is a transfer of electrons Oxidation-Reduction (Redox) Reactions • These are reactions in which there is a transfer of electrons. You can use the acronym LEO/GER to identify oxidation and reduction processes • LEO – loss of electrons is oxidation • GER – gain of electrons is reduction • Oxidation and reduction must occur together Redox Reactions • Oxidizing agent – the electron accepting substance • The oxidizing agent is REDUCED • Reducing agent – electron donating substance • The reducing agent is OXIDIZED Oxidation-Reduction Reactions Clorox bleach contains a strong oxidizing agent, NaClO, which oxidizes this red dye to a colorless product. Oxidation Numbers • A convenient method of bookkeeping for electrons • Oxidation numbers do not imply ionic charge on an atom • A change in the oxidation number implies a transfer of electrons • Redox reactions can be recognized by the change in the oxidation numbers Assigning Oxidation Numbers • 1. An atom in its elemental state has an oxidation number of 0 • 2. A monatomic ion has an oxidation number equal to its charge • 3. The sum of the oxidation numbers of the atoms must equal the net charge for a molecule or polyatomic ion. • 4. Group 1A and 2A have oxidation states equal to their group – except in metalic form Assigning Oxidation Numbers • 5. Fluorine has an ox. # of -1 in compounds • 6. Hydrogen has an ox. # number of +1 in most compounds (occasionally -1) • 7. Oxygen has an ox. # of -2 in most compounds • 8. Halogens usually have an oxidation number of -1 • Except when Cl, Br, or I is bonded to oxygen. Example • C2O42– • C2 O42– x -2 assign x for C because it is unknown and -2 for O based on rule 7 x2 x4 multiply by 2 and 4 (the number of carbons and oxygens) 2x -8 = -2 equation is equal to the charge of the polyatomic x = +3 So, the oxidation number of carbon in C2O42– is +3 Assign Oxidation Numbers to each atom below. 1. Fe 6. Na2Cr2O7 2. O2 7. ClO4– 3. CO2 8. BrO– 4. MnO4– 9. IO2– 5. Fe2O3 10. SO42– When conflicts occur • When conflicts in the rules occur, assign the first rule and ignore the later, conflicting, rule Oxidation Numbers Molybdenum disulfide, MoS2, is a black powder used as a lubricant. Hydrogen peroxide, H2O2, shown as a 3% aqueous H2O2 is a strong oxidizing agent. What are the oxidation numbers of each element in these two compounds? Oxidation Numbers in a Reaction • Identify which species is oxidized and which is reduced in the following reaction. • 2 Fe2O3(s) + 3C (s) ––> 4 Fe(s) + 3 CO2(g) • -2 0 0 -2 • x3 x2 • -6 -4 • 2x – 6 = 0 x-4=0 • x = +3 x = +4 Oxidation Numbers in a Reaction • Identify which species is oxidized and which is reduced in the following reaction. • 2 Fe2O3(s) + 3C (s) ––> 4 Fe(s) + 3 CO2(g) • +3 -2 0 0 +4 -2 Oxidation Numbers in a Reaction • Identify which species is oxidized and which is reduced in the following reaction. • 2 Fe2O3(s) + 3C (s) ––> 4 Fe(s) + 3 CO2(g) • +3 -2 0 0 +4 -2 • Iron gains electrons, thus, is reduced Oxidation Numbers in a Reaction • Identify which species is oxidized and which is reduced in the following reaction. • 2 Fe2O3(s) + 3C (s) ––> 4 Fe(s) + 3 CO2(g) • +3 -2 0 0 +4 -2 • Iron gains electrons, thus, is reduced • C loses electrons, thus, is oxidized Oxidation Numbers in a Reaction • Identify which species is oxidized and which is reduced in the following reaction. • 2 Fe2O3(s) + 3C (s) ––> 4 Fe(s) + 3 CO2(g) • +3 -2 0 0 +4 -2 • Iron gains electrons, thus, is reduced • C loses electrons, thus, is oxidized • Oxidation and reduction must occur together Balance a redox reaction using the half-reaction method Balancing Redox Reactions • 1. Write the two unbalanced half reactions separately • 2. Balance each half reaction for all atoms other than H and O • 3. Balance both half-reactions for oxygen by adding H2O to the side with less O Balancing Redox Reactions • 4. Balance for H by adding H+ to the side with less H • 5. Balance the charge in both half-reactions by adding electrons to the side of the equation which is more positive • 6. Multiply each half reaction by a scalar to make the number of electrons gained in one half reaction equal to the number lost in the other half reaction. Balancing Redox Reactions • 7. Add the two reactions together and cancel electrons and molecules which occur on both sides of the equation. • 8. In a basic solution, add OH- in the final step to neutralize the H+ ions. • 9. Combine H+ with OH– to form water • 10. Cancel any water possible Example problem • H2C2O4 + MnO4– ––> Mn2+ + CO2 Example problem • H2C2O4 + MnO4– ––> Mn2+ + CO2 • 1. Write two unbalanced half reactions Example problem • H2C2O4 + MnO4– ––> Mn2+ + CO2 • 1) Unbalanced Half reactions: • H2C2O4 ––> ___ CO2 • MnO4– ––> Mn2+ Example problem • 2) Balance the Half reactions for elements other than H and O: • H2C2O4 ––> ___ CO2 • MnO4– ––> Mn2+ Example problem • 2) Balance the Half reactions: • H2C2O4 ––> _2_ CO2 • MnO4– ––> Mn2+ Example problem • 3 & 4) Balance the Half reactions (O & H): • H2C2O4 ––> _2_ CO2 + needs 2 H • MnO4– ––> Mn2+ Example problem • 3 & 4) Balance the Half reactions (O & H): • H2C2O4 ––> _2_ CO2 + 2 H+ • MnO4– ––> Mn2+ Example problem • 3 & 4) Balance the Half reactions (O & H): • H2C2O4 ––> _2_ CO2 + 2 H+ • MnO4– ––> Mn2+ + (needs 4 O atoms) Example problem • 3 & 4) Balance the Half reactions (O & H): • H2C2O4 ––> _2_ CO2 + 2 H+ • MnO4– ––> Mn2+ + 4 H2O Example problem • 3 & 4) Balance the Half reactions (H & O): • H2C2O4 ––> _2_ CO2 + 2 H+ • needs 8H + MnO4– ––> Mn2+ + 4 H2O Example problem • 3 & 4) Balance the Half reactions (H & O): • H2C2O4 ––> _2_ CO2 + 2 H+ • 8 H+ + MnO4– ––> Mn2+ + 4 H2O Example problem • 5) Balance the Half reactions (Charge): • H2C2O4 ––> _2_ CO2 + 2 H+ • 8 H+ + MnO4– ––> Mn2+ + 4 H2O Example problem • 5) Balance the Half reactions (Charge): • H2C2O4 ––> _2_ CO2 + 2 H+ + 2 e – • 5 e + 8 H+ + MnO4– ––> Mn2+ + 4 H2O Example problem • 6) Balance the Half reactions (Charge): • 5[H2C2O4 ––> _2_ CO2 + 2 H+ + 2 e –] • 2[5 e + 8 H+ + MnO4– ––> Mn2+ + 4 H2O] Example problem • 6) Balance the Half reactions (Charge): • • 5H2C2O4 ––> 10 CO2 +10H+ +10 e – 10 e + 16 H+ + 2MnO4– –> 2 Mn2+ +8H2O Example problem • 6 & 7) Cancel and add the two reactions • 5H2C2O4 ––> 10 CO2 +10H+ +10 e – • 10 e + 16 H+ + 2MnO4– –> 2 Mn2+ +8H2O __________6_remain________________________ 5H2C2O4 + 6 H+ + 2MnO4– –> 10 CO2 + 2 Mn2+ + 8H2O Balance the following reactions in an acidic + solution (not all will need H ions) • 1. • 2. • 3. • 4. • 5. Cr (s) + Ni2+ → Cr 3+ (aq) + Ni (s) Cl2 (aq) + Br– → Br2(aq) + Cl– (aq) Zn (s) + NO3- (aq) → Zn2+ (aq) + N2O (aq) MnO4– (aq) + C2H5OH (aq) –> Mn2+ (aq) + C2H4O (aq) H2S (aq) + Cr2O72- (aq) → S (s) + 3 Cr3+ (aq) 1046L Poster guidelines • • General o 1 slide with all the info/sections ▪ Change slide dimensions to 48×36 inches o Fonts/sizes must be readable (minimum font size 20pt.) o Templates available online/on PowerPoint o Example can be found in canvas or google o Proper grammar and full sentences ▪ Must be in third person and passive wherever applies (i.e. methods) o Make sure all figures, charts, tables, and sections are properly labelled o Any copying among students will result in a zero for BOTH students Format o Basic sections ▪ Introduction • Explain theoretical background of the project, purpose, how it will be applied, key terms. ▪ Materials and methods ▪ Results/data • Table 1, Figure 1, PantherAde calculations ▪ Discussions • Restate important data/ results and use them to support conclusions • Sources of error and how to correct for better results. ▪ Conclusion • Future research ▪ References • 0 (for the ENTIRE poster) if none • In proper format/style (ACS) • Minimum of 3 references of which at least one must be a peer reviewed paper hexosaminidase Properties of of Matter and Separations Title Poster Here Samantha Abud, Javier Alfonso, Aalia Thameem, Gabriela Perez-Alvarez Doctors, Researchers, PhDs, etc… Department of Something Science, The Rockefeller University, New York, NYFL CHM 1046L General Chemistry 2 LabFlorida International University, Miami, INTRODUCTION: ● The uncertainty of the different compounds found in the landfill have caused a team of troubleshooters to come and assess the situation. With the help of the analytical team, the unknown compounds will be investigated by taking an in-depth look into the compound’s chemical and physical properties. ● The separation of the different compounds will identify the physical and chemical properties of each element within the substance. Identifying whether something is naturally a solid, liquid, or gas can give an indication of how the compound was formed and if it will be easily separated. ● In this experiment, our team was given an unknown compound to separate the different components that make up the unknown compound. The compound (unknown A) that was given to our team was composed of NaCl, Naphthalene, and Benzoic acid. To separate the compound, our team used vacuum filtration, separating a solid from a liquid with a vacuum beaker(Eni Generalic n.d.) . ● Our team was also asked to determine each components solubility, the ability to dissolve in a solvent at a given temperature, which is an important property of an element(Science daily, n.d). Solubility will be determined by adding H20, HCl, and NaOH to each element in the compound in order to determine if it dissolves. PROCEDURES: Part 1 ● The components of our unknown (Bag A) were given: NaCl, Benzoic acid, and Napthalene. ● NaCl was placed into 3 test tubes, Benzoic acid was placed into 3 separate test tubes, & Naphthalene was placed into 3 separate test tubes, ● In one test tube of each compound, DI water was put into the test tube and swirled. ● In another test tube for each compound, 1 M NaOH was put into the test tube and swirled. ● In the last test tube for each compound, 3 M HCl was put into the test tube and swirled Part 2 NaCl, Benzoic Acid, Naphthalene DI Water/ Vacuum Filtration Insoluble Soluble Benzoic Acid & Naphthalene NaCl 1 M NaOH/ Vacuum Filtration Soluble Benzoic Acid 3 M HCl/ Vac. Fil. Insoluble Naphthalene (Dried) Evaporate NaCl (s) ● Water was first placed into a 250 mL beaker and set on a hot plate to boil. ● 20 mL of DI water was mixed in a beaker with the unknown (bag A). ● The mixture was vacuum filtrated and the supernatant was placed into an evaporating dish and placed on top of the beaker with boiling water leaving NaCl (s). ● The remaining solid was mixed with 1 M NaOh and vacuum filtrated. ● The precipitate remained insoluble as dried Naphthalene. ● The supernatant (Benzoic Acid) was mixed with 3 M HCl and vacuum filtrated resulting in dried Benzoic Acid and a HCl and NaOH in solution. DATA & RESULTS: DISCUSSION: ● Masses of 2.34 g, 1.324 g and 1.666 g were acquired for NaCl, Naphthalene and Benzoic Acid, respectively. The initial mass of the mixture was 5.064 g, therefore, the percent composition of each of the components results in 46.21% NaCl, 26.15% Naphthalene, and 32.90% Benzoic Acid. ● However, the percentages add up to approximately 105%, because the total mass obtained was 5.33 g, which in comparison to the original mass provides an overall 5.25% error. ● Such value resulted from errors throughout the experiment, such as using different weight balances, not weighing the filter papers beforehand, thus using a different mass and not placing the filter paper correctly on the filter funnel. Week 1: Small Scale Tests CONCLUSION/FUTURE RESEARCH Compounds Solubility in Water NaCl Soluble Slightly Soluble Insoluble Benzoic Acid Insoluble Soluble Insoluble Naphthalene Insoluble Insoluble Insoluble ● The purpose of this experiment was to come up with a method of separating each component in the compound. In addition, our team had to determine the solubility of each component. The components found in our unknown were identified as NaCl, Benzoic Acid, and Naphthalene. ● Through the small scale test our group determined that NaCl was soluble in water. However, Benzoic acid and Naphthalene form precipitates. Therefore, the NaCl component could be isolated by adding water. The reason why NaCl is soluble is because it is a polar ion; the Na is positively charged and is attracted to the negative O ions. In addition, Cl is a negative ion and is therefore attracted to the positive H ions (American Chemical Society). ● Benzoic acid forms a precipitate because even though it contains a polar carboxylic group, the rest of the acid is a nonpolar (World of Chemicals). This leads to little to no dissociation water. Naphthalene is also nonpolar which leads to its insolubility in water. ● Our group determined that Benzoic acid is soluble in 1M NaOH. However, Naphthalene on the other hand was not. Naphthalene was insoluble in 3M HCl as well. ● By identifying the compounds in unknown A and determining how they can be separated from the unknown, the team has beneficial information that they can use to navigate the landfill and clean it without detrimental effects. The objectives accomplished in this experiment also contribute to the public sector. Benzoic acid is used in athlete’s foot ointments because it restricts bacterial growth (World of Chemicals).Naphthalene is used in the manufacturing of plastics ( U.S National Library of Medicine). Knowing the solubility and isolation procedures for these compounds allows for them to be safely and efficiently used in various production capabilities. Solubility in 1 M Solubility in 3 M NaOH HCl Week 2: Separation & Percent Composition % Comp. Initial weight of mixture 5.064 g Amount DI water used for mixture 20.0 mL Weight of Evaporating Dish 42.975 g Weight of Filter Paper 0.378 g Weight of Dried Naphthalene + filter paper 1.702 g Weight of Dried Benzoic Acid + filter paper 2.044 g Weight of NaCl + Evaporating Dish 45.315 g Final Weight of dried Naphthalene 1.324 g 46.21% Final Weight of Dried Benzoic Acid 1.666 g 26.15% Final Weight of NaCl 2.34 g 32.90% Total Final Mass 5.33 g RESULTS: Percent Composition: NaCl: (2.34 g / 5.064 g) x 100 = 46.21 % REFERENCES: Naphthalene: (1.324 g / 5.064 g ) x 100 = 25.15 % Soluble HCl & NaOH http://chemphys.armstrong.edu/P1/polar/polarity.html Insoluble Benzoic Acid (Dried) Benzoic Acid: (1.666 g / 5.064 g ) x 100 = 32.90 % Total % = 104.26 % https://www.nlm.nih.gov/

Do you have a similar assignment and would want someone to complete it for you? Click on the ORDER NOW option to get instant services at essayloop.com. We assure you of a well written and plagiarism free papers delivered within your specified deadline.