Kaplan MCAT General Chemistry Review

Solution: All these species are neutral, so the oxidation numbers of each compound must add up to zero. In SnCl₂, because there are two chlorines present and chlorine has an oxidation number of -1, Sn must have an oxidation number of +2. Similarly, the oxidation number of Sn in SnCl₄ is +4; the oxidation number of Pb is +4 in PbCl₄ and +2 in PbCl₂. Notice that the oxidation number of Sn goes from +2 to +4; it loses electrons and thus is oxidized, making it the reducing agent. Because the oxidation number of Pb has decreased from +4 to +2, it has gained electrons and been reduced. Pb is the oxidizing agent. The sum of the charges on both sides of the reaction is equal to zero, so charge has been conserved.

BALANCING REDOX REACTIONS

Okay, okay. We know you don’t like balancing redox reactions. We know that there are many steps involved and that it can sometimes be difficult to remember how to balance not only mass but charge as well. And we recognize the unfortunate fact that we’ve just presented you with seven rules of assigning oxidation numbers to understand and remember. And now we’re giving you five steps to remember. And—hey, don’t think we didn’t just catch you rolling your eyes! Sigh all you want, but the truth of the matter is, the process of balancing redox reactions is tested on the MCAT. We might suggest that balancing redox reactions is sort of like balancing a checkbook, but we are hesitant to do so for at least two reasons: (1) balancing a checkbook isn’t exactly “fun,” and (2) aside from accountants (for whom it is simply part of their nature), we don’t know anyone who actually balances his checkbook.

By assigning oxidation numbers to the reactants and products, you can determine how many moles of each species are required for conservation of charge and mass, which is necessary to balance the equation. To balance a redox reaction, both the net charge and the number of atoms must be equal on both sides of the equation. The most common method for balancing redox equations is the half-reaction method, also known as the ion-electron method, in which the equation is separated into two half-reactions—the oxidation part and the reduction part. Each half-reaction is balanced separately, and they are then added to give a balanced overall reaction. As we review the steps involved, let’s consider a redox reaction between KMnO₄ and HI in an acidic solution.

MCAT Expertise

Methodical, step-by-step approaches like this one are great for the MCAT. Most often, you will not have to get through all of these steps before you can narrow down your answer choices. Often you will be able to find the correct answer halfway through the steps.

MnO₄^- + I^- I₂ + Mn²⁺

Step 1: Separate the two half-reactions.

I^- I₂

MnO₄^- Mn²⁺

Step 2: Balance the atoms of each half-reaction. First, balance all atoms except H and O. Next, in an acidic solution, add H₂O to balance the O atoms and then add H⁺ to balance the H atoms. (In a basic solution, use OH^- and H₂O to balance the O’s and H’s.)

To balance the iodine atoms, place a coefficient of 2 before the I^- ion.

2 I^- I₂

For the permanganate half-reaction, Mn is already balanced. Next, balance the oxygens by adding 4 H₂O to the right side.

MnO₄^- Mn²⁺ + 4 H₂O

Finally, add H⁺ to the left side to balance the 4 H₂Os. These two half-reactions are now balanced.

MnO₄^- + 8 H⁺ Mn²⁺ + 4 H₂O

Step 3: Balance the charges of each half-reaction. The reduction half-reaction must consume the same number of electrons as are supplied by the oxidation half. For the oxidation reaction, add 2 electrons to the right side of the reaction:

2 I^- I₂ + 2 e^-

For the reduction reaction, a charge of +2 must exist on both sides. Add 5 electrons to the left side of the reaction to accomplish this:

5 e^- + 8 H⁺ + MnO₄^- Mn²⁺ + 4 H₂O

Next, both half-reactions must have the same number of electrons so that they will cancel. Multiply the oxidation half by 5 and the reduction half by 2.

5(2I^- I₂ + 2e^-)

2(5e^- + 8H⁺ + MnO₄^- Mn²⁺ + 4 H₂O)

Step 4: Add the half-reactions:

10 I^- 5 I₂ + 10 e^-

16 H⁺ + 2 MnO₄^- + 10 e^- 2 Mn²⁺ + 8 H₂O

The final equation is this:

10 I^- + 10 e^- + 16 H⁺ + 2 MnO₄^- 5 I₂ + 2 Mn²⁺ + 10 e^- + 8 H₂O

To get the overall equation, cancel out the electrons and any H₂Os, H⁺s, or OH^-s that appear on both sides of the equation.

10 I^- + 16 H⁺ + 2 MnO₄^- 5 I₂ + 2 Mn²⁺ + 8 H₂O

Step 5: Finally, confirm that mass and charge are balanced. There is a +4 net charge on each side of the reaction equation, and the atoms are stoichiometrically balanced.

Electrochemical Cells