Problem 16 - Chemistry: An Atoms-Focused Approach

Question

Balance the following half-reactions by adding the appropriate number of electrons. Which are oxidation half-reactions and which are reduction half- reactions? a. \(\mathrm{TiO}^{2+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ti}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\) b. \(\mathrm{S}_{4}^{2+}(a q) \rightarrow 2 \mathrm{S}_{2}^{2-}(a q)\) c. \(\mathrm{VO}_{2}{\underline{\phantom{xx}}}^{+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\) d. \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} \rightarrow \mathrm{Fe}(\mathrm{CN})_{6}{\underline{\phantom{xx}}}^{4-}(a q)\)

Step-by-Step Solution

Answer

Question: Identify the oxidation and reduction half-reactions among the following: a) TiO²⁺(aq) + 2H⁺(aq) → Ti³⁺(aq) + H₂O(l) b) S₄²⁺(aq) → 2S₂²⁻(aq) c) VO₂⁺(aq) + 2H⁺(aq) → VO²⁺(aq) + H₂O(l) d) Fe(CN)₆³⁻ → Fe(CN)₆⁴⁻(aq) Answer: Half-Reaction a is an oxidation half-reaction. Half-Reaction b is a reduction half-reaction. Half-Reaction c is an oxidation half-reaction. Half-Reaction d is a reduction half-reaction.

Step 1: Balance elements

We have one Ti atom on both sides, so it is already balanced. **Step 2: Balance O by adding water**

Step 2: Balance oxygen

To balance the one oxygen atom on the left, we will add one water molecule to the right side of the equation: \(\mathrm{TiO}^{2+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ti}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\) **Step 3: Balance H by adding H+ ions**

Step 3: Balance hydrogen

We have two H atoms on the left and two on the right, so it is balanced. **Step 4: Balance charges by adding electrons**

Step 4: Balance charges

The net charge on the left is +1, and on the right, it is +3. Add two electrons to the right side to balance the charges: \(\mathrm{TiO}^{2+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ti}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 e^-\) #atatype#redox

Step 5: Determine oxidation or reduction

Since we added electrons to the product side, this is an oxidation half-reaction. ## Half-Reaction b ## **Step 1: Balance elements other than H and O**

Step 6: Balance elements

We have four S atoms on both sides, so it is already balanced. **Step 2: Balance O by adding water**

Step 7: Balance oxygen

There are no oxygen atoms in this half-reaction. **Step 3: Balance H by adding H+ ions**

Step 8: Balance hydrogen

There are no hydrogen atoms in this half-reaction. **Step 4: Balance charges by adding electrons**

Step 9: Balance charges

The net charge on the left is +2, and on the right, it is -4. Add six electrons to the left side to balance the charges: \(6 e^-+ \mathrm{S}_{4}^{2+}(a q) \rightarrow 2 \mathrm{S}_{2}^{2-}(a q)\) #atype#redox

Step 10: Determine oxidation or reduction

Since we added electrons to the reactant side, this is a reduction half-reaction. ## Half-Reaction c ## **Step 1: Balance elements other than H and O**

Step 11: Balance elements

We have one V atom on both sides, so it is already balanced. **Step 2: Balance O by adding water**

Step 12: Balance oxygen

To balance the two oxygen atoms on the left side, we will add one water molecule to the right side of the equation: \(\mathrm{VO}_{2}{\underline{\phantom{xx}}}^{+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\) **Step 3: Balance H by adding H+ ions**

Step 13: Balance hydrogen

We have two H atoms on the left and two on the right, so it is balanced. **Step 4: Balance charges by adding electrons**

Step 14: Balance charges

The net charge on the left is +1, and on the right, it is +2. Add one electron to the right side to balance the charges: \(\mathrm{VO}_{2}{\underline{\phantom{xx}}}^{+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+ e^-\) #atype#redox

Step 15: Determine oxidation or reduction

Since we added electrons to the product side, this is an oxidation half-reaction. ## Half-Reaction d ## **Step 1: Balance elements other than H and O**

Step 16: Balance elements

The Fe and CN atoms are already balanced on both sides. **Step 2: Balance O by adding water**

Step 17: Balance oxygen

There are no oxygen atoms in this half-reaction. **Step 3: Balance H by adding H+ ions**

Step 18: Balance hydrogen

There are no hydrogen atoms in this half-reaction. **Step 4: Balance charges by adding electrons**

Step 19: Balance charges

The net charge on the left is -3, and on the right, it is -4. Add one electron to the left side to balance the charges: \(e^-+\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\rightarrow \mathrm{Fe}(\mathrm{CN})_{6}{\underline{\phantom{xx}}}^{4-}(a q)\) #atype#redox

Step 20: Determine oxidation or reduction

Since we added electrons to the reactant side, this is a reduction half-reaction. #Summary# Half-Reaction a: Oxidation Half-Reaction b: Reduction Half-Reaction c: Oxidation Half-Reaction d: Reduction

Key Concepts

Half-Reaction Balancing

Balancing half-reactions is a crucial part of understanding redox reactions. To balance a half-reaction, you need to ensure that all elements and charges are balanced on both sides of the equation.

• First, balance all atoms except for hydrogen and oxygen.
• Next, balance the oxygen atoms by adding water molecules.
• Then, balance the hydrogen atoms by adding hydrogen ions (H⁺).
• Finally, balance the charges by adding electrons (e^-).

This step-by-step process ensures that both mass and charge conservation are maintained.

Oxidation and Reduction

Oxidation and reduction are two fundamental processes in redox reactions. They involve the transfer of electrons between substances. Oxidation occurs when a substance loses electrons, resulting in an increase in oxidation state. In contrast, reduction occurs when a substance gains electrons, decreasing its oxidation state.
Identifying which substance is oxidized and which is reduced helps in understanding how electrons move during the reaction.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical reactions. In electrochemical reactions, oxidation and reduction processes occur simultaneously. These reactions can convert chemical energy into electrical energy and vice versa.
Applications of electrochemistry include batteries, fuel cells, and electrolysis, making it vital for both scientific and practical purposes.

Electron Transfer

The transfer of electrons is the core of redox reactions. During these reactions, electrons move from one reactant to another, changing their oxidation states. Electron transfer governs the energy changes in a chemical reaction.
Understanding electron flow is essential for predicting reaction behavior and for applications in designing electrochemical cells and circuits.

Chemical Equations Balancing

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is satisfied. This involves making sure there are equal numbers of each type of atom on both sides of the reaction equation.
For redox reactions, balancing also involves ensuring that the charge is balanced. This is crucial for reactions taking place in aqueous solutions, where ions are present, such as in electrochemical reactions.