Half Reactions

Balancing Redox using 1/2 reactions Often times, Redox reactions are more complicated and require a specific balancing procedure. This is often the case when balancing in acidic or basic solutions. It’s going to be best if we have a list of rules to follow as we go through these problems. Consider these the 7 steps to highly successful Chemistry II Students!!

Step 1: Write the separate half reactions for Oxidation and Reduction. Figure out which elements are changing oxidation numbers. Be sure to include the entire compounds and molecules. Step 2: Balance “other” atoms in the half-reactions. Put correct coefficients for everything that is NOT an Oxygen or Hydrogen.

These reactions happen in aqueous solutions These reactions happen in aqueous solutions. So to balance Oxygens and Hydrogens, you can not simply add those elements to the equation. Step 3: Balance Oxygen atoms by adding H2O molecules to the opposite side. Step 4: Balance Hydrogen atoms by adding Hydrogen ions (H+ ions) to the opposite side.

Step 5: Balance the charges in each half reaction by adding electrons as needed. Step 6: If necessary, multiply the half reactions by integers so electrons will be equal to each other in the half reactions. Step 7: Add the half reactions together and simplify the equation.

MnO2(s)+HCl(aq)MnCl2(aq)+Cl2(g)+H2O(l) This equation could be balanced without using the half reaction method, however, you will not always have all the ions available in a problem. Step 1: Write the separate half reactions for Oxidation and Reduction. Look for elements that are changing oxidation numbers.

MnO2(s)+HCl(aq)MnCl2(aq)+Cl2(g)+H2O(l) List of Substances: (Ions separate) (4+)(2-) MnO2(s), H1+, Cl1-  Mn2+,Cl1-,Cl20, H1+, O2- What element change oxidation numbers? Those are the half reaction members! Chlorine has to be part of the redox reaction, so it has to change oxidation numbers.

MnO2(s)+HCl(aq)MnCl2(aq)+Cl2(g)+H2O(l) The reaction has been broken down into the two half reactions. MnO2(s)  Mn2+ Cl1-  Cl2 Didn’t we cancel out oxygen? It has to stay b/c MnO2 is a solid. (Don’t worry too much here.) Step 2: Balance “other” atoms in the half-reactions. (Anything not H or O) Need to balance the chlorine atoms…

MnO2(s)  Mn2+ 2 Cl1-  Cl2 MnO2(s)  Mn2+ + 2 H2O Step 3: Balance Oxygen atoms by adding H2O molecules to the other side. Only Manganese Equation! MnO2(s)  Mn2+ + 2 H2O The oxidation side doesn’t change. Step 4: Balance Hydrogen atoms by adding H+ ions to the other side.

4 H+ + MnO2(s)  Mn2+ + 2 H2O Step 5: Balance the charge in each half reaction by adding electrons as needed. Reactant side = +4 Product side = +2 If I add electrons, I can only reduce the charge. Which side gets the electrons? 2 e- + 4 H+ + MnO2(s)  Mn2+ + 2 H2O

2 Cl1-  Cl2 Step 5: Balance the charge in each half reaction by adding electrons as needed. Reactant side = -2 Product side = +0 Which side gets the electrons? 2 Cl1-  Cl2 + 2 e-

2 e- + 4 H+ + MnO2(s)  Mn2+ + 2 H2O 2 Cl1-  Cl2 + 2 e- Step 6: If necessary, multiply the half reactions so the electrons will be the same in the half reactions. 2 e- + 4 H+ + MnO2(s)  Mn2+ + 2 H2O 2 Cl1-  Cl2 + 2 e- The electrons in the reduction half reaction match the electrons in the oxidation half reaction! Hooray!

2 e- + 4 H+ + MnO2(s) + 2 Cl1-  Mn2+ + 2 H2O + Cl2 + 2 e- Step 7: Add the half reactions together and simply the equation. 2 e- + 4 H+ + MnO2(s) + 2 Cl1-  Mn2+ + 2 H2O + Cl2 + 2 e- Simplify the equation by crossing the exact same things on both sides. Just the electrons in this case.

4H++MnO2(s)+2 Cl1-Mn2++2H2O+ Cl2 This is our answer! Doesn’t look balanced but check the reaction and make sure it is balanced in both mass and charge.

S + NO3-  SO2 + NO

Balancing Redox in Basic Solutions These can be completed the same way, except one additional step. In acidic solutions, we add H+ ions to balance the hydrogen atoms in water molecules. H+ ions are readily available in acidic solutions, but not in basic.

Basic Problems = Simple Solutions! So after balancing the H+ ions, you must add the same number of OH- ions to both sides. This step neutralizes the H+ ions to form water on one side, and the other side ends up with the hydroxide ions, which are abundant in basic solutions.

Zn(s) + NO2-(aq)  NH3(g) + Zn(OH)42-(aq) Step 1: Write the separate half reactions: Oxidation: Zn(s)  Zn(OH)42-(aq) Reduction: NO2-(aq)  NH3(g)

Step 2: Same # of atoms Step 3: Balance Oxygen by adding H2O Ox: Zn + 4 H2O  Zn(OH)42- Red: NO2-  NH3 + 2 H2O Step 4: Balance Hydrogen by adding H+ Ox: Zn + 4 H2O  Zn(OH)42- + 4 H+ Red: 7 H+ + NO2-  NH3 + 2 H2O

Step 5: Balance the charge in each half reaction by adding electrons as needed. Ox: Zn + 4 H2O  Zn(OH)42- + 4 H+ + 2 e- Red: 6 e- + 7 H+ + NO2-  NH3 + 2 H2O *New Step – Because this reacts in a basic solution, need to neutralize H+ ions.

New Step: Neutralize H+ by adding OH- ions to both sides! 4 OH- + Zn + 4 H2O  Zn(OH)42- + 4 H+ + 2 e-+ 4 OH- 7 OH- + 6 e- + 7 H+ + NO2-  NH3 + 2 H2O + 7 OH- How does that help us? Are we just going to cancel it out? NO! What will H+ and OH- ions make?

New Step: Neutralize H+ by adding OH- Together they make water! (H2O) 4 OH- + Zn + 4 H2OZn(OH)42- + 2 e-+ 4 H2O 7 H2O + 6 e- + NO2-  NH3 + 2 H2O + 7 OH- Step 6: If necessary, multiply the half reactions. (Multiply top reaction by 3)

Multiply top reaction by three! 12 OH- + 3 Zn + 12 H2O 3 Zn(OH)42- + 6 e-+ 12 H2O 7 H2O + 6 e- + NO2-  NH3 + 2 H2O + 7 OH- Step 7: Combine and Simplify! Pay attention to all compounds/molecules

Simplify! 12 OH- + 3 Zn + 12 H2O + 7 H2O + 6 e- + NO2-  3 Zn(OH)42- + 6 e- + 12 H2O + NH3 + 2 H2O + 7 OH-

I came . . . I saw . . . I RedOxed