Documents
Resources
Learning Center
Upload
Plans & pricing Sign in
Sign Out
Get this document free

Lesson 4 - Balancing half reactions

VIEWS: 12 PAGES: 2

									                                 Electrochemistry ISU

                        Lesson 4 – Balancing by Half reaction


               The Half Reaction Method for Balancing Redox Reactions
Note: This method assumes the reaction occurs in aqueous, acidic solution where H2O
        +
and H are plentiful. If the reaction occurs in a basic solution, this method allows you to
correct for that fact at the end (see step 8 below).
1. Write two half-reactions, one for oxidation and one for reduction. If you have multiple
    oxidations, put them all together in one half reaction. Do the same if you have
    multiple reductions.
2. Balance each half-reaction as follows:
        a. Balance all elements other than oxygen and hydrogen.
        b. Balance oxygen by adding the appropriate number of water molecules (H2O) to
            the side of the equation that needs more oxygen atoms.
                                                                                   +
        c. Balance hydrogen by adding the appropriate number of hydrogen ions (H ) to
            the side of the equation that needs more hydrogen atoms.
        d. Balance the charge by adding the appropriate number of electrons to the side of
            the equation with the greater overall positive charge.
3. Multiply each half-reaction by a whole number so that the electrons lost in the
    oxidation half-reaction equal the number of electrons gained in the reduction half-
    reaction.
4. Add the two half reactions together, keeping all of the reactants together on the left
    side of the reaction arrow and all of the products together on the right side of the
    reaction arrow. The electrons will cancel so they are not shown in the final equation.
5. Cancel any substances that appear on both sides of the equation. Check to make sure
   that the equation is balanced. If you removed spectator ions before you wrote the
   half-reactions, put them back in and balance by inspection.
6. If your reaction occurs in a basic solution, you must also do the following. Remove any
    +                                                       -
   H from the equation by adding an equal number of OH ions to both sides of the
                                +      -                                    +          -
   equation. Remember that H + OH forms H2O so you can replace the H and OH that
   occur on the same side with an equal number of H2O. Now cancel any water
   molecules that appear on both sides of the equation.
Example:
            2-
   CrO4 (aq) + Cu (s)  Cr(OH)3 (s) + Cu(OH)2 (s) (basic)

Half Reactions:
Oxidation: Cu (s)  Cu(OH)2 (s)
                  2-
Reduction: CrO4 (aq)  Cr(OH)3 (s)
1. Balance each half reaction: Oxidation: Cu (s)  Cu(OH)2 (s)
a. Cu is already balanced.
b. Add 2 H20 to the left to balance O:            2 H20 (l) + Cu (s)  Cu(OH)2 (s)
             +                                                                                    +
c. Add 2 H to the right to balance H:             2 H20 (l) + Cu (s)  Cu(OH)2 (s) + 2 H (aq)
            -                                                                                         +       -
d. Add 2 e to the right to balance the charge: 2 H20 (l) + Cu (s) Cu(OH)2 (s) + 2 H (aq) + 2 e


2. Balance each half reaction: Reduction: CrO42- (aq)  Cr(OH)3 (s)
a. Cr is already balanced.
                                                                   2-
b. Add 1 H20 to the right to balance O:                  CrO4 (aq)  Cr(OH)3 (s) + H20 (l)
             +                                                +                 2-
c. Add 5 H to the left to balance H:                     5 H (aq) + CrO4 (aq)  Cr(OH)3 (s) + H20 (l)
            -                                                 -         +                2-
d. Add 3 e to the left to balance the charge:            3 e + 5 H (aq) + CrO4 (aq)  Cr(OH)3 (s) + H20 (l)


3. Multiply the oxidation half-reaction by 3, and the reduction half-reaction by 2 to make the number of
     electrons equal in the two half-reactions.
                                                                        +            -
     Oxidation: 6 H20 (l) + 3 Cu (s)  3 Cu(OH)2 (s) + 6 H (aq) + 6 e
     Reduction: 6 e- + 10 H+ (aq) + 2 CrO42- (aq)  2 Cr(OH)3 (s) + 2 H20 (l)
4. Add them together.
                                                         +
5. Note that the electrons cancel, as do 6 of the H ions and 2 of the water molecules:
                                   +                          2-
   4 H20 (l) + 3 Cu (s) + 4 H (aq) + 2 CrO4 (aq)  3 Cu(OH)2 (s) + 2 Cr(OH)3 (s)
6. If this reaction occurred in acidic solution, we would now be done. However, since this reaction occurs
                                                                            -
     in basic solution, we have one more step to do. Add 4OH to both sides so it can combine with the
        +
     4H on the left to form 4 H2O: Always add the same amount of OH- ions as H+ ions in the
     equation on both sides.

                         +               -                   2-                                                   -
4 H20 (l) + 3 Cu (s) + 4 H (aq) + 4OH (aq) + 2 CrO4 (aq)  3 Cu(OH)2 (s) + 2 Cr(OH)3 (s) + 4OH (aq)
                                             2-                                                           -
4 H20 (l) + 3 Cu (s) + 4 H2O (l) + 2 CrO4 (aq)  3 Cu(OH)2 (s) + 2 Cr(OH)3 (s) + 4OH (aq)
                             2-                                                               -
8 H20 (l) + 3 Cu (s) + 2 CrO4 (aq)  3 Cu(OH)2 (s) + 2 Cr(OH)3 (s) + 4OH (aq)

								
To top