Complex Redox Eqn

REDOX EQUATIONS AND REACTIONS OF COMPLEXES (Questions) Your feedback on these self-help problems is appreciated. Click here to send an e-mail. For on-line tutorial help with redox questions, see the chemCAL module, Electrochemistry.		Shortcut to Questions Q: 1 2 3 4 5 6 7
1	Write an equation for the reaction which occurs when excess 4 M ammonia solution is added to each of the following. If no reaction occurs, write "no reaction". (1.) a water solution of silver nitrate (2.) a water solution of copper(II) nitrate (3.) a water solution of zinc chloride (4.) a water solution of tetraamminecopper(II) sulfate (5.) a water solution of diammonium copper(II) sulfate
2	Arsenic(III) oxide is amphoteric. Write a balanced equation for each of two reactions that establish this fact.
3	What is the oxidation number of the underlined central atom in each of the following species? For on-line help with the concept of oxidation numbers, see the chemcal screen titled "Oxidation Numbers A", and in particular, the hint to that screen where the rules for deriving oxidation numbers are set out. (1.) Cs⁺, (2.) Bi³⁺, (3.) N^3-, (4.) SO₄^2-, (5.) NH₃, (6.) N₂H₄, (7.) NO₂, (8.) NO₂^-, (9.) PbO₂, (10.) BaO, (11.) H₂O₂, (12.) NaH, (13.) HClO₄, (14.) HClO, (15.) HBrO₃, (16.) [Fe(CN)₆]^4-, (17.) [Cu(CN)₄]^3-, (18.) [SnCl₆]^2-, (19.) K₃[Fe(CN)₆], (20.) K₂[PtCl₆]
4	Using the ion-electron half-reaction method balance the following "skeleton" redox equations: (1.) Cr₂O₇^2- + H₂S(aq) → Cr³⁺ + S(s) (acidic solution) (2.) Cr₂O₇^2- + I^- → Cr³⁺ + I₂(s) (acidic solution) (3.) Fe³⁺ + I^- → Fe²⁺ + I₂(s) (4.) NO₂^- + MnO₄^- → NO₃^- + MnO₂(s) (basic solution) (5.) Br₂(l) + OH^- → BrO₃^- + Br^- (hot basic solution) (6.) H₂S(aq) + NO₃^- → S(s) + NO₂(g) (acidic solution) (7.) PbS(s) + NO₃^- → Pb²⁺ + S(s) + NO(g) (dilute acidic solution) (8.) MnO₄^- + Cl^- → Mn²⁺ + Cl₂(g) (acidic solution) (9.) SO₂(g) + NO₃^- → SO₄^2- + NO(g) (acidic solution) (10.) ClO₃^- → ClO₂ + ClO₄^- (acidic solution) (11.) H₂O₂ + I₂(s) → HIO₃ (12.) S₂O₃^2- + I₂(s) → S₄O₆^2- + I^- (13.) Al(s) + OH^- → Al(OH)₄^- + H₂(g) (strongly basic solution) (14.) Ag(s) + H₂S(g) → Ag₂S(s) + H₂(g) (15.) Pb(s) + PbO₂(s) → Pb²⁺ (acidic solution)
5	Using the ion-electron half-reaction method balance the following redox equations for which all reactants and products are given, and underline all reductants. (1.) H₃AsO₄ + Sn²⁺ + H⁺ → H₃AsO₃ + Sn⁴⁺ + H₂O (2.) NO₃^- + Fe²⁺ + H⁺ → NO(g) + Fe³⁺ + H₂O (3.) N₂O₄(g) + Br^- + OH^- → NO₂^- + BrO₃^- + H₂O
6	Write a balanced equation for each of the following redox reactions (1.) The oxidation of iron(II) ion to iron(III) ion by chlorine. (2.) The oxidation of nitrogen(II) oxide (NO) to nitrogen dioxide by concentrated nitric acid. (3.) The oxidation of tin(II) ion by potassium dichromate in dilute aqueous sulfuric acid solution. (4.) The reduction of iron(III) ion to iron(II) ion by hydriodic acid. (5.) The oxidation of iron(II) ion by dichromate ion in dilute sulfuric acid solution. (6.) The oxidation of hydrogen peroxide by permanganate ions in dilute sulfuric acid solution. (7.) The action of concentrated nitric acid on copper giving nitrogen dioxide. (8.) The action of excess warm dilute sodium hydroxide solution on zinc.
7	(1.) Give five oxidation states of manganese. (2.) When dilute hydrochloric acid is added to a water solution of potassium manganate, the products are manganese dioxide and permanganate ion. Write an equation for this redox reaction.

REDOX EQUATIONS (Answers)
1	(1.) Ag⁺ + NH₃(aq) + H₂O → AgOH(s) + NH₄⁺, followed by: AgOH(s) + 2NH₃(aq) → [Ag(NH₃)₂⁺] + OH^-. This is an example of a reaction producing a product which can react further with the added ammonia (competing equilibria). In the first reaction ammonia acts as a base (ie a source of hydroxide ions), whilst in the second it acts as a ligand, coordinating to the silver atom. (2.) Cu²⁺ + 2NH₃(aq) + 2H₂O → Cu(OH)₂(s) + 2NH₄⁺, followed by: Cu(OH)₂(s) + 4NH₃(aq) → [Cu(NH₃)₄²⁺] + 2OH^- (3.) Zn²⁺ + 2NH₃(aq) + 2H₂O → Zn(OH)₂(s) + 2NH₄⁺, followed by: Zn(OH)₂(s) + 4NH₃(aq) → [Zn(NH₃)₄²⁺] + 2OH^- (4.) No reaction. Note that tetraamminecopper(II) is the ion produced in part (2.), and is the final product between copper(II) ions and NH₃. (5.) Cu²⁺ + 2NH₃(aq) + 2H₂O → Cu(OH)₂(s) + 2NH₄⁺, followed by: Cu(OH)₂(s) + 4NH₃(aq) → [Cu(NH₃)₄²⁺] + 2OH^-
2	(1.) As₂O₃(s) + 6H⁺ → 2As³⁺ + 3H₂O (2.) As₂O₃(s) + 3H₂O + 2OH^- → 2[As(OH)₄^-] An amphoteric compound is one that can react with both H⁺ and OH^-. In the first equation As₂O₃ reacts with H⁺ to form the arsenic(III) cation. In the second equation As₂O₃ reacts with OH^- to form the tetrahydroxoarsenate(III) anion.
3	(1.) +I, (2.) +III, (3.) -III, (4.) +VI, (5.) -III, (6.) -II, (7.) +IV, (8.) +III, (9.) +IV, (10.) +II, (11.) -I, (12.) -I, (13.) +VII, (14.) +I, (15.) +V, (16.) +II, (17.) +I, (18.) +IV, (19.) +III, (20.) +IV
4	A worked solution is given for part (1.) - the other problems are approached by applying the same steps. (1.) There are five steps which provide the easiest way to balance redox equations: Step 1: Divide the unbalanced equation Cr₂O₇^2- + H₂S → Cr³⁺ + S into two half-reactions involving the species oxidised and reduced: Cr₂O₇^2- → Cr³⁺ H₂S → S Step 2: Balance atoms and charges in each half-reaction. The order for balancing is all atoms other than O and H, then O (by adding H₂O), then H (by adding H⁺), and finally charges: Cr₂O₇^2- + 14H⁺ +6e^- → 2Cr³⁺ + 7H₂O H₂S → S + 2H⁺ + 2e^- Step 3: Multiply each half-reaction through by an integer to make electrons lost in one half-reaction equal the number gained in the other (if necessary): Cr₂O₇^2- + 14H⁺ +6e^- → 2Cr³⁺ + 7H₂O 3H₂S → 3S + 6H⁺ + 6e^- .... multiplied through by 3 to give 6e^- lost. Step 4: Add the two half-reactions together, cancelling any common terms, and adding states of matter: Cr₂O₇^2-(aq) + 3H₂S(aq) + 8H⁺(aq) → 2Cr³⁺(aq) + 3S(s) + 7H₂O(l) Step 5: An extra step here to check for errors is a good idea, especially with complicated reactions - simply check that the atoms on each side actually balance. This procedure may seem long, but a systematic way to balance redox equations is essential. It is easy to see that with so many reactants, products, and coefficients, balancing a whole redox equation by trial and error would be very frustrating - without taking electron transfer into account finding the actual, correct equation is practically impossible! (2.) Cr₂O₇^2- + 6I^- + 14H⁺ → 2Cr³⁺ + 3I₂(s) + 7H₂O (3.) 2Fe³⁺ + 2I^- → 2Fe²⁺ + I₂(s) (4.) 3NO₂^- + 2MnO₄^- + H₂O → 3NO₃^- + 2MnO₂(s) + 2OH^- (5.) 3Br₂(l) + 6OH^- → BrO₃^- + 5Br^- + 3H₂O (6.) H₂S(aq) + 2NO₃^- + 2H⁺ → S(s) + 2NO₂(g) + 2H₂O (7.) 3PbS(s) + 2NO₃^- + 8H⁺ → 3Pb²⁺ + 3S(s) + 2NO(g) + 4H₂O (8.) 2MnO₄^- + 10Cl^- + 16H⁺ → 2Mn²⁺ + 5Cl₂(g) + 8H₂O (9.) 3SO₂(g) + 2NO₃^- + 2H₂O → 3SO₄^2- + 2NO(g) + 4H⁺ (10.) 3ClO₃^- + 2H⁺ → 2ClO₂ + ClO₄^- + H₂O (11.) 5H₂O₂ + I₂(s) → 2HIO₃ + 4H₂O (12.) 2S₂O₃^2- + I₂(s) → S₄O₆^2- + 2I^- (13.) 2Al(s) + 2OH^- + 6H₂O → 2Al(OH)₄^- + 3H₂(g) (14.) 2Ag(s) + H₂S(g) → Ag₂S(s) + H₂(g) (15.) Pb(s) + PbO₂(s) + 4H⁺ → 2Pb²⁺ + 2H₂O
5	(1.) H₃AsO₄ + Sn²⁺ + 2H⁺ → H₃AsO₃ + Sn⁴⁺ + H₂O (2.) NO₃^- + 3Fe²⁺ + 4H⁺ → NO(g) + 3Fe³⁺ + 2H₂O (3.) 3N₂O₄(g) + Br^- + 6OH^- → 6NO₂^- + BrO₃^- + 3H₂O
6	(1.) 2Fe²⁺ + Cl₂(g) → 2Fe³⁺ + 2Cl^- (2.) NO(g) + 2NO₃^- + 2H⁺ → 3NO₂(g) + H₂O (3.) 3Sn²⁺ + Cr₂O₇^2- + 14H⁺ → 3Sn⁴⁺ + 2Cr³⁺ + 7H₂O (4.) 2Fe³⁺ + 2I^- → 2Fe²⁺ + I₂(s) (5.) 6Fe²⁺ + Cr₂O₇^2- + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O (6.) 5H₂O₂ + 2MnO₄^- + 6H⁺ → 5O₂(g) + 2Mn²⁺ + 8H₂O Note that the above equation is the only correct one since it is derived from a consideration of electrons gained and lost. Other equations can be written, derived by trial and error - these are incorrect! (7.) Cu(s) + 2NO₃^- + 4H⁺ → Cu²⁺ + 2NO₂(g) + 2H₂O (8.) Zn(s) + 2OH^- + 2H₂O → Zn(OH)₄^2- + H₂(g)
7	(1.) Any five of: 0, +II, +III, +IV, +VI, +VII (2.) 3MnO₄^2- + 4H⁺ → 2MnO₄^- + MnO₂(s) + 2H₂O. In this reaction MnO₄^2- acts as both oxidant and reductant.