CHE140 Introduction to Redox Reactions and Stoichiometry

Final Exam and Unit Administration

• Examination Details:

  • The final exam for CHE140 is scheduled for Monday, June 8.

  • The exam will take place in the morning session. Students with My Access plans specifying a preference for afternoon exams will be contacted by Access and Inclusion.

  • The format is a face-to-face, on-campus, closed-book examination. No personal notes are permitted; however, a data sheet will be provided.

  • A practice exam, frequently asked questions (FAQs), and exam-style practice problems will be released two weeks before the exam, during Week 14.

• Assessment Strategy and Weighting:

  • Total coursework (Mastering Chemistry, workshop quizzes, and lab reports) contributes 50% of the final grade.

  • Students can potentially pass the course before the final exam by maximizing marks in these during-semester assessments.

  • This approach, termed the "drip drip technique," involves continuous study through small, frequent assessments to reduce the performance pressure required on the final exam.

  • There is no hurdle requirement for the final exam in this unit.

• Upcoming Resources:

  • A reaction guide (2-3 pages) will be provided to compare precipitation, acid-base, and redox reactions side-by-side to assist in identifying reaction types out of context.

  • Updates for Week 9 lecture slides (mathematical derivations from the document camera), Workshop 3 answers, and workshop lecture question answers will be released by early next week.

• MyExperience Surveys:

  • Surveys are open for unit feedback. Students are entered into a draw for a $100$ gift card upon completion. Submissions can be started now and submitted after exams.

Stoichiometry and Yield Calculations Recap

• Core Definition: Stoichiometry refers to the use of reaction equations to measure chemicals or elements. It utilizes the coefficients in front of compounds to establish molar relationships.

  • Example: In the combustion of octane, every $2$ moles of octane produce $16$ moles of carbon dioxide ($CO_2$).

• Primary Conversion Equations:

  • Mass and Moles: $n = \frac{m}{M}$

  • Concentration and Moles: $C = \frac{n}{V}$

  • Ideal Gas Law: $PV = nRT$

• Limiting Reagents and Yields:

  • Limiting Reagent: The reactant that is completely consumed first, limiting the amount of product formed. This is determined by comparing the moles of product possible from each available reactant; the lower outcome identifies the limiting reagent.

  • Theoretical Yield: The maximum amount of product that can be produced based on the limiting reagent.

  • Percentage Yield: A ratio of the actual yield over the theoretical yield.

  • $\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

Solution Stoichiometry and Ion Concentration

• Process for Solution Problems:

  • Step 1: Ensure the chemical equation is balanced.

  • Step 2: Convert given information (e.g., concentration and volume) into moles ($n = C \times V$). Volume must be in liters for calculations ($1000\,mL = 1\,dm^3$).

  • Step 3: Identify the limiting reagent using molar ratios.

  • Step 4: Determine excess reagents and products formed using a table of initial, change, and final molar amounts.

• Example Case: Mixing $NaOH(s)$ and $HCl(aq)$:

  • Reaction: $NaOH(s) + HCl(aq) \rightarrow H_2O(l) + NaCl(aq)$

  • Initial Conditions: $10\,g$ of $NaOH$ ($M = 39.997\,g\,mol^{-1}$) mixed with $100\,mL$ of $0.5\,M\,HCl$.

  • Moles of $HCl$: $0.5\,mol\,dm^{-3} \times 0.1\,dm^3 = 0.05\,mol$.

  • Moles of $NaOH$: $10\,g / 39.997\,g\,mol^{-1} = 0.25\,mol$.

  • Limiting Reagent: Since the ratio is $1:1$, $HCl$ ($0.05\,mol$) is the limiting reagent, and $NaOH$ ($0.25\,mol$) is in excess.

• Ion Concentrations Post-Reaction:

  • After the reaction, species split into ions in aqueous solution.

  • $Na^+$ comes from both the excess $NaOH$ and the product $NaCl$. Total moles of $Na^+ = 0.25\,mol$.

  • $OH^-$ remains from the excess $NaOH$. Moles of $OH^- = 0.25 - 0.05 = 0.20\,mol$.

  • $Cl^-$ comes from the product $NaCl$. Moles of $Cl^- = 0.05\,mol$.

  • Final concentration for each ion is calculated as $C = \frac{n}{V_{total}}$, where $V_{total} = 0.1\,dm^3$.

Ionic Compounds and Polyatomic Ions

• Ionic Bonds: Characterized by the transfer of electrons from a metal (left side of periodic table) to a non-metal (right side).

  • Potassium Chloride ($KCl$): Potassium ($K$) in Group 1 loses one electron to become $K^+$; Chlorine ($Cl$) in Group 17 gains an electron to become $Cl^-$.

• Structure: Ionic compounds exist as extended 3D networks called lattices, rather than discrete molecules. The formula unit (e.g., $KCl$ or $MgF_2$) is the smallest ratio of ions that results in a net neutral charge.

• Common Charges:

  • Group 1A: $+1$

  • Group 2A: $+2$

  • Group 3A: $+3$ (e.g., Aluminium)

  • Group 7A (17): $-1$

  • Group 6A (16): $-2$

  • Group 5A (15): $-3$

• Transition Metals: Can exhibit multiple charges (e.g., Iron can be $Fe^{2+}$ or $Fe^{3+}$).

• Polyatomic Ions to Memorize:

  • Hydroxide ($OH^-$)

  • Nitrate ($NO_3^-$)

  • Sulfate ($SO_4^{2-}$)

  • Carbonate ($CO_3^{2-}$)

  • Ammonium ($NH_4^+$)

Introduction to Redox Reactions

• Definition: Reduction-Oxidation (Redox) reactions involve the transfer of electrons between species.

• OIL RIG Mnemonic:

  • Oxidation Is Loss of electrons.

  • Reduction Is Gain of electrons.

• Simultaneity: Oxidation and reduction must occur together. If one species loses electrons, another must gain them.

• Biological and Practical Applications:

  • Batteries: Lithium-ion batteries function via the oxidation of lithium metal ($Li \rightarrow Li^+ + e^-$), creating a flow of electrons (electricity).

  • Cellular Respiration: Glucose is oxidized and oxygen is reduced to generate energy via the movement of ions.

  • Environmental Reactions: Iron rusting, bleaching hair, and fruit browning (oxidation of apples).

Rules for Assigning Oxidation Numbers

• Purpose: Oxidation numbers are used for "electron counting" to track the movement of electrons in a reaction.

• The Rules (Applied in order):

  1. Elements in their neutral, elemental form have an oxidation state of $0$ (e.g., $Cu(s)$, $O_2(g)$, $I_2(s)$).

  2. Monoatomic ions have an oxidation state equal to their charge (e.g., $Ca^{2+}$ is $+2$).

  3. Oxygen in compounds is $-2$ (Exceptions include peroxides or compounds with Fluorine).

  4. Hydrogen is $+1$ in most compounds (Exception: Hydrides like $NaH$, where it is $-1$).

  5. Neutral compounds must have a sum of oxidation states equal to $0$. Polyatomic ions have a sum equal to the ion's charge.

  6. In molecular compounds, the more electronegative element is assigned the charge it would have as an anion.

• Examples:

  • Ammonia ($NH_3$): $H = +1$. Sum must be zero: $N + 3(+1) = 0 \rightarrow N = -3$.

  • Carbon Dioxide ($CO_2$): $O = -2$. Sum must be zero: $C + 2(-2) = 0 \rightarrow C = +4$.

  • Nitric Acid ($HNO_3$): $H = +1$, $O = -2$. Sum: $(+1) + N + 3(-2) = 0 \rightarrow N = +5$.

Identifying Redox Agents

• Oxidized Species: The atom whose oxidation number increases (becomes more positive). This species is the Reducing Agent.

• Reduced Species: The atom whose oxidation number decreases (becomes more negative). This species is the Oxidizing Agent.

The Half-Equation Method for Balancing Redox Reactions

1. Identify and Split: Determine which species are being oxidized and reduced. Write individual half-equations.

2. Balance Atoms: Balance all atoms except Oxygen and Hydrogen.

3. Balance Oxygen: Add $H_2O$ to the side lacking oxygen.

4. Balance Hydrogen: Add $H^+$ (protons) to the side lacking hydrogen.

5. Balance Charge: Add electrons ($e^-$) to the more positive side of each half-equation.

6. Equalize Electrons: Multiply the half-equations by integers so the number of electrons lost in oxidation equals the number gained in reduction.

7. Combine and Cancel: Add the equations together and cancel species appearing on both sides (typically electrons, and sometimes $H_2O$ or $H^+$).

• Worked Example: Iodide and Dichromate:

  • Oxidation: $2I^- \rightarrow I_2 + 2e^-$

  • Reduction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

  • Normalization: Multiply the iodine equation by $3$ to reach $6$ electrons.

  • Final Combined Equation: $6I^- + Cr_2O_7^{2-} + 14H^+ \rightarrow 3I_2 + 2Cr^{3+} + 7H_2O$