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Balancing Redox Equations a. Use the following steps to balance the redox reaction...

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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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llltkl's profile pic

Posted (Answer #2)

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(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)

Adding,

 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)

Adding,

2MnO4–(aq) + 4H2O + 3NO2–(aq) + 6OH–(aq) + 6e-  → 3NO3–(aq) + 3H2O + 2MnO2(s) + 8OH–(aq) + 6e-

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

2MnO4–(aq) + H2O + 3NO2–(aq) + → 3NO3–(aq) + 2MnO2(s) + 2OH–(aq)

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karythcara's profile pic

Posted (Answer #1)

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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.

Sources:

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