Chemical Equations, continued from page 1

Balancing Net Ionic Equations:  Charge-Balancing

Consider this net ionic reaction:

Cu (s)  +  Ag⁺ (aq)   ––�?gt;   Ag (s)  +  Cu²⁺ (aq)

It is balanced with respect to elements, but not with respect to charge.  There is a total of +1 charge of ions on the left but +2 on the right.  We have to put a coefficient of 2 in front of Ag⁺ (aq) (and in front of Ag (s) also to keep the silvers balanced) in order balance this reaction correctly:

Cu (s)  +  2 Ag⁺ (aq)   ––�?gt;   2 Ag (s)  +  Cu²⁺ (aq)

Now the charge of the ions on both sides is the same, +2.

This will make more sense if we look at the original formula equation (with spectator nitrate ions present) before simplifying it to the net ionic equation:

Cu (s)  +  AgNO₃ (aq)   ––�?gt;   Ag (s)  +  Cu(NO₃)₂ (aq)

Now we can easily see that the nitrate ions are not balanced; we need to have two NO₃^{�?/SUP> on each side:}

Cu (s)  +  2 AgNO₃ (aq)   ––�?gt;   2 Ag (s)  +  Cu(NO₃)₂ (aq)

Balancing Oxidation-Reduction ("Redox") Reactions

These kinds of reactions often require a more methodical approach to balancing.  In aqueous solution, these can also be balanced in acidic solution or basic solution.  They are part of the general topic of oxidation and reduction, oxidation numbers, half-reactions, and electrochemistry which we won't go into here, except to outline the steps.

In the method of half-reactions, you first break down the reaction into the unbalanced oxidation and reduction half-reactions.  Then you balance each half-reaction separately by the following procedure:

First do steps 1-4 to each half-reaction:
1. Balance the elements other than O and H.
2. Balance the oxygens by adding H₂Os to the appropriate side.
3. Balance the hydrogens by adding H⁺ to the appropriate side.
4. Balance the charge by adding electrons (e^{�?/SUP>) to the appropriate side.}

Then you are ready to do steps 5 & 6:
5. Balance the electrons lost and gained by multiplying one or both half-reactions through by an integer (so the electrons cancel when the half-reactions are added together).
6. Add the half-reactions together and combine like terms, if necessary.

By default, this gives the balanced reaction in acidic aqueous solution (since H⁺ ions are used to balance the hydrogens).  To convert to basic solution, if asked for, simply add the same number of hydroxide ions, OH^{�?/SUP>, as you have H+ ions in the balanced reaction in acidic solution, to both sides of the equation (so it remains balanced).  Convert each H+ + OH�?/SUP> to H₂O and simplify the number of water molecules in the reaction to get the final reaction in basic solution.  You could also do this to each balanced half-reaction before adding them together in step 6 if you prefer.}

Unbalanced reaction:
MnO₄^{�?/SUP> (aq)  +  ClO₂�?/SUP> (aq)   ––�?gt;   MnO₂ (s)  +  OH�?/SUP> (aq)  +  ClO₄�?/SUP> (aq)}

Balanced reaction in acidic aqueous solution:
3 ClO₂^{�?/SUP> (aq)  +  4 MnO₄�?/SUP> (aq)  +  4 H+ (aq)   ––�?gt;   3 ClO₄�?/SUP> (aq)  +  4 MnO₂ (s)  +  2 H₂O (l)}

Balanced reaction in basic aqueous solution:
4 MnO₄^{�?/SUP> (aq)  +  3 ClO₂�?/SUP> (aq)  +  2 H₂O (l)   ––�?gt;   4 MnO₂ (s)  +  4 OH�?/SUP> (aq)  +  3 ClO₄�?/SUP> (aq)}

(This example is from a post here at http://groups.msn.com/ChemistryCorner/general.msnw?action=get_message&mview=0&ID_Message=1918&LastModified=4675570793624195831.)




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