Problem 54 - Chemistry: Principles and Reactions

Question

Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{3+}(a q)\) (b) \(\mathrm{Zn}^{2+}(a q) \longrightarrow \mathrm{Zn}(s)\) (c) \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}(a q) \longrightarrow \mathrm{N}_{2}(g)\) (d) \(\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{CH}_{2} \mathrm{O}(a q)\)

Step-by-Step Solution

Answer

Question: Classify each of the given half-reactions as either oxidation or reduction: (a) Ti in TiO2 has an oxidation state of +4 and later changes to an oxidation state of +3. (b) Zn with an initial oxidation state of +2 changes to an oxidation state of 0. (c) N is present in NH4+ with an oxidation state of -3 and later changes to an oxidation state of 0 in N2(g). (d) C is initially present in CH3OH with an oxidation state of -2 and changes to an oxidation state of 0 in CH2O. Answer: (a) Reduction (b) Reduction (c) Oxidation (d) Oxidation

Step 1: (a) Determine the oxidation state of Ti before and after reaction

Initially, Ti is present in \(\mathrm{TiO}_{2}\). In such a compound, the oxidation state of O is always -2 (except in peroxides, peroxy acis and in \(\mathrm{F}_2 \mathrm{O}\).). Therefore, the oxidation state of Ti in \(\mathrm{TiO}_{2}\) can be calculated using the following equation: $$ \mathrm{Ti} + 2(-2) = 0 $$ Solving for Ti, we get: $$ \mathrm{Ti} = +4$$ After the reaction, the oxidation state of Ti is given as +3.

Step 2: (a) Classify the half-reaction as oxidation or reduction

Since the oxidation state of Ti decreases from +4 to +3, it gained an electron. Thus, the half-reaction is a reduction reaction.

Step 3: (b) Determine the oxidation state of Zn before and after the reaction

Initially, Zn is present with +2 oxidation state. After the reaction, the elemental form of Zn has an oxidation state of 0.

Step 4: (b) Classify the half-reaction as oxidation or reduction

Since the oxidation state of Zn decreases from +2 to 0, it gained two electrons. Therefore, this half-reaction is a reduction reaction.

Step 5: (c) Determine the oxidation state of N before and after the reaction

Initially, N is present in \(\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}\). In this compound, the oxidation state of H is +1. We can calculate the oxidation state of N using the following equation: $$ \mathrm{N} + 4(+1) = +1 $$ Solving for N, we get: $$ \mathrm{N} = -3$$ After the reaction, N is present in \(\mathrm{N}_{2}(g)\), where the oxidation state of N is 0.

Step 6: (c) Classify the half-reaction as oxidation or reduction

Since the oxidation state of N increases from -3 to 0, it lost three electrons. Therefore, this half-reaction is an oxidation reaction.

Step 7: (d) Determine the oxidation state of C before and after the reaction

Initially, C is present in \(\mathrm{CH}_{3} \mathrm{OH}\). In this compound, O has an oxidation state of -2, and H has an oxidation state of +1. We can calculate the oxidation state of C using the following equation: $$ \mathrm{C} + 3(+1) + (-2) + (+1) = 0 $$ Solving for C, we get: $$ \mathrm{C} = -2$$ After the reaction, C is present in \(\mathrm{CH}_{2} \mathrm{O}\). Using the same reasoning as before, the oxidation state of C can be calculated using: $$ \mathrm{C} + 2(+1) + (-2) = 0 $$ Solving for C, we get: $$ \mathrm{C} = 0$$

Step 8: (d) Classify the half-reaction as oxidation or reduction

Since the oxidation state of C increases from -2 to 0, it lost two electrons. Therefore, this half-reaction is an oxidation reaction. In summary: (a) Reduction (b) Reduction (c) Oxidation (d) Oxidation

Key Concepts

Oxidation States

Understanding oxidation states is crucial in identifying redox reactions. An oxidation state, also known as oxidation number, represents the degree of oxidation of an atom in a compound. It indicates the number of electrons an atom gains or loses when forming compounds.

Here's a quick guide to determine oxidation states:

• In a pure element, the oxidation state is always 0. For example, the oxidation state of Zn in Zn(s) is 0.
• For monoatomic ions, the oxidation state equals the ion charge. For Zn²⁺, it's +2.
• Oxygen generally has an oxidation state of -2 in most compounds (exceptions exist like peroxides).
• Hydrogen is usually +1, except in metal hydrides where it's -1.

To find an element's oxidation state in a compound, consider all known oxidation states first, then solve for the unknown element, using the fact that the overall charge of a molecule or ion must balance to zero or its given charge. This helps us analyze changes that occur during a reaction to identify electron transfer.

Half-Reactions

Half-reactions break complex redox reactions into two simpler steps—one representing oxidation and the other reduction. This method helps to see explicitly the movement of electrons and changes in oxidation states.

Consider the reaction:

• Example (a)
  • The half-reaction shows \(\mathrm{TiO}_{2}(s) \rightarrow \mathrm{Ti}^{3+}(aq)\).This indicates titanium's oxidation state changing from +4 to +3, showing a gain of electrons, hence a reduction process.
• Example (b)
  • The half-reaction is \(\mathrm{Zn}^{2+}(aq) \rightarrow \mathrm{Zn}(s)\),showing zinc reducing from +2 to 0 by gaining electrons.

Breaking reactions into half-reactions clarifies which elements lose electrons (oxidation) and which gain electrons (reduction), making the overall electron transfer more understandable.

Oxidation and Reduction Classification

Classifying reactions as either oxidation or reduction is integral to understanding redox chemistry. To make this classification:

• Oxidation involves an increase in oxidation state, signifying a loss of electrons. For example:
  • In example (c), \(\mathrm{NH}_{4}^{+}(aq) \rightarrow \mathrm{N}_{2}(g)\), the oxidation state of nitrogen goes from -3 to 0, indicating a loss of electrons.
  • Likewise, in example (d), \(\mathrm{CH}_{3}\mathrm{OH}(aq) \rightarrow \mathrm{CH}_{2}\mathrm{O}(aq)\), carbon's oxidation state increases from -2 to 0, showing oxidation.
• Reduction involves a decrease in oxidation state, indicating a gain of electrons. For example:
  • In example (a) and (b), where both titanium and zinc demonstrate reduced oxidation states, from +4 to +3 and +2 to 0, respectively, characterizing reduction.

Recognizing these patterns helps in predicting and balancing redox reactions effectively, fostering a deeper grasp of chemical changes in reactions.