Question

Balance each of the following oxidation-reduction reactions by using the oxidation states method. a. \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a a)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Pb}(s) \rightarrow \mathrm{Pb}(\mathrm{OH})_{2}(s)\) c. \(\mathrm{H}^{+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

The balanced oxidation-reduction reactions are: a. \(2\mathrm{Cl}_{2}(g) + 2\mathrm{Al}(s) \rightarrow 2\mathrm{Al}^{3+}(a q) + 6\mathrm{Cl}^{-}(a q)\) b. \(1/2\mathrm{O}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{Pb}(s) \rightarrow \mathrm{Pb}(\mathrm{OH})_{2}(s)\) c. \(8\mathrm{H}^{+}(a q) + \mathrm{MnO}_{4}^{-}(a q) + 5\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q) + 5\mathrm{Fe}^{3+}(a q) + 4\mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-step solution

Identify the oxidation states

Before the reaction, Cl in Cl2 has an oxidation state of 0, and Al has an oxidation state of 0. After the reaction, the oxidation state of Al becomes +3 (in Al3+), and the oxidation state of Cl becomes -1 (in Cl-).

Balance the electrons lost and gained

Al loses 3 electrons (going from 0 to +3) while each Cl gains 1 electron (going from 0 to -1). Since there are 2 Cl atoms in Cl2, we need 2 Cl- ions to balance the electron gain. Therefore, we need to multiply the Al species by 2 to balance the electron loss. The balanced reaction is: 2Cl2(g) + 2Al(s) → 2Al3+(aq) + 6Cl-(aq) b. \(\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Pb}(s) \rightarrow \mathrm{Pb}(\mathrm{OH})_{2}(s)\)

Identify the oxidation states

Before the reaction, O in O2 has an oxidation state of 0, H in H2O has an oxidation state of +1, and Pb has an oxidation state of 0. After the reaction, the oxidation state of O in Pb(OH)2 is -2, and the oxidation state of Pb in Pb(OH)2 is +2.

Balance the electrons lost and gained

O gains 2 electrons (going from 0 to -2) while Pb loses 2 electrons (going from 0 to +2). We need to multiply the O2 species by 1/2 to balance the electron gain. The balanced reaction is: 1/2O2(g) + H2O(l) + Pb(s) → Pb(OH)2(s) c. \(\mathrm{H}^{+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Identify the oxidation states

Before the reaction, the oxidation state of H is +1, Mn in MnO4- is +7, and Fe is +2. After the reaction, the oxidation state of Mn in Mn2+ is +2, Fe in Fe3+ is +3, and O in H2O is -2.

Balance the electrons lost and gained

Mn loses 5 electrons (going from +7 to +2) and Fe gains 1 electron (going from +2 to +3). To balance the electron loss for Mn, we need to multiply the Fe species by 5. The balanced reaction is: 8H+(aq) + MnO4^-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

Key concepts

Balancing Chemical Equations

To understand balancing chemical equations, it is important to realize that chemical reactions are like math equations needing balance on both sides. This means the number of atoms for every element involved in the reaction remains unchanged through the process. For Oxidation-Reduction (Redox) Reactions, this becomes vital because it also ensures that the electric charge before and after the reaction matches.

In the given exercises, you observe atoms of elements like chlorine, aluminum, oxygen, lead, hydrogen, manganese, and iron undergoing changes. Balancing these reactions involves determining how many molecules of each substance are necessary to maintain equilibrium.

• Start by listing each element involved in the reaction.
• Count how many atoms of each element are present on both sides of the equation.
• Add coefficients to molecules as needed to balance the atoms for each element on both sides.

Remember, you can only add coefficients (whole numbers placed before molecules) and not alter the chemical formula itself, ensuring atoms—and charge—balance accordingly.

Oxidation States

Oxidation states help us track how electrons are transferred in a reaction. They are theoretical charges that atoms would have if the compound was composed of ions. In our exercises, recognizing oxidation states is key to understand which species are oxidized and which are reduced.

Here's a simple guide to identifying oxidation states:

• Elements in their pure form, like Cl₂ or Al, have an oxidation state of 0.
• For ions, the oxidation state equals the charge of the ion, like Al³⁺ has an oxidation state of +3.
• H is typically +1, and O is -2 in most compounds.

In the exercises provided, you calculate these states before and after the reaction to see how they change. This change in oxidation states indicates the transfer of electrons, recording what substances are oxidized or reduced.

Electron Transfer

Electron transfer is at the heart of oxidation-reduction reactions. It is the process where electrons move from one reactant (the electron donor) to another (the electron acceptor). Understanding this can reveal much about the nature of redox reactions and what it means for substances to be oxidized or reduced.

Let's break this down further:

• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.
• During redox reactions, you often find one reactant giving up electrons and another gaining electrons, accomplishing both oxidation and reduction simultaneously.

Using the provided exercises, the number of electrons lost by the oxidized substance has to balance with the number gained by the reduced substance. This is seen in the calculated equations, where coefficients are adjusted to achieve this balance.

Redox Reactions

Redox reactions are a specific type of chemical reaction involving the transfer of electrons between two substances. They are fundamental to many chemical processes, including metabolism in the body and corrosion of metals.

In a redox reaction:

• The substance that loses electrons is said to be oxidized.
• The substance that gains electrons is said to be reduced.
• For example, in the reaction between Cl₂ and Al, Cl gains electrons and is reduced, while Al loses electrons and is oxidized.

Each exercise examines these shifts in oxidation states to illustrate how electrons are transferred. By identifying these shifts, we balance the redox equations, maintaining the integrity of both mass and charge across the reaction—all hallmark signs of a balanced redox process.