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

First, assign oxidation numbers to all elements. PBr3: P: +3 (as it is combined with three bromine atoms, each with -1 oxidation state) Br: -1 H2O: H: +1 O: -2 H3PO3: H: +1 P: +3 (as it is combined with three oxygen atoms, each with -2 oxidation state, and three hydrogen atoms, each with +1 oxidation state) O: -2 HBr: H: +1 Br: -1 Now, let's compare the oxidation states: P: +3 in PBr3 and +3 in H3PO3 (No change) Br: -1 in PBr3 and -1 in HBr (No change) H: +1 in H2O and +1 in HBr and H3PO3 (No change) O: -2 in H2O and -2 in H3PO3 (No change) As there is no change in oxidation state for any element, this reaction does not involve oxidation-reduction.

First, assign oxidation numbers to all elements. NaI: Na: +1 I: -1 HNO3: H: +1 N: +5 (as it is combined with three oxygen atoms, each with -2 oxidation state, and one hydrogen atom with +1 oxidation state) O: -2 NaIO3: Na: +1 I: +5 (as it is combined with three oxygen atoms, each with -2 oxidation state) O: -2 HNO2: H: +1 N: +3 (as it is combined with two oxygen atoms, each with -2 oxidation state, and one hydrogen atom with +1 oxidation state) O: -2 Now, let's compare the oxidation states: Na: +1 in NaI and +1 in NaIO3 (No change) I: -1 in NaI and +5 in NaIO3 (Change) H: +1 in HNO3 and +1 in HNO2 (No change) N: +5 in HNO3 and +3 in HNO2 (Change) O: -2 in HNO3 and -2 in NaIO3 and HNO2 (No change) As there is a change in the oxidation state for I and N, this reaction involves oxidation-reduction. Iodine (I) is oxidized from -1 to +5, and Nitrogen (N) is reduced from +5 to +3 in the process.

First, assign oxidation numbers to all elements. SO2: S: +4 (as it is combined with two oxygen atoms, each with -2 oxidation state) O: -2 HNO3: H: +1 N: +5 (as it is combined with three oxygen atoms, each with -2 oxidation state, and one hydrogen atom with +1 oxidation state) O: -2 H2O: H: +1 O: -2 H2SO4: H: +1 S: +6 (as it is combined with four oxygen atoms, each with -2 oxidation state, and two hydrogen atoms, each with +1 oxidation state) O: -2 NO: N: +2 (as it is combined with one oxygen atom, with -2 oxidation state) O: -2 Now, let's compare the oxidation states: S: +4 in SO2 and +6 in H2SO4 (Change) O: -2 in SO2, HNO3, H2O, H2SO4, and NO (No change) H: +1 in HNO3, H2O, and H2SO4 (No change) N: +5 in HNO3 and +2 in NO (Change) As there is a change in the oxidation state for S and N, this reaction involves oxidation-reduction. Sulfur (S) is oxidized from +4 to +6 in the process, and Nitrogen (N) is reduced from +5 to +2.