Example



Example

EXAMPLE Balancing Redox Equations for Reactions Run in Acidic Conditions:  Balance the following redox equation using the half-reaction method.

      Cr2O72-(aq)  +  HNO2(aq)  -->  Cr3+(aq)  +  NO3-(aq)   (acidic)

Solution:

Step #1:   Write the skeletons of the oxidation and reduction half-reactions.

You will usually be given formulas for two reactants and two products. In such cases, one of the reactant formulas is used in writing one half-reaction, and the other reactant formula is used in writing the other half-reaction. (In most cases, you do not need to know which reactant is oxidized and which is reduced.) The product formula in each half-reaction must include all of the elements in the reactant formula except hydrogen and oxygen. There are circumstances that make this step more complicated, but we will stick to simpler examples at this stage.

Cr2O72-  -->  Cr3+

HNO2  -->  NO3-

Step #2:    Balance all elements other than H and O.

To balance the chromium atoms in our first half-reaction, we need a two in front of Cr3+.

Cr2O72-  -->  2Cr3+

HNO2  -->  NO3-

Step #3:    Balance the oxygen atoms by adding H2O molecules on the side of the arrow where O atoms are needed.

The first half-reaction needs seven oxygen atoms on the right, so we add seven H2O molecules.

Cr2O72-  -->  2Cr3+  +  7H2O

The second half-reaction needs one more oxygen atom on the left, so we add one H2O molecule.

HNO2    +  H2O  -->  NO3- 

Step #4:    Balance the hydrogen atoms by adding H+ ions on the side of the arrow where H atoms are needed.

The first half-reaction needs 14 hydrogen atoms on the left to balance the 14 hydrogen atoms in the 7 H2O molecules, so we add 14 H+ ions to the left. 

Cr2O72-  +  14H+  -->  2Cr3+  +  7H2O

The second half-reaction needs three hydrogen atoms on the right to balance the three hydrogen atoms on the left, so we add 3 H+ ions to the right. 

HNO2    +  H2O  -->  NO3-  +  3H+

Step #5:    Balance the charge by adding electrons, e-.

The electrons go on the side of the equation with the highest charge (most positive or least negative). We add enough electrons make the charge on that side of the equation equal to the charge on the other side of the equation.

The sum of the charges on the left side of the chromium half-reaction is +12 (-2 for the Cr2O72- plus +14 for the 14 H+). The sum of the charges on the right side of the chromium half-reaction is +6 (for the 2 Cr3+). If we add six electrons to the left side, the sum of the charges on each side of the equation becomes +6.

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

The sum of the charges on the left side of the nitrogen half‑reaction is zero. The sum of the charges on the right side of the nitrogen half-reaction is +2 (-1 for the nitrate plus +3 for the 3 H+). If we add two electrons to the right side, the sum of the charges on each side of the equation becomes zero.

HNO2    +  H2O  -->  NO3-  +  3H+  +  2e-

(Although it is not necessary, you can check that you have added the correct number of electrons by looking to see whether the net change in oxidation number for each half-reaction is equal to the number of electrons gained or lost. Because the two Cr atoms in Cr2O72- are changing from +6 to +3, the net change in oxidation number is 2(-3) or -6. This would require six electrons, so we have added the correct number of electrons to the first half-reaction. The N atom in HNO2 changes from +3 to +5, so the net change is +2. Two electrons would be lost in this change, so we have added the correct number of electrons to the second half-reaction.)

Step #6:    If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction half-reaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number lost.

For the chromium half-reaction to gain six electrons, the nitrogen half-reaction must lose six electrons. Thus, we multiply the coefficients in the nitrogen half-reaction by 3.

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

3(HNO2    +  H2O  -->  NO3-  +  3H+  +  2e-)

or

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

3HNO2    +  3H2O  -->  3NO3-  +  9H+  +  6e-

Step #7:    Add the 2 half-reactions as if they were mathematical equations.

The 3 H2O in the second half-reaction cancel three of the 7 H2O in the first half-reaction to yield 4 H2O on the right of the final equation.

The 9 H+ on the right of the second half-reaction cancel nine of the 14 H+ on the left of the first half-reaction leaving 5 H+ on the left of the final equation.

     Cr2O72-  +  3HNO2   +  5H+  -->  2Cr3+  +  3NO3-   +  4H2O

Step #8:    Check to make sure that the atoms and the charge balance.

The atoms in our example balance and the sum of the charges is +3 on each side, so our equation is correctly balanced.

Cr2O72-(aq)    +  3HNO2(aq)    +  5H+(aq)    
                   -->  2Cr3+(aq)    +  3NO3- (aq)    +  4H2O(l)

Click here to see a Sample Study Sheet for this task.

Click here to see an exercise that will allow you to try this task yourself. 

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