, for balancing charge,

2MnO4–(aq) + 4H2O +6e- → 2MnO2(s) + 8OH–(aq)