Question
For oxidation-reduction reactions, often it is easier to balance the net ionic form of the equation first and then to derive the chemical equation from the net ionic equation. The following is a method for this procedure: Write the skeleton equation from the information given for the reactants and products. Assign oxidation numbers to every element (including the elements of any acids and bases). The elements that are spectator ions do not change oxidation numbers. However, sometimes an ion can be involved as both a spectator ion and in oxidation-reduction. An example of this will be the nitrate ion in the copper chemistry experiment. Balance the net ionic equation following the rules given in the previous section. Remember to add in the spectator ions on the side needing them when you balance the atoms other than oxygen and hydrogen. If needed, divide to reduce the coefficients to the lowest whole-number ratio. At this point you will need to add in the counter ion for the acid or base used. Add one counter ion for each H^+ (for sulfuric acid you will add HSO_4^-) or OH^- in the equation to each side. The result of this step is the ionic equation. Check to make sure that the total charge on each side of the reaction is zero. Combine anions and cations to create the balanced chemical equation. No uncombined ions should remain. Check to make sure the number of atoms is still balanced and that the coefficients are in the lowest whole-number ratio. An example of this method of deriving a chemical equation from the balanced net ionic form of an equation for a redox reaction will now be shown. Practice problems will be found in the Copper Chemistry laboratory experiment and questions. Potassium permanganate reacts with chromium (III) chloride to produce manganese (IV) oxide and the chromate ion in potassium hydroxide. KMnO_4 + CrCl_3 rightarrow MnO_2 + CrO_4^2- + 1 + 7 - 2 + 3 - 1 + 4 - 2 + 6 - 2 3 e^- + 4 H_2 O + KMnO_4 rightarrow MnO_2 + K^+ + 2 H_2 O +4 OH 80H + 4 H_2 O + CrCl_3 rightarrow CrO_4^2- + 3 Cl^- + 8 H_2 O + 3 e^- 4 OH + KMnO_4 + CrCl_3 rightarrow MnO_2 CrO_4^2- + 3 Cl^- + K^+ + 2 H_2 O 4 KOH + KMnO_4 + CrCl_3 rightarrow MnO_2 + K_2 CrO_4 + 3 KCl + 2 H_2 O Balance atom other than H or O. Balance O by adding water. Balance H by adding H^- Balance charge by adding e^-.
Explanation / Answer
Ans. KMnO4 + CrCl3 = CrO42- + MnO2
KMnO4 + 2H2O + 3e = MnO2 + 4 OH- + K +
CrCl3 + 8 OH- = CrO42- + 4H2O + 3cl- + 4H2O
KMnO4 + CrCl3 + 4OH- = MnO2 + CrO42- + K+ + 2H2O + Cl-
OH- + K+ = KOH
K+ + Cl- = KCl
2K+ + CrO42- = K2CrO4
KMnO4 + CrCl3 + 4KOH = MnO2 + K2CrO4 + 3KCl + 2H2O
