BALANCING REDOX REACTIONS IN BAS

BALANCING REDOX REACTIONS IN BASE SOLUTION

You can balance redox reactions in base solution in almost the same way as if they were in acid solution. In fact, initially we will assume the reaction is taking place in acid solution but eventually switch over to base. We'll go through the sequence of steps one at a time using a sample equation.

Suppose you wanted to balance the following equation:

MnO₄^- + CN^- --> MnO₂ + CNO^- (in base)

The first step is to split this equation into two half reactions. One half reaction will show the reduction and the other will show the oxidation. It is not important to initially know which half reaction is which.

MnO₄^- --> MnO₂

CN^- --> CNO^-

We'll deal with one half reaction at a time, starting with the half reaction for permanganate ion.

Check the balance of any atom other than O or H first. In this case the manganese atoms are balanced as written:

MnO₄^- --> MnO₂

Next, balance O's using H₂O as if in acid solution by adding H₂O's as necessary to the side deficient in O. In this case, 2 H₂O's are needed on the right hand side of the half reaction:

MnO₄^- --> MnO₂ + 2 H₂O

To balance H's, add as many H⁺ ions as needed to the side deficient in H, as if in acid solution. In this case, 4 H⁺ are needed on the left hand side of the half reaction:

4 H⁺ + MnO₄^- --> MnO₂ + 2 H₂O

Now "reality" comes into the picture. We should switch over to base solution by adding the same number of OH^- ions as we have H⁺ ions to both sides of the half reaction. In this case, 4 OH^- ions are added to both sides of the half reaction:

4 OH^- + 4 H⁺ + MnO₄^- --> MnO₂ + 2 H₂O + 4 OH^-

The H⁺ is now "removed" from the equation by taking the H⁺ ions and OH^- ions found on one side of the half reaction and forming H₂O's. In this case, 4 H⁺ ions and 4 OH^- ions make 4 H₂O molecules:

4 H₂O ~~4 OH^- + 4 H⁺~~ + MnO₄^- --> MnO₂ + 2 H₂O + 4 OH^-

Notice how some water molecules may be cancelled. There are 4 H₂O's on the left hand side of the equation, and 2 H₂O's on the right hand side of the reaction. The 2 H₂O's on the right can cancel 2 of the 4 H₂O on the left:

2 4 H₂O + MnO₄^- --> MnO₂ + ~~2 H₂O~~ + 4 OH^-

Mass balance has now been achieved. Charge balance comes next. Determine the net charge on each side of the half reaction and add electrons to the more positive side so that the charge becomes the same on both sides. In this case, the net charge on the left and side of the half reaction is -1; on the right hand side, the net charge is -4. A charge of -1 is more positive than a charge of -4, therefore, add 3 electrons to the left hand side so that the charge is -4 on both sides:

3 e^- + 2 H₂O + MnO₄^- --> MnO₂ + 4 OH^-

This half reaction, showing the reduction, is now both mass and charge balanced. You can tell it's a reduction because electrons are being gained (they are a "reactant"). We'll put this half reaction aside for now.

Repeat the previous steps to obtain mass and charge balance for the second half reaction. First, check the balance of any atom other than O or H. In this case, both the C and N are balanced as written:

CN^- --> CNO^-

Next, pretending we are carrying out the reaction in acid solution, balance the O using H₂O. In this case, 1 H₂O is needed on the left hand side to balance the O:

H₂O + CN^- --> CNO^-

To balance H's, add H⁺ to the side deficient in H. In this case, add 2 H⁺ to the right hand side of the half reaction:

H₂O + CN^- --> CNO^- + 2 H⁺

Now it is time to switch over to base. Recall this is done by adding the same number of OH^- as you have H⁺ to both sides of the half reaction. In this case 2 OH^- must be added to both sides:

2 OH^- + H₂O + CN^- --> CNO^- + 2 H⁺ + 2 OH^-

Convert the 2 H⁺ and 2 OH^- on the right hand side to H₂O's:

2 OH^- + H₂O + CN^- --> CNO^- + ~~2 H⁺ + 2 OH^-~~ 2 H₂O

Cancel H₂O where possible. In this case, the H₂O on the left hand side of the half reaction will cancel one of the H₂O's on the right hand side of the half reaction.

2 OH^- + ~~H₂O~~ + CN^- --> 2 H₂O + CNO^-

The half reaction is now mass balanced. Next, balance the charge as before by adding the electrons to the more positive side so that the charges are the same on both sides. In this case, the net charge on the left hand side of the half reaction is -3 ( 2 hydroxides and 1 cyanide); the net charge on the right hand side is -1 (1 cyanate). A charge of -1 is more positive than a charge of -3, so add 2 electrons to the right hand side of the half reaction:

2 OH^- + CN^- --> CNO^- + H₂O + 2 e^-

This half reaction showing the oxidation is now both mass and charge balanced. Next, we will retrieve the first half reaction and proceed to follow the steps needed to combine them to for the overall reaction. Before we can add the two half reactions to obtain the overall net ionic reaction, the number of electrons lost must equal the number of electrons gained. In this case, 2 electrons are lost and 3 electrons are gained:

2 H₂O + MnO₄^- + 3 e^- --> MnO₂ + 4 OH^-

2 OH^- + CN^- --> CNO^- + H₂O + 2 e^-

We need to find the least common multiple (lcm) of 2 and 3. The lcm, in this case, is 6. Multiply through each half reaction by the factor necessary to obtain the least common multiple. Here, the first half reaction must be multiplied by 2 and the second by 3:

2 x ( 2 H₂O + MnO₄^- + 3 e^- --> MnO₂ + 4 OH^- )

3 x ( 2 OH^- + CN^- --> CNO^- + H₂O + 2 e^- )

The resulting half reactions are:

4 H₂O + 2 MnO₄^- + 6 e^- --> 2 MnO₂ + 8 OH^-

6 OH^- + 3 CN^- --> 3 CNO^- + 3 H₂O + 6 e^-

Sum the two half reactions together, cancelling any like terms to obtain the balanced, net ionic reaction. In this case, the 6 electrons will cancel, the 3 H₂O's on the right hand side will cancel 3 of the 4 H₂O's on the left hand side, and the 6 OH^-on the left hand side will cancel 6 of the 8 OH^- on the right hand side:

1 4 H₂O + 2 MnO₄^- + 6 e^- --> 2 MnO₂ + 2 8 OH^-

~~6 OH^-~~ + 3 CN^- --> 3 CNO^- + ~~3 H₂O~~ + 6 e^-

What remains should be the balanced, net ionic equation. The most important step is to check this final equation for mass and charge balance:

H₂O + 2 MnO₄^- + 3 CN^- --> 2 MnO₂ + 2 OH^- + 3 CNO^-

The mass balances: 2 H, 2 Mn, 9 O, 3 C, and 3 N are found on each side of the reaction. For charge balance, the net charge on the left hand side is -5 ( -2 from two permanganate ions and -3 form three cyanide ions), and also -5 on the right hand side ( -2 from two hydroxide ions and -3 form three cyanate ions). Since mass and charge balance has been achieved, the reaction must be correctly balanced.

Summary of Steps

1. Divide the equation into two half reactions.

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

3. Balance O using H ₂O. Add as many H₂O's as necessary to the side deficient in O.

4. As if with acid solution, balance the H's by adding H⁺ to the side deficient in H.

5. Switch over to base by adding the same number of OH^- as there are H⁺ to both sides of the half reaction.

6. One side of the half reaction will contain H⁺ and OH^-. Convert the n H⁺ and n OH^- to n H₂O's.

7. Cancel H₂O molecules where possible.

8. Balance the charge by adding electrons to the more positive side to equal the less positive side.

9. Repeat these steps for the second half reaction.

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

11. Add the two half reactions together, cancelling the electrons, and any OH^- and/or H₂O where possible.

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

Created With HTML Assistant Pro - 05/11/2001 © Copyrght, 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.