Balance Reduction and Oxidation (Redox) Reactions

• Balance reduction-oxidation (Redox) equations till you have developed a logical method without having to memorize all the steps needed to balance redox equations.

  Balance Oxidation and Reduction Reaction equations

  You should have studied the two modules or pages Oxidation states and Half reactions before attempting this module. The steps for balancing Redox equations are:
  1. Identify the elements that are oxidized and reduced from their oxidation states. See Oxidation States.
  2. Write the oxidation and reduction half-reactions and balance them. See Half Reactions.
  3. Add the two half-reactions algebraically such that the electrons in the two half-reaction equations cancel completely. Cancel other species such as H⁺, OH^-, and H₂O common to the two sides, if necessary.
  4. Check your equation and make certain that numbers of atoms and charge are equal on both sides.

  Balance the equation from the following two half-reactions:

  Cr₂O₇^2- + 14 H⁺ + 6 e^- --> 2 Cr³⁺ + 7 H₂O
  H₂C₂O₄ --> 2 CO₂ + 2 H⁺ + 2 e^-

  Multiply the second equation by 3 and then add them algebraically so that the electrons in the two half-reaction equations cancel completely.

  Cr₂O₇^2- + 14 H⁺ + 6 e^- --> 2 Cr³⁺ + 7 H₂O
  3 H₂C₂O₄ --> 6 CO₂ + 6 H⁺ + 6 e^-
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  Cr₂O₇^2- + 8 H⁺ + 3 H₂C₂O₄ --> 2 Cr³⁺ + 7 H₂O + 6 CO₂

  Make sure you check your balanced equation.

  Balance the equation from the two half-reactions:

  Cd --> Cd²⁺ + 2 e^-
  4 H⁺ + NO₃^- + 3 e^- --> NO + 2 H₂O

  The first half-reaction has 2 electrons, whereas the second one has 3. The lowest common multiple of 2 and 3 is 6. Thus, you multiply the first equation by 3 and the second one by 2. The half-reaction equations become

  3 Cd --> 3 Cd²⁺ + 6 e^-
  8 H⁺ + 2 NO₃^- + 6 e^- --> 2 NO + 4 H₂O

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  3 Cd + 8 H⁺ + 2 NO₃^- --> 3 Cd²⁺ + 2 NO + 4 H₂O

  Since the last step in balancing oxidation and reduction reactions is simple, it will be more productive simply to practice the determination of oxidation states and balancing of half-reactions again in the dialogue.

  In a basic solution, Fe(OH)₂ and Fe(OH)₃ are solids. The former may be oxidized by H₂O₂.

  Fe(OH)₂ + H₂O₂ --> Fe(OH)₃ + H₂O.

  Balance this equation.

  The balanced half-reactions are:

  Fe(OH)₂ --> Fe(OH)₃ + e^-
  H₂O₂ + 2 e^- --> 2 OH^-

  Thus, the balanced equation is

  2 Fe(OH)₂ + H₂O₂ --> 2 Fe(OH)₃

  Note that the balanced equation does not have an H₂O in it.

  Let us balance the following reaction, which is carried out in an acidic solution:

  I^- + IO₃^- --> I₂

  The half-reactions are:

  2 I^- --> I₂ + 2 e^- (oxidized)
  2 IO₃^- + 10 e^- --> I₂ (reduced)

  The balanced equation is

  10 I^- + 2 IO₃^- --> 6 I₂

  You may study Electrochemistry to gain better insight to the oxidation reduction process.

  Confidence Building Questions

  • How many electrons should there be in the half-reaction?
    Al_(s) --> Al³⁺ + ? e^-

    If you get... lost

    Try... Al³⁺ has 3 positive charges, and each electron has 1 negative charge.

    If you get... 3

    Excellent... Al is a group III element.

  • How many electrons should there be in the half-reaction.
    2 H₂O₂ --> 2 H₂O + O₂ + ? e^-
    This is one of three modes of reaction for H₂O₂.

    If you get... 2

    Try... Both O atoms in H₂O₂ change oxidation states in this reaction.

    If you get... lost

    Try... Neither side of the equation requires the presence of ionic species.

    If you get... 0

    Excellent... This is an example of disproportionation. In which the change of oxidation state of one atom is compensated by the change of oxidation state of another atom.
    2 H₂O₂ --> 2 H₂O + O₂

  • How many electrons should there be in the half reaction
    H₂O₂ --> 2 H⁺ + O₂ + ? e^-
    Note that H₂O₂ is being oxidized in this reaction.

    If you get... 0

    Try... There is a change of oxidation state for O in this reaction.

    If you get... lost

    Try... There are 2 positive charges on the right-hand of the equation.

    If you get... 2

    Excellent... In this reaction, H₂O₂ is a reducing agent.

  • How many electrons should there be in the half reaction.
    H₂O₂ + ? e^- --> 2 OH^-
    Note thatH₂O₂ is being reduced in this reaction.

    If you get... 0

    Try... The oxidation state of O atoms changes from -1 to -2.

    If you get... 1

    Try... There are two O atoms whose oxidation states have changed.

    If you get... lost

    Try... There are 2 negative charges on the right-hand-side of the equation.

    If you get... 2

    Excellent... The oxidation states of 2 O atoms change from -1 to -2.

  • How many electrons should there be in the half reaction
    S₈ + ? e^- --> 8 S^2-?

    If you get... 2

    Try... There are 8 S atoms in the equation.

    If you get... lost

    Try... Eight S atoms changed from 0 to -2.

    If you get... 16

    Excellent... 8 * (-2) = -16.

  • What is the oxidation state of In in In(OH)₃?

    If you get... 3

    Try... Oxidation state is a signed number. Our convention is '+3'.

    If you get... lost

    Try... Remember the OH^- ion?

    If you get... +3

    Excellent... Read OH as OH^-.

  • What is the oxidation state of Fe in FeAsO₄?

    If you get... 3

    Try... Use a signed number '+3'.

    If you get... lost

    Try... Since As belongs to group V A, it has an oxidation state of +5. The common oxidation states of Fe are +2 and +3.

    If you get... +3

    Excellent... The common oxidation states for Fe are +2 and +3.

  • In the reaction
    2 Cr³⁺ + H₂O + 6 ClO₃^- --> Cr₂O₇^2- + 6 ClO₂ + 2 H⁺,
    identify the reducing agent.

  If you get... Cr

  Try... Please be specific, Cr³⁺ ion is the reducing agent.

  If you get... lost

  Try... A reducing agent is the one that has been oxidized.

  If you get... Cr³⁺

  Excellent... Since Cr³⁺ is oxidized, it is a reducing agent.

  • Identify the reducing agent in the reaction
    Cr₂O₇^2- + H₂C=O --> HCOOH + Cr³⁺.
    The formula H₂C=O with a double bond between C and O is an aldehyde, and it may also be written as H₂CO. HCOOH is an acid. Write it formally as H₂CO₂ to determine the oxidation state of C.

    If you get... H₂CO

    Try... Yes, this is equal to H₂C=O. But now enter H₂C=O to obtain a little more information.

    If you get... lost

    Try... Since Cr₂O₇^2- is reduced, it is an oxidant. It can never be a reductant.

    If you get... H₂C=O

    Excellent... Oxidation of an alcohol gives an aldehyde. Its oxidation gives an acid, which can further be oxidized to give CO₂ .

  • Identify the reducing agent in the reaction:
    7 CN^- + 2 OH^- + 2 Cu(NH₃)₄²⁺ --> 2 Cu(CN)₃^2- + 8 NH₃ + CNO^- + H₂O. (give the formula)

    If you get... CN

    Try... It should really be CN^-.

    If you get... lost

    Try... Consider CN^- + 2 OH^- --> CNO^- + H₂O + 2 e^-

    If you get... CN^-

    Excellent... You have just answered a fairly difficult question.

  • Identify the oxidant in the reaction:
    7 CN^- + 2 OH^- + 2 Cu(NH₃)₄²⁺ --> 2 Cu(CN)₃^2- + 8 NH₃ + CNO^- + H₂O.

    If you get... Cu

    Try... You should give Cu(NH₃)₄²⁺. The charge on the species is important.

    If you get... lost

    Try... The oxidation state for Cu is +2 in Cu(NH₃)₄²⁺, +1 in Cu(CN)₃^2-. Justify this statement yourself.

    If you get... Cu(NH₃)₄²⁺

    Excellent... This is a difficult question.