# Balancing redox reactions in acid solution

## Balancing redox reactions in acid solution

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# Balancing redox reactions in acid solution

BALANCING REDOX REACTIONS IN ACID SOLUTION

Balancing redox reactions in acid can be easy, if you follow the rules. We'll work through an example step by step introducing each rule in turn.

Suppose you want to balance the following equation:

Cr2O72- + NO2- --> Cr3+ + NO3- (acid)

First, split the reaction into two half reactions. One half reaction will show the oxidation; the other half reaction will show the reduction. It is not necessary at this point to know in advance which is which:

Cr2O72- --> Cr3+

NO2- --> NO3-

Balance any atom other than oxygen or hydrogen first. Work with one half reaction at a time. In the first half reaction, there are 2 Cr on the left hand side and only 1 Cr on the right hand side. Balance the Cr first:

Cr2O72- --> 2 Cr3+

Next, balance the O by adding as much H2O as needed to the side deficient in O. In this case, 7 H2O is needed on the right hand side to balance the 7 O on the left hand side:

Cr2O72- --> 2 Cr3+ + 7 H2O

Since this reaction occurs in acid solution, balance the H by using H+ ions. Add as many H+ ions as needed to the side deficient in H. In this case, 14 H+ ions are needed on the left hand side of the half reaction:

14 H+ + Cr2O72- --> 2 Cr3+ + 7 H2O

Mass balance has now been achieved for this half reaction.

All that remains is charge balance. To balance the charge, find the net charge on each side of the half reaction, and add as many electrons as needed to the more positive side so as to equal the charge on the more negative side. In this case the net charge on the left hand side of the half reaction is +12 ( +14 for the 14 hydrogen ions and -2 for the dichromate ion). The net charge on the right hand side is +6 (+6 due to two Cr3+ ions and 0 for the 7 water molecules). The net charge on the left hand side (+12) is more positive than the charge on the right hand side (+6). By adding 6 electrons (6e-) to the left hand side, the net charge will be the same on both sides of the half reaction:

6 e- + 14 H+ + Cr2O72- --> 2 Cr3+ + 7 H2O

This half reaction is now mass and charge balanced, and shows the reduction. Notice how electrons are on the reactant side which means they are being gained (reduction involves gaining electrons).

Repeat the same steps for the second half reaction:

NO2- --> NO3-

Balance the oxygen by using water. Add 1 H2O to the left hand side of the half reaction:

H2O + NO2- --> NO3-

Balance the hydrogens by adding H+ to the side deficient in H. In this case, add 2 H+ to the right hand side of the half reaction:

H2O + NO2- --> NO3- + 2 H+

Mass balance has now been achieved. To balance the charge, add electrons to the more positive side to equal the less positive side of the half reaction. In this case, the net charge on the left hand side of the half reaction is -1; the net charge on the right hand side of the half reaction is +1. The right hand sides of the half reaction is more positive, so add 2 electrons to the right hand side to make the net charge on both sides of the half reaction -1:

H2O + NO2- --> NO3- + 2 H+ + 2 e-

This half reaction is now both mass and charge balanced, and shows the oxidation. Notice how electrons are on the product side of the half reaction which means they are being lost (oxidation is a loss of electrons).

Before you can add up the two half reactions to the balanced net ionic reaction, the number of electrons lost must equal the number of electrons gained. Examination of the two half reactions shows 2 electrons lost and 6 electrons gained:

6 e- + 14 H+ + Cr2O72- --> 2 Cr3+ + 7 H2O

H2O + NO2- --> NO3- + 2 H+ + 2 e-

You need to find the least common multiple (lcm) of 2 and 6. In this case, the lcm is 6. Next, multiply each of the half reactions by the factor needed to obtain the 1 cm. Therefore, multiply the first half reaction by 1 and the second half reaction by 3:

1 x ( 6 e- + 14 H+ + Cr2O72- --> 2 Cr3+ + 7 H2O )

3 x ( H2O + NO2- --> NO3- + 2 H+ + 2 e- )

This yields the following half reactions:

6 e- + 14 H+ + Cr2O72- --> 2 Cr3+ + 7 H2O

3 H2O + 3 NO2- --> 3 NO3- + 6 H+ + 6 e-

Next, cancel any terms which are common to both sides of the half reactions. For instance, the 6 electrons may be cancelled, the 3 H2O on the left hand side of the second half reaction will cancel 3 of the 7 H2O on the right hand side of the first half reaction, and the 6 H+ on the right hand side of the second half reaction will cancel 6 of the 14 H+ on the left hand side of the first half reaction. These cancellations are shown below:

6 e-+ 8 14 H+ + Cr2O72- --> 2 Cr3+ + 4 7 H2O

3 H2O + 3 NO2- --> 3 NO3- + 6 H+ + 6 e-

Sum the two half reactions together to obtain the balanced net ionic equation:

8H+ + Cr2O72- + 3 NO2- --> 2 Cr3+ + 3 NO3- + 4 H2O

Finally, the most important step: check the final equation for mass and charge balance. If mass and charge balance has been achieved, the equation cannot be wrong! In this case, mass balances (8H, 2Cr, 13 O, 3N are found on both sides of the equation), and charge balances (a net charge of +3 is found on both sides of the equation).

The steps are summarized on the next page and are followed by a few problems for you to try. The most common step omitted by students is step # 2. Omitting this step will often lead to errors in the number of O's or lack of charge balance.

If you find your final equation is not balanced, it is usually best to start over. Finding your error may be time consuming on a test. Go work another problem and then come back to the balancing so that you won't repeat your previous errorwhile balancing the equation.

Summary of Steps:

1. Split the equation into two half reactions.

2. Balance any atom other than O or H first.

3. Balance O's using H2O. Add as many waters as needed to balance the oxygens on the side deficient in oxygen.

4. Balance H's using H+. Add as many H+ ions as needed to the side deficient in H.

5. The mass should now be balanced. To balance the charge, determine the charge on each side of the reaction, and add as many electrons to the more positive side as needed to make the charge the same on both sides.

6. Repeat steps 1 through 4 for both half reactions.

7. If the number of electrons lost does not equal the number of electrons gained, find the common multiple and multiply each half reaction by the necessary factor.

8. Sum both half reactions to obtain the balanced, net ionic reaction. In many cases, look for H+ ions and H2O molecules to cancel.

9. Check the final equation for mass and charge balance.

Note: If you wish to express H+ as H3O+ then change the number of H+ into the same number of H3O+ and add that same number of H2O to the opposite side of the equation. For example:

8 H+ + Cr2O72- + 3NO2- --> 2 Cr3+ + 3 NO2- + 4 H2O

becomes

8 H3O+ + Cr2O72- + 3 NO2- --> 2 Cr3+ + 3 NO2- + 12 H2O

© Copyright, 2001, L. Ladon. Permission is granted to use and duplicate these materials for non-profit educational use, under the following conditions: No changes or modifications will be made without written permission from the author. Copyright registration marks and author acknowledgement must be retained intact.

Source : http://pages.towson.edu/ladon/redoxa.doc

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