Question

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g)-\mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow\) \(2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) (c) \(\begin{aligned} 3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) & \\ & 3 \mathrm{~S}(s)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned}\) (d) \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q)--\rightarrow\) \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Short answer

In each oxidation-reduction equation: a) Iodine (I) decreases its oxidation number by 5 and Carbon (C) increases its oxidation number by 2. b) Mercury (Hg) decreases its oxidation number by 2 and Nitrogen (N) increases its oxidation number by 2. c) Sulfur (S) increases its oxidation number by 2 and Nitrogen (N) decreases its oxidation number by 3. d) Oxygen (O) in H₂O₂ increases its oxidation number by 1, while the other elements' oxidation numbers remain unchanged.

Step-by-step solution

a) I₂O₅(s) + 5 CO(g) ⟶ I₂(s) + 5 CO₂(g)

Determine the oxidation numbers of the elements: - I₂O₅: Iodine(I) = +5, Oxygen(O) = -2 - CO: Carbon(C) = +2, Oxygen(O) = -2 - I₂: Iodine(I) = 0 - CO₂: Carbon(C) = +4, Oxygen(O) = -2 Now compare the changes in oxidation numbers: - Iodine(I): From +5 to 0, a decrease of 5. - Carbon(C): From +2 to +4, an increase of 2. The other elements' oxidation numbers remain unchanged.

b) 2 Hg^(2+)(aq) + N₂H₄(aq) ⟶ 2 Hg(l) + N₂(g) + 4 H^(+)(aq)

Determine the oxidation numbers of the elements: - Hg^(2+): Mercury(Hg) = +2 - N₂H₄: Nitrogen(N) = -2, Hydrogen(H) = +1 - Hg: Mercury(Hg) = 0 - N₂: Nitrogen(N) = 0 - H^(+): Hydrogen(H) = +1 Now compare the changes in oxidation numbers: - Mercury(Hg): From +2 to 0, a decrease of 2. - Nitrogen(N): From -2 to 0, an increase of 2. - Hydrogen(H) remains unchanged.

c) 3 H₂S(aq) + 2 H^(+)(aq) + 2 NO₃^(-)(aq) ⟶ 3 S(s) + 2 NO(g) + 4 H₂O(l)

Determine the oxidation numbers of the elements: - H₂S: Hydrogen(H) = +1, Sulfur(S) = -2 - H^(+): Hydrogen(H) = +1 - NO₃^(-): Nitrogen(N) = +5, Oxygen(O) = -2 - S: Sulfur(S) = 0 - NO: Nitrogen(N) = +2, Oxygen(O) = -2 - H₂O: Hydrogen(H) = +1, Oxygen(O) = -2 Now compare the changes in oxidation numbers: - Sulfur(S): From -2 to 0, an increase of 2. - Nitrogen(N): From +5 to +2, a decrease of 3. - The other elements' oxidation numbers remain unchanged.

d) Ba^(2+)(aq) + 2 OH^(-)(aq) + H₂O₂(aq) + 2 ClO₂(aq) ⟶ Ba(ClO₂)₂(s) + 2 H₂O(l) + O₂(g)

Determine the oxidation numbers of the elements: - Ba^(2+): Barium(Ba) = +2 - OH^(-): Oxygen(O) = -2, Hydrogen(H) = +1 - H₂O₂: Oxygen(O) = -1, Hydrogen(H) = +1 - ClO₂: Chlorine(Cl) = +3, Oxygen(O) = -2 - Ba(ClO₂)₂: Barium(Ba) = +2, Chlorine(Cl) = +3, Oxygen(O) = -2 - H₂O: Oxygen(O) = -2, Hydrogen(H) = +1 - O₂: Oxygen(O) = 0 Now compare the changes in oxidation numbers: - Oxygen(O) in H₂O₂: From -1 to 0, an increase of 1. - Chlorine(Cl): From +3 to +3, no change. - The other elements' oxidation numbers remain unchanged.

Key concepts

Changes in Oxidation Number

Understanding changes in oxidation number is crucial when studying oxidation-reduction (redox) reactions. The oxidation number, often referred to as oxidation state, can be considered the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic.

An increase in oxidation number signifies oxidation and typically involves the loss of electrons. Conversely, a decrease in oxidation number indicates reduction, corresponding with a gain of electrons. Detecting these changes allows for the identification of which elements are oxidized and which are reduced in a redox reaction.

For example, in the reaction between I₂O₅ and CO to form I₂ and CO₂, iodine is reduced (oxidation number decreases from +5 to 0), and carbon is oxidized (oxidation number increases from +2 to +4). By tracking these changes, we get insights into the electron transfer processes underlying redox reactions.

Balancing Redox Equations

Balancing redox equations involves ensuring that both mass and charge are conserved during the reaction. This can be more complex than balancing standard chemical equations, as it requires balancing not just the number of atoms but also the electrons lost in oxidation and gained in reduction.

To balance a redox equation, one can use the half-reaction method, which separately balances the reduction and oxidation reactions before combining them. It's important to adjust coefficients to balance the number of electrons transferred and to ensure the same number of electrons are involved in both half-reactions.

Tips for Balancing Redox Equations

• Assign oxidation states to determine which species are oxidized and reduced.
• Separate the reaction into half-reactions for oxidation and reduction.
• Balance each half-reaction for mass and charge, adding H₂O, H⁺, and electrons as necessary.
• Combine the half-reactions, making sure the number of electrons cancels out.
• Verify that both mass and charge are balanced in the final equation.

Properly balancing redox equations is imperative not just for academic exercises, but also for practical applications in chemistry and industry.

Oxidation States

Oxidation states are a fundamental concept in the study of chemistry, particularly in redox reactions. An element's oxidation state is a numerical indicator of its degree of oxidation or reduction; it essentially represents the number of electrons an atom can gain, lose, or share when it forms chemical compounds.

Determining Oxidation States:

• The oxidation state of an atom in its elemental form is always zero.
• For a simple (monoatomic) ion, the oxidation state is equal to the charge of the ion.
• Oxygen generally has an oxidation state of -2, except in peroxides where it is -1, and in compounds bonded to fluorine, where it can be positive.
• Hydrogen generally has an oxidation state of +1, except when bonded to metals in hydrides where it is -1.
• The sum of the oxidation states for all atoms in a neutral molecule or formula unit equals zero; for an ion, it equals the charge of the ion.

Mastering the assignment of oxidation states is essential, as it enables students to understand electron transfer reactions and to predict the outcomes of chemical reactions.