Problem 79 - Chemistry Matter and Change

Question

Balance these ionic redox equations by any method. a. \(\mathrm{Sb}^{3+}+\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{SbO}_{4}^{3-}+\mathrm{Mn}^{2+}(\text { in acid solution })\) b. \(\mathrm{N}_{2} \mathrm{O}+\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{NO}_{2}^{-}\) (in basic solution)

Step-by-Step Solution

Answer

The balanced ionic redox equations are: a. \(Sb^{3+} + MnO_{4}^{-} + 8 H^{+} \rightarrow SbO_{4}^{3-} + Mn^{2+} + 4 H_2O\) b. \(2 N_2O + ClO^{-} + \frac{1}{2} H_2O \rightarrow Cl^{-} + 2 NO_2^{-} + OH^{-} + 2 e^-\)

Step 1: Assign oxidation numbers

: Sb: +3 (given as Sb^3+) Mn: +7 (Mn is in MnO4^- with O being -2) O (in MnO4^-): -2 O (in SbO4^3-): -2 Sb (in SbO4^3-): +5 Mn (in Mn^2+): +2

Step 2: Determine the change in oxidation numbers, oxidizing agent, and reducing agent

: Sb: from +3 to +5, it is being oxidized (+2 change) Mn: from +7 to +2, it is being reduced (-5 change)

Step 3: Balance the equation

: 1. Balance atoms other than O and H. Sb and Mn are both balanced as they are appearing once. 2. Balance O using H2O: - There are 4 O atoms in MnO4^- and SbO4^3-. Add 4 H2O on the product side. Sb^3+ + MnO4^- -> SbO4^3- + Mn^2+ + 4 H2O 3. Balance H using H+: - There are 8 H atoms in 4 H2O. Add 8 H+ ions to the reactant side. Sb^3+ + MnO4^- + 8 H+ -> SbO4^3- + Mn^2+ + 4 H2O 4. Balance charges by adding electrons (e-): - The overall charge on the reactant side is -1 and on the product side is -1, so the charges are balanced. The final balanced equation is: Sb^3+ + MnO4^- + 8 H+ -> SbO4^3- + Mn^2+ + 4 H2O b. $\mathrm{N}_{2} \mathrm{O}+\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{NO}_{2}^{-}$ (in basic solution)

Step 1: Assign oxidation numbers

: N (in N2O): +1 (O is -2, total charge is 0) O (in N2O): -2 Cl (in ClO^-): +1 (O is -2, total charge is -1) O (in ClO^-): -2 Cl (in Cl^-): -1 N (in NO2^-): +3 (O is -2, total charge is -1) O (in NO2^-): -2

Step 2: Determine the change in oxidation numbers, oxidizing agent, and reducing agent

: Cl: from +1 to -1, it is being reduced (-2 change) N: from +1 to +3, it is being oxidized (+2 change)

Step 3: Balance the equation

: 1. Balance atoms other than O and H. N: coefficient is 2 for N2O. 2 N2O + ClO^- -> Cl^- + 2 NO2^- 2. Balance O by using OH- and H2O: - Reactant side has 5 O, product side has 4 O. Add 1 OH- to the product side. 2 N2O + ClO^- -> Cl^- + 2 NO2^- + OH- 3. Balance H using H2O: - Reactant side has no H, product side has 1 H. Add 1/2 H2O to the reactant side. 2 N2O + ClO^- + 1/2 H2O -> Cl^- + 2 NO2^- + OH- 4. Balance charges by adding electrons (e-): - Reactant side has a charge of -1, and the product side has a charge of -3. Add 2 e- to the product side. 2 N2O + ClO^- + 1/2 H2O -> Cl^- + 2 NO2^- + OH- + 2 e- 5. The final balanced equation in basic solution is: 2 N2O + ClO^- + 1/2 H2O -> Cl^- + 2 NO2^- + OH- + 2 e-

Key Concepts

Oxidation Numbers

Oxidation numbers are essential in understanding redox reactions. They help you track the flow of electrons during chemical reactions. Think of them as planet Earth's way of keeping score of which atom owns how many electrons before and after a reaction.

To determine oxidation numbers, you follow these general rules:

• Oxidation numbers of atoms in a pure element are zero.
• For monoatomic ions, the oxidation number equals the ion's charge.
• Usually, oxygen is assigned an oxidation number of -2, except in peroxides where it is -1, and in OF2 where it is +2.
• Hydrogen generally has an oxidation number of +1, but in metal hydrides, it becomes -1.

In the exercise, you determined the oxidation numbers of each atom to identify which elements were oxidized and reduced. Sb went from +3 to +5, indicating oxidation. Mn went from +7 to +2, indicating reduction. This identification is crucial in setting the stage for balancing the equation.

Acidic Solution

Balancing redox equations in acidic solutions involves a few extra steps. In such solutions, you have hydronium ions ( H^+ ) available to help balance the equation.

Typically, you:

• First balance atoms other than oxygen and hydrogen.
• Balance oxygen by adding water molecules.
• Balance hydrogen by adding H^+ ions.
• Finally, balance the charges by adding electrons.

For instance, in the redox reaction involving Sb and MnO4^- in acid, you added 8 H^+ to balance hydrogen, which naturally appears when balancing oxygen was achieved through adding H2O molecules.

The key here is recognizing the environmental condition of the solution (acidic, indicated by the presence of H^+ ), which assists in methodically balancing the overall equation.

Basic Solution

Balancing equations in basic solutions is similar to acidic solutions, but with a twist. Here, hydroxide ions (OH^-) are prevalent.

Here's how you approach it:

• Start by balancing all atoms other than oxygen and hydrogen.
• To balance oxygen, add water molecules where needed.
• Since you're in a basic solution, balance hydrogens by adding OH^-, which differs from adding H^+ in acidic conditions.
• Finish by balancing charges with electrons.

In the given exercise, you started with balancing nitrogen and chlorine before using additional OH^- to counterbalance extra hydrogens after incorporating water molecules. The OH^- were strategically placed to neutralize excess H^+ ions that generally emerge while balancing oxygen atoms.

Electron Transfer

The concept of electron transfer is at the heart of redox reactions. It involves analyzing how electrons move from one species to another, and this movement is quantified using oxidation numbers.

During electron transfer:

• Oxidation occurs when a molecule, atom, or ion loses electrons.
• Reduction occurs when it gains electrons.
• The oxidizing agent gains electrons and is reduced, while the reducing agent loses electrons and is oxidized.

In the exercise, for instance, Mn was reduced, showing it gained electrons, and acted as the oxidizing agent. Meanwhile, Sb was oxidized, losing electrons, acting as the reducing agent. Recognizing how electrons transfer helps confirm whether you've correctly balanced overall charges, as seen when you add e^- to match charges on both sides of the equation.