Question

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. $$ \begin{array}{l} \text { (a) } \mathrm{PBr}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HBr}(a q) \\ \text { (b) } \mathrm{NaI}(a q)+3 \mathrm{HOCl}(a q) \longrightarrow \mathrm{NaIO}_{3}(a q)+3 \mathrm{HCl}(a q) \\ \text { (c) } 3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ 3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g) \\ \text { (d) } 2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \\ \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$

Short answer

(a) Not an oxidation-reduction reaction, as there is no change in oxidation numbers. (b) Oxidation-reduction reaction. Iodine is oxidized (from -1 to +5) and chlorine is reduced (from +1 to -1). (c) Oxidation-reduction reaction. Sulfur is oxidized (from +4 to +6) and nitrogen is reduced (from +5 to +2). (d) Oxidation-reduction reaction. Sulfur is reduced (from +6 to +4) and bromine is oxidized (from -1 to 0).

Step-by-step solution

(a) Assigning oxidation numbers

: First, let's assign oxidation numbers to all elements in the compounds on both sides of the equation \(PBr_{3}(l) + 3 H_{2}O(l) \longrightarrow H_{3}PO_{3}(aq) + 3 HBr(aq)\). P: +3 Br: -1 H: +1 O: -2

(a) Identifying changes in oxidation numbers

: Now, let's compare the oxidation numbers of elements on both sides of the equation: P: +3 → +3 (no change) Br: -1 → -1 (no change) H: +1 → +1 (no change) O: -2 → -2 (no change) No change in oxidation numbers, so this is not an oxidation-reduction reaction.

(b) Assigning oxidation numbers

: Let's assign oxidation numbers to all elements in the compounds on both sides of the equation \(NaI(aq) + 3 HOCl(aq) \longrightarrow NaIO_{3}(aq) + 3 HCl(aq)\). Na: +1 I: -1 O: -2 H: +1 Cl: -1 (in NaCl and HCl) and +1 (in HOCl)

(b) Identifying changes in oxidation numbers

: Now, let's compare the oxidation numbers of elements on both sides of the equation: Na: +1 → +1 (no change) I: -1 → +5 (change) O: -2 → -2 (no change) H: +1 → +1 (no change) Cl: -1 → +1 (change in HOCl) and -1 → -1 (no change in HCl) There is a change in oxidation numbers for I and Cl, so this is an oxidation-reduction reaction. Iodine is being oxidized (from -1 to +5) and chlorine is being reduced (from +1 to -1).

(c) Assigning oxidation numbers

: Let's assign oxidation numbers to all elements in the compounds on both sides of the equation \(3 SO_{2}(g) + 2 HNO_{3}(aq) + 2 H_{2}O(l) \longrightarrow 3 H_{2}SO_{4}(aq) + 2 NO(g)\). S: +4 O: -2 H: +1 N: +5

(c) Identifying changes in oxidation numbers

: Now, let's compare the oxidation numbers of elements on both sides of the equation: S: +4 → +6 (change) O: -2 → -2 (no change) H: +1 → +1 (no change) N: +5 → +2 (change) There is a change in oxidation numbers for S and N, so this is an oxidation-reduction reaction. Sulfur is being oxidized (from +4 to +6) and nitrogen is being reduced (from +5 to +2).

(d) Assigning oxidation numbers

: Let's assign oxidation numbers to all elements in the compounds on both sides of the equation \(2 H_{2}SO_{4}(aq) + 2 NaBr(s) \longrightarrow Br_{2}(l) + SO_{2}(g) + Na_{2}SO_{4}(aq) + 2 H_{2}O(l)\). H: +1 S: +6 (in H2SO4 and Na2SO4) and +4 (in SO2) O: -2 Na: +1 Br: -1 (in NaBr) and 0 (in Br2)

(d) Identifying changes in oxidation numbers

: Now, let's compare the oxidation numbers of elements on both sides of the equation: H: +1 → +1 (no change) S: +6 → +4 (change in H2SO4 and SO2) and +6 → +6 (no change in Na2SO4) O: -2 → -2 (no change) Na: +1 → +1 (no change) Br: -1 → 0 (change) There is a change in oxidation numbers for S and Br, so this is an oxidation-reduction reaction. Sulfur is being reduced (from +6 to +4) and bromine is being oxidized (from -1 to 0).

Key concepts

Changes in Oxidation Numbers

Understanding the changes in oxidation numbers is a fundamental aspect of analyzing oxidation-reduction (redox) reactions in chemistry. An oxidation number is a value assigned to an atom that represents its degree of oxidation or reduction. In essence, it's a bookkeeping tool for electrons during chemical reactions.

During a redox reaction, elements undergo changes in oxidation number indicating the transfer of electrons between species. For example, when an element's oxidation number increases, we say it has been oxidized; this means it loses electrons. Conversely, when the oxidation number decreases, the element is reduced, indicating that it gains electrons.

In the context of the exercise, when comparing the oxidation numbers before and after the reaction, the changes signify whether a redox reaction has occurred. For instance, in reaction (b), iodine (I) changes its oxidation number from -1 to +5, suggesting it has lost electrons and has been oxidized, while chlorine (Cl) reduces its number from +1 to -1, indicating a gain of electrons and therefore reduction.

Assigning Oxidation Numbers

Assigning oxidation numbers to the elements in a chemical compound is essential for identifying redox reactions. It helps in determining how electrons are distributed in a molecule or ion and which atoms have undergone a change in their oxidation state.

To assign oxidation numbers, follow several rules: The oxidation number of a free element is always zero, for a monatomic ion, it is equal to the ion's charge, and for hydrogen it is +1 (except when bonded to metals in hydrides where it is -1). In compounds, oxygen generally has an oxidation number of -2 (except in peroxides and with fluorine), and halogens usually have an oxidation number of -1, unless they are combined with oxygen or other halogens.

By applying these rules, we can assign oxidation numbers to all atoms in a reaction and track how they change. This is the first step taken in solving the problem from the exercise, and it is crucial for diagnosing whether a reaction is a redox process.

Balancing Chemical Equations

Balancing chemical equations is a critical skill in chemistry that ensures the law of conservation of mass is respected. The atoms of each element should have the same number on both the reactant and product sides of the equation. Balancing becomes more complex when dealing with redox reactions, as both mass and charge must balance.

One approach to balancing redox equations is the half-reaction method, which involves separating the reaction into oxidation and reduction half-reactions, balancing each for mass and charge, and then combining them back together. Alternatively, the oxidation number method, which relates to the changes in oxidation numbers, can also be used wherein the changes in oxidation states are used to balance the reaction.

In the exercise provided, each balanced equation was checked for changes in oxidation numbers as part of determining whether it represented a redox reaction, showing the interdependent nature of these concepts.