Principles 2 Week 13

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Monday April 8, 2024 Day 58 REDOX Review

Textbook Readings 4.9: Oxidation-Reduction Reactions	Course Lectures 20.1 pdf Video* Oxidation Numbers 20.2 pdf Video* Oxidation Reduction Rxns
Objectives 1. Determine oxidation states of elements appearing in elemental form, as ions and in chemical compounds. 2. Use oxidation states to determine what species is oxidized and reduced in chemical REDOX reactions. 3. For oxidized and reduced elements, correctly determine the total number of electrons lost or gained for the REDOX reaction. 4. In REDOX settings, identify the oxidizing and reducing agents	Introduction to REDOX reactions and Oxidation States Introduction to REDOX reactions and Oxidation States
Homework Problems 51.1 Determine the oxidation states of all elements in each of the following: a. H₂CO₃ b. N₂ c. Zn(OH)₄^2- d. NO₂^- e. LiH f. Fe₂O₃ g. K⁺ h. SO₃^2- i. H₂O₂ j. O₂ k. PbO₂ l. MnO₄^- 51.2 Identify the element being oxidized and reduced in the following reactions. a. Cr⁺ + Sn⁴⁺ → Cr³⁺ + Sn²⁺ b. 2 ClO₂ + 2 I ^- → 2 ClO₂^- + I₂ c. 2 As + 3 Cl₂ → 2 AsCl₃ d. O₂ + 4 H⁺ + 2 Zn → 2 H₂O + 2 Zn²⁺ e. 6 KOH + 3Cl₂ → KClO₃ + 5 KCl + 3 H₂O f. 5 NaClO₃ + 3 H₂O + 3 I₂ → 6 HIO₃ + 5 NaCl 51.3 Identify the oxidizing agent and the reducing agent for each of the reactions above. 51.4 For each species identified in 51.2, write the oxidation or reduction as balanced chemical equations that include electrons as products (oxidation) or reactants (reduction) 51.5 All of the above REDOX reactions are properly balanced. To be balanced, the TOTAL electrons produced in oxidation must equal the TOTAL electrons consumed by the reduction. Examine the reactions above and their coefficients/subscripts carefully. Determine the TOTAL number of electrons produced and consumed. Click and drag the region below for correct answers 51.1 a. H: +1 O: -2 C: +4 b. N₁: 0 N₂: 0 c. Zn: +2 O:-2 H: +1 d. N: +3 O: -2 e. Li: +1 H: -1 (Hydrides: Exception to the hydrogen guidelines) f. Fe: +3 O: -2 g. K: +1 h. S: +4 O: -2 i. H: +1 O: -1 (Peroxides: Exception to oxygen guidelines) j. O: 0 O: 0 k. Pb: +4 O: -2 l. Mn: +7 O: -2 51.2 a. Cr⁺ is oxidized to form Cr³⁺ Sn⁴⁺ is reduced to form Sn²⁺ b. I^- is oxidized to form I⁰ Cl⁺⁴ is reduced to form Cl⁺³ c. As⁰ is oxidized to form As⁺³ Cl⁰ is reduced to form Cl^-1 d. Zn⁰ is oxidized to form Zn²⁺ O⁰ is reduced to form O^-2 e. Cl⁰ is oxidized to form Cl⁵⁺ Cl⁰ is reduced to form Cl^-1 f. I⁰ is oxidized to form I⁵⁺ Cl⁵⁺ is reduced to form Cl^-1 51.3 a. Oxidizing agent: Sn⁴⁺ Reducing agent: Cr⁺ b. Oxidizing agent: ClO₂ Reducing agent: I- c. Oxidizing agent: Cl₂ Reducing agent: As d. Oxidizing agent: O₂ Reducing agent: Zn e. Oxidizing agent: Cl₂ Reducing agent: Cl₂ f. Oxidizing agent: NaClO₃ Reducing agent: I₂ ***Note the entire chemical formula is used when naming oxidizing and reducing agents.d. N: +3, O: -2 e. Li: +1 H: -1 f. Fe: +8/3 O: -2 51.4 a. Oxidation: Cr⁺ → Cr³⁺ + 2 e^- Reduction: Sn⁴⁺ + 2 e^- → Sn²⁺ b. Oxidation: I^- → I⁰ + 1 e^- Reduction: Cl⁴⁺ + 1 e^- → Cl³⁺ c. Oxidation: As⁰ → As⁺³ + 3 e^- Reduction: Cl⁰ + 1 e^- → Cl^-1 d. Oxidation: Zn⁰ → Zn⁺² + 2 e^- Reduction: O⁰ + 2 e^- → O^-2 e. Oxidation: Cl⁰ → Cl⁺⁵ + 5 e^- Reduction: Cl⁰ + 1 e^- → Cl^-1 f. Oxidation: I⁰ → I⁺⁵ + 5 e^- Reduction: Cl⁺⁵ + 6 e^- → Cl^-1 (Danger: Make sure you understand why)⁺ is being oxidized Sn⁴⁺ is being reduced 51.5 a. 2 electron produced by oxidation and 2 electron consumed by reduction b. 2 electrons produced by oxidation and 2 electrons consumed by reduction c. 6 electrons produced by oxidation and 6 electrons consumed by reduction d. 4 electrons produced by oxidation and 4 electrons consumed by reduction e. 5 electrons produced by oxidation and 5 electrons consumed by reduction f. 30 electrons produced by oxidation and 30 electrons consumed by reduction. Fe is being oxidized Hg²⁺ is being reduced c. As is being oxidized Cl₂ is being reduced 3. a. Cr⁺ is the reducing agent Sn⁴⁺ is the oxidizing agent b. Fe is the reducing agent Hg²⁺ is the oxidizing agent c. As is the reducing agent Cl₂ is the oxidizing agent

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Tuesday April 9, 2024 Day 59 Balancing REDOX Reactions

Textbook Readings 19.2: Balancing REDOX Equations	Course Lectures 20.3 pdf Video* Balancing REDOX reactions
Balancing REDOX reactions (Acidic) Balancing REDOX reactions (Acidic)	Balancing REDOX reactions (Basic) Balancing REDOX reactions (Basic)
Objectives 1. Correctly separate a REDOX reaction into ½ reactions 2. Balance ½ reactions with respect to a. Other elements b. Oxygen (with H₂O) c. Hydrogen with H⁺ d. Electrons with e^- 3. Adjust ½ reactions such that electrons gained (reduction) equal electrons lost (oxidation) 4. Combine ½ reactions and simplify 5. Adjust with OH^- for a basic environment as needed.
Homework Problems 52.1 a. For the following unbalanced REDOX reaction write the two unbalanced ½ cell reaction HCOOH + MnO₄^- → CO₂ + Mn²⁺ (Basic solution) b. Balance the ½ cell reactions from part "a" with respect to all elements except oxygen and hydrogen. c. Balance the ½ cell reaction from part "b" with respect to oxygen d. Balance the ½cell reactions from part "c" with respect to hydrogen e. Balance the ½ cell reactions from part "d" with respect to electrons f. Multiply both ½ cell reactions as needed to get the electrons on the product side of the oxidation reaction to equal the number of electrons on the reactant side of the reduction reaction. g. Combine the two, balanced ½ reactions and cancel terms as necessary. This is the balanced "acidic" form of this reaction. h. To balance as a "basic" reaction, add OH- to both sides of the reaction to nuetralize the H+ ions that appear. Their reaction forms water. Cancel product and reactant H₂O as needed This is the balanced "basic " form of this reaction. 51.2 Balance the following three REDOX reactions: a. ClO₂^-1 → ClO₂ + Cl^-1 (Acidic solution) b. Cr(OH)₃ + Br₂ → CrO₄^2- + Br^-1 (Basic Solution) c. H₂O₂ + ClO₂ → O₂ + ClO₂^- (Basic Solution) Answers: Click and drag in the space below 52.1 a. MnO₄^- → Mn²⁺ & HCOOH → CO₂ b. MnO₄^- → Mn²⁺ & HCOOH → CO₂ c. MnO₄^- → Mn²⁺+ 4H₂O & HCOOH → CO₂ d. 8 H⁺ + MnO₄^- → Mn²⁺+ 4H₂O & HCOOH → CO₂ + 2 H⁺ e. 8 H⁺ + MnO₄^- + 5e^- → Mn²⁺+ 4H₂O & HCOOH → CO₂ + 2 H⁺ + 2e^- f. 8 H⁺ + MnO₄^- + 5e^- → Mn²⁺+ 4H₂O & HCOOH → CO₂ + 2 H⁺ + 2e^- multiply by X2 Multiply X5 16 H⁺ + 2 MnO₄^- + 10e^- → 2 Mn²⁺+ 8 H₂O & 5 HCOOH → 5 CO₂ + 10 H⁺ + 10e^- g. 6 H⁺ + 2 MnO₄^- + 5 HCOOH → 2 Mn²⁺+ 8 H₂O + 5 CO₂ h. 6 H⁺ + 2 MnO₄^- + 5 HCOOH → 2 Mn²⁺+ 8 H₂O + 5 CO₂ add 6 OH^- add 6 OH^- 6 H₂O + 2 MnO₄^- + 5 HCOOH → 2 Mn²⁺+ 8 H₂O + 5 CO₂ + 6 OH^- Cancel H₂O 2 MnO₄^- + 5 HCOOH → 2 Mn²⁺+ 2 H₂O + 5 CO₂ + 6 OH^- That's it! :) 51.2 a. 5 ClO₂^- + 4 H⁺ → 4 ClO₂ + Cl^- + 2 H₂O b. 10 OH^- + 2 Cr(OH)₃ + 3 Br₂ → 2 CrO₄^2- + 8 H₂O + 6 Br^- c. H₂O₂ + 2ClO₂ + 2 OH^- →2 ClO₂^- + O₂ + 2 H₂O

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Wednesday April 10, 2024 Day 60 Reaction Spontaneity and Electrochemical Cells

Textbook Readings 19.3: Voltaic Cells- Generating Electricity from Spontaneous Chemical Reactions P2: Table of Standard Reduction Potentials _{NOTE: Oxidation is ABOVE reduction in this table.}	Course Lectures 21.1 pdf Video* Spontaneous REDOX reactions 21.2 pdf Video* Electrochemical Cells: Intro
Objectives 1. Determine spontaneous and non-spontaneous versions of REDOX reactions using the table of Standard Reduction Potentials. 2. Given a galvanic cell, identify the oxidized and reduced species. Also identify Anode Cathode and describe all operational aspects of the cell.	Introduction to Galvanic Cells Introduction to Galvanic Cells
Homework Problems 53.1 The table of Standard Reduction Potentials is a list of ½ cell reactions in an order that let's us determine how to write spontaneous REDOX reactions. _{NOTE: The organization of the Standard Reduction tables used on this website are the top/bottom reverse of those refered to in my lecture videos. Oxidation is ABOVE reduction in this table}
In the abbreviated tableabove, both copper and zinc are identified. Because copper appears below zinc in the table, the spontaneous reaction will be one where copper is reduced and zinc is oxidized: Zn_(s) + Cu²⁺_(aq) → Cu_(s) + Zn²⁺_(aq) This is the spontaneous form of the REDOX reaction and it required us to reverse the copper ½ reaction as it appears on the table to reflect copper's oxidation. For each of the following combinations, use the Standard Reduction Potentials Table to write the spontaneous and non-spontaneous versions of the REDOX reaction. a. Zn & Pb b. Cd & Ni c. Fe & Sn d. Mg & Al e. Na & I₂ 53.2 For each of the following situations, determine if a spontaneous REDOX reaction occurs. If a spontaneous reaction does occur, write the reaction. If a spontaneous reaction does NOT occur, indicate this by "striking through" the reaction. a. Solid magnesium placed in a solution of 1.0 M PbCl₂ b. Solid nickel placed in a solution of 1.0 M FeCl₂ c. Solid zinc placed in a solution of 1.0 M CuCl₂ d. Solid lead placed in a solution of 1.0 M AlCl₃
53.3 Gavanic (a.k.a. Voltaic) cells are clever arrangements of REDOX ½ reactions. The oxidation reaction occurs on one side and the reduction reaction occurs on the other side. The entire assembly is known as a "cell". Referring to the cell at right, answer the following questions: a. Which species is oxidized and which is reduced. b. Which electrode is the anode? Which electrode is the cathode?
c. Which electrode gains mass as the cell operates? Which electrode loses mass as the cell operates? d. In which direction do electrons flow through the wire? e. What are the oxidation and reduction half reactions? f. What is the balanced total cell reaction? g. How many electrons flow according to the balanced total cell reaction? h. How do the NO₃^- and K⁺ ions move as the cell operates? 53.4 A voltaic cell is constructed using the following two ½ cells and a KBr salt bridge: Half cell #1: Nickel rod in 1.00 M Nickel chloride solution (NiCl₂) Half cell #2: Copper rod in 1.00 M copper chloride solution (CuCl₂). Draw the cell (beakers, electrodes, wires, etc) and then answer the same questions as in problem 53.3 Answers: Click and drag in the space below 53.1 a. Spontaneous: Zn_(s) + Pb²⁺_(aq) → Zn²⁺_(aq) + Pb_(s) non- Spontaneous: Zn²⁺_(aq) + Pb_(s) → Zn_(s) + Pb²⁺_(aq) b. Spontaneous: Cd_(s) + Ni²⁺_(aq) → Cd²⁺_(aq) + Ni_(s) non- Spontaneous: Cd²⁺_(aq) + Ni_(s) → Cd_(s) + Ni²⁺_(aq) c. Spontaneous: Fe_(s) + Sn⁴⁺_(aq) → Fe²⁺_(aq) + Sn²⁺_(aq) non- Spontaneous: Fe²⁺_(aq) + Sn²⁺_(aq) → Fe_(s) + Sn⁴⁺_(aq) d. Spontaneous: 2Al³⁺_(aq) + 3 Mg_(s) → 2 Al_(s) + 3 Mg²⁺_(aq) non- Spontaneous: 2 Al_(s) + 3 Mg²⁺_(aq) → 2Al³⁺_(aq) + 3 Mg_(s) e. Spontaneous: 2 Na_(s) + I_2(s) → 2 Na⁺_(aq) + 2 I^-_(aq) non- Spontaneous: 2 Na⁺_(aq) + 2 I^-_(aq) → 2 Na_(s) + I_2(s) 53.2 a. This question is asking if the following reaction is spontaneous: Mg_(s) + Pb²⁺_(aq) → Mg²⁺_(aq) + Pb_(s) Where Mg is oxidized and Pb²⁺ is reduced. This is consistant with Mg being lower on the Standard Reduction Table. The reaction is spontaneous and we would observe magnesium metal dissolving and lead metal forming. b. This question is asking if the following reaction is spontaneous: Ni_(s) + Fe²⁺_(aq) → Ni²⁺_(aq) + Fe_(s) Where Ni is oxidized and Fe²⁺ is reduced. This is NOT consistant with Fe being lower on the Standard Reduction Table The reaction is NON spontaneous and we would observe no reaction taking place. c. This question is asking if the following reaction is spontaneous: Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s) Where Zn is oxidized and Cu²⁺ is reduced. This is consistant with Zn being lower on the Standard Reduction Table. The reaction is spontaneous and we would observe zinc metal dissolving and copper metal forming. d. This question is asking if the following reaction is spontaneous: 2 Al³⁺_(aq) + 3 Pb_(s) → 2 Al_(s) + 3 Pb²⁺_(aq) Where Pb is oxidized and Al³⁺ is reduced. This is NOT consistant with Al being lower on the Standard Reduction Table. The reaction is NON spontaneous and we would observe no reaction taking place. 53.3 a. Al is oxidized and Zn²⁺ is reduced based upon their positions on the reduction table b. Oxidation = Anode (-) ... Al electrode Reduction = Cathode (+) ... Zn electrode c. Zn electrode gains mass (...we say that the zinc "plates out") Al electrode loses mass (...we say the aluminum electrode dissolves) d. Electrons flow from the Al electrode (Anode) to the Zn electrode (Cathode) through the wire. e. Oxidation Al_(s) → Al³⁺_(aq) + 3e^- Reduction Zn²⁺_(aq) + 2e^-→ Zn_(s) f. 2Al_(s) + 3Zn²⁺_(aq) → 2Al³⁺_(aq) + 3 Zn_(s) g. 6 electrons flow for every cycle of the reaction. h. The NO₃^-ions move out of the salt bridge and into the Al(NO₃)₃ solution to replace the negative charge that leaves as electrons flow through the wire. The K⁺ ions move out of the salt bridge and into the Zn(NO₃)₂ solution to neutralize the negative charge that develops as electrons flow through the wire and into the Zn ½ cell. 53.4 a. Ni is oxidized and Cu²⁺ is reduced based upon their positions on the reduction table b. Oxidation = Anode (-) ... Ni electrode Reduction = Cathode (+) ... Cu electrode c. Cu electrode gains mass (...we say that the copper "plates out") Ni electrode loses mass (...we say the nickel electrode dissolves) d. Electrons flow from the Ni electrode (Anode) to the Cu electrode (Cathode) through the wire. e. Oxidation Ni_(s) → Ni²⁺_(aq) + 2e^- Reduction Cu²⁺_(aq) + 2e^-→ Cu_(s) f. Ni_(s) + Cu²⁺_(aq) → Ni³⁺_(aq) + 3Cu_(s) g. 2 electrons flow for every cycle of the reaction. h. The Br^-ions move out of the salt bridge and into the NiCl₂ solution to replace the negative charge that leaves as electrons flow through the wire. The Na⁺ ions move out of the salt bridge and into the CuCl₂ solution to neutralize the negative charge that develops as electrons flow through the wire and into the Cu ½ cell.

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Thursday April 11, 2024 Day 61 Abbreviated Electrochemical Cell Diagrams
Textbook Readings 19.3: Voltaic Cells- Generating Electricity from Spontaneous Chemical Reactions P2: Table of Standard Reduction Potentials _{NOTE: Oxidation is ABOVE reduction in this table.}	Course Lectures 21.1 pdf Video* Spontaneous REDOX reactions _{**Note: The REDOX table used in this video is the top to bottom reverse of the one we use in this course. In other words, oxidation is ABOVE reduction in tables like the one at left or available here.} 21.2 pdf Video Electrochemical Cells: Intro
Objectives 1. Construct abbreviated electrochemical cell diagrams including: * Oxidation (left) and Reduction (right) * (aq), (s) and (g) designations * Solution concentrations * Phase barriers \| * Salt bridge \|\| * Conductive electrodes 2. Given an abbreviated electrochemical cell diagram write the oxidation and reduction ½ reactions and the net cell reaction.	Cell Diagrams Cell Diagrams
Homework Problems. 54.1 Consider the following abbreviated cell diagram: Cd_(s) \| Cd(NO₃)_2(aq) (1.0 M) \|\| AgNO_3(aq) (1.0 M) \| Ag_(s) a. Which species is oxidized? Which species is reduced? b. Identify the cathode. Identify the anode. c. What are the respective half cell reactions? d. What is the net cell reaction e. Which electrode increases in mass as the cell operates Which electrode decreases in mass as the cell operates f. Which solution increases in concentration as the cell operates Which solution decreases in concentration as the cell operates g. If a wire were connected across the cell, in what direction would the electrons flow? 54.2 If the following spontaneous REDOX reaction were used in a galvanic cell, what would be the abbreviated cell diagram? Cr_(s) + Fe³⁺_(aq) (2.0 M) → Cr³⁺_(aq) (3.0 M) + Fe_(s) 54.3 In the following abbreviated cell diagram, a platinum electrode is used because the reduction reaction doesn't involve anything that can serve as an electrical conductor. Although the Pt electrode conducts electrical current in/out of the ½ cell, the electrode is completely unreactive and isn't oxidized or reduced. It does serve as a catalytic surface that facilitates the reduction of H⁺ into H₂. Pb_(s) \| Pb²⁺_(aq)(2.5 M) \|\| H⁺_(aq) (1.0 M) \|H_2(g) \| Pt What are the oxidation, reduction and net reactions for this cell? 54.4 If the following REDOX reaction were used in a galvanic cell, what would be the abbreviated cell diagram? Pb_(s) + 2FeCl_3(aq) (1.0 x 10^-3 M) → 2FeCl_2(aq) (3.5 M) + PbCl_2(aq) Answers: Click and drag in the space below 54.1 a. Oxidized: Cd_(s) Reduced: Ag⁺_(aq) b. Anode (-): Cd_(s) Cathode (+): Ag_(s) c. Oxidation: Cd_(s) → Cd²⁺_(aq) + 2 e^- Reduction: Ag⁺_(aq) + 1 e^- → Ag_(s) d. Cd_(s) + 2 Ag⁺_(aq) → Cd²⁺_(aq) + 2 Ag_(s) e. The Cd electrode dissolves (via oxidation) and loses mass The silver electrode gains mass (via reduction) as aqueous Ag⁺ plates out. f. Cd(NO₃)_2(aq) solution increases in concentration AgNO_3(aq) solution decreases in concentration g. Electrons would flow through the wire from the Cd electrode and into the Ag electrode. 54.2 Cr_(s₎ \| Cr³⁺_(aq)(3.00 M) \|\| Fe³⁺_(aq) (2.00 M) \| Fe_(s) 54.3 Oxidation: Pb_(s) → Pb²⁺_(aq) + 2 e^- Reduction: 2H⁺_(aq) + 2 e^- → H_2(g) Net Reaction: Pb_(s) + 2H⁺_(aq) → H_2(g) + Pb²⁺_(aq) 54.4 Pb_(s) \| Pb²⁺_(aq) \|\| Fe³⁺_(aq) (1.0 x 10^-3 M), Fe²⁺_(aq) (3.5 M) \| Pt_(s) Platinum electrode required since neither form of iron is metallic and can serve as an electrode

Friday April 12, 2024 Day 62 Standard Reduction Potentials and the Standard Hydrogen Electrode (SHE)

Textbook Readings 19.4: Standard Reduction Potentials P2: Table of Standard Reduction Potentials _{NOTE: Oxidation is ABOVE reduction in this table.}	Course Lectures 21.3 pdf Video Standard Reduction Potentials _{***Note: The REDOX table used in this video is the top to bottom reverse of the one we use in this course. In other words, oxidation is ABOVE reduction in tables like the one at left or available here.} 21.4 pdf Video Predicting Cell Potentials
Standard Reduction Potentials and the Standard Hydrogen Electrode Standard Reduction Potentials and the Standard Hydrogen Electrode	Standard Reduction Potentials and the Standard Hydrogen Electrode Standard Reduction Potentials and the Standard Hydrogen Electrode
Objectives 1. Navigate the Table of Standard Reduction potentials, identify ½ reactions and determine combinations that occur spontaneously. 2. Use E^o_red values to determine the E^o_cell values for REDOX reactions and electro- chemical cells. 3. Determine if metals dissolve in HCl or HNO₃. 4. Identify a reaction as spontaneous or not based uponits E^o_cell value .	Strong Acids as Oxidizing Agents Strong Acids as Oxidizing Agents
Homework Questions 55.1 What is the standard reduction potential for the standard hydrogen electrode (SHE)? 55.2 Half reactions with negative standard reduction potentials will exhibit oxidation when paired with an SHE. This means that reduction occurs at the SHE. a. What should be seen at the SHE as this happens? b. How should the concentration of H⁺ change and what effect will this have on the pH of the SHE solution? 55.3 What is E°_cell for the following balanced reaction? Zn_(s) + Pb²⁺_(aq) → Zn²⁺_(aq) + Pb_(s) Given: Zn²⁺_(aq)+ 2e^– → Zn_(s) E°_red= –0.7618V Pb²⁺_(aq) + 2e^– → Pb_(s) E°_red₌–0.126V 55.4 What is E°_cell for the following balanced reaction? Al_(s) + Fe³⁺_(aq) → Al³⁺_(aq) + Fe_(s) Given: Fe³⁺_(aq) + 2e^– → Fe_(s) E°_red₌ +0.771V Al³⁺_(aq) + 2e^– → Al_(s) E°_red= –1.676V 55.5 What is E°_cell for the following balanced reaction? Fe³⁺_(aq) + NO_(aq) + 2H₂O_(l) → 4H⁺ _(aq) + Fe_(s) + NO₃^-_(aq) Given: NO₃^–_(aq)+ 4H⁺_(aq) + 3e^– → NO_(aq) + 2H₂O E°_red = +0.960 V Fe³⁺_(aq) + 3e^– → Fe _(s) E°_red = +0.771 V 55.6 What is E°_cell for the following cell configuration? What is the net cell reaction? Cu_(s) \| Cu²⁺_(aq) (1.0 M) \|\| Ag⁺ _(aq) (1.00 M) \| Ag_(s) 55.7 Some metals dissolve in acid and others don't and what happens depends on the identities of both the metal and the acid. For example, Zn_(s) dissolves in HCl_(aq) via the following REDOX reaction: Zn_(s) + 2H⁺_(aq) → Zn²⁺_(aq) + H_2(g) This reaction is spontaneous because Zn is above H₂ in the table of standard reduction potentials. However, copper metal won't dissolve in HCl(aq). In other words, the following reaction does NOT take place (i.e. isn't spontaneous): Cu_(s) + 2H⁺_(aq) → Cu²⁺_(aq) + H_2(g) The reason the reaction isn't spontaneous is that copper is below H₂ in the table of standard reduction potentials. Interestingly, copper metal WILL dissolve in nitric acid (HNO₃) but for different reasons. Although Cu_(s) won't react with H⁺_(aq), it will react with NO₃^-_(aq)via the following NO₃^- reduction ½ reaction: NO₃^-_(aq) + 4H⁺_(aq) + 3e^- --> NO(g) + 2H₂O_(l) E_red = 0.96V Because Cu is higher than NO₃^- on the reduction table, copper is oxidized and dissolves. In other words, Cu(s) dissolves in HNO₃ not because of a reaction with H⁺, but rather because of a reaction with NO₃^-. Find the following metals on the Standard Reduction Table and compare their positions to H⁺ and NO₃^-. For each, say whether they would dissolve in HCl, HNO₃ , both or none. a. Au b. Fe c. Ag d. Pb 1. Which of the following metals would dissolve in nitric acid (HNO₃) but not in HCl? a. Au b. Cu c. Fe d. Ag e. Co f. Mg 2. Nitric acid is known as an "oxidizer" and is stored away from all other chemicals. What is an oxidizer and why are the special storage conditions required? Answers: Click and drag in the space below 55.1 zero volts 55.2 a. Reduction at the SHE can be written as 2H⁺_(aq) + 2e^- → H_2(g) Since hydrogen gas is a product, we should observe additional bubbles of H₂. b. H+ is used up and concentration of H⁺ goes down. With fewer H⁺ ions in solution, the pH of the solution goes UP. 55.3 + 0.637 V 55.4 + 2.431 V 55.5 - 0.189 Volts (note that it is negative. This tells us that the reaction is non spontaneous as written. The reverse reaction would have a E°_cell = +0.189 V and therefore be spontaneous) 55.6 + 0.4577 Volts Cu_(s) + 2 Ag⁺_(aq) → Cu²⁺_(aq) + 2 Ag_(s) 55.7 a. Au (E_red = 1.50 V) No reaction with either HCl or HNO₃ b. Fe (E_red = -0.45 V) Reacts with both HCl and HNO₃ c. Ag (E_red = 0.80 V) Will react with HNO₃ but not HCl d. Pb (E_red = -0.13 V) Will react with both HCl and HNO₃

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