Solutions Labs Dry Lab 4 Oxidation-Reduction Equations

Solutions Labs Dry Lab 4 Oxidation-Reduction Equations
#9 A Volumetric Analysis #15 Bleach Analysis Chemical equations Chapters 8-11

Will the substances mix?
NaNO3 and H2O Miscible-ionic + polar C6H14 and H2O Immiscible-nonpolar + polar I2 and C6H14 Miscible-nonpolar + nonpolar I2 and H2O immiscible-nonpolar + polar 3/25/2017

bent molecule 104.5o Aqueous solutions- solvent is water
Polar molecule-unequal charge distribution gives water ability to dissolve many compounds 3/25/2017

Hydration Positive ends of water molecules are attracted to negatively charged anions Negative ends of water molecules are attracted to positively charged cations Strong forces in positive and negative ions of solid are replaced by strong water-ion interactions 3/25/2017

Solution homogeneous mixture of two or more substances
Solute- If it and solvent in same phase, one in lesser amount If it and solvent in different phases, one that changes phase One dissolved Solvent- If it and solute in same phase, one in greater amount If it and solute in different phases, one retaining phase One doing dissolving 3/25/2017

Electrolytes and Nonelectrolytes
Many aqueous solutions of ionic compounds conduct electricity whereas water itself essentially does not conduct-electrolyte solutions Dissociation-ionic substances break apart completely into ions when dissolved in water Ionization-some molecular compounds (nonionic) break apart into ions when dissolved in water 3/25/2017

Solvation of ionic compounds
Polarity of water Ions surrounded by water molecules prevent recombining of ions (solvated) Ions/shells free to move around, so dispersed uniformly throughout solution 3/25/2017

Electrolyte Conductivity Degree Examples of Dissociation
Strong high total strong acids (HCl, HNO3) many ionic salts (NaCl, Sr(NO3)3-ions) strong bases (NaOH, Ba(OH)2, other IA/IIA hydroxides) Weak low to partial weak organic acids moderate (tap water, HCO2H formic acid, C2H3O2) weak bases (NH3) Non none close to zero sugar, sugar solution, AgCl, Fe2O3 (neutral) 3/25/2017

Complete following dissociation equations: (all in water)
CaCl2(s)Ca2+(aq) + 2Cl-(aq) Fe(NO3)3(s)  KBr(s)  (NH4)2Cr2O7(s)  Strong, Weak, or Nonelectrolytes HClO4 C6H12 LiOH NH3 CaCl2 HC2H3O2 3/25/2017

Molarity M = moles of solute Liter of solution
Moles = millimoles = micromoles liter milliliter microliter BUT Moles does not equal millimoles or moles liter liter microliter

Convert from mass of NaCl to moles of NaCl.
What is the molarity of a solution in which 3.57 g of sodium chloride, NaCl, is dissolved in enough water to make 25.0 ml of solution? Convert from mass of NaCl to moles of NaCl. 3.57 g NaCl x 1 mol NaCl = mol NaCl 58.44 g NaCl Convert 25.0 mL to L and substitute these two quantities into the defining equation for molarity. Molarity = mol NaCl = M NaCl l L solution We read this as 2.44 molar NaCl. It is important to understand that molarity means moles of solute per liter of solution, and not per liter of solvent. 3/25/2017

How many mL of solution are necessary if we are to have a 2
How many mL of solution are necessary if we are to have a 2.48 M NaOH solution that contains g of the dissolved solid? 31.52 g NaOH mol NaOH = mol NaOH 40.00 g NaOH 2.48 M NaOH = mol NaOH = .318 L = 318.mL NaOH x L 3/25/2017

Determine the molarity of Cl- ion in a solution prepared by dissolving 9.82 g of CuCl2 in enough water to make 600 mL of solution. 9.82 g CuCl2 1 mol CuCl2 = mol CuCl2 g CuCl2 M = mol CuCl2 = M CuCl2 .600 L Ion = 2 mol Cl- solute 1 mol CuCl2 M CuCl2 x 2 mol Cl = M L solution mol CuCl2 3/25/2017

Calculate the mass of NaCl needed to prepare 175 mL of a 0
Calculate the mass of NaCl needed to prepare 175 mL of a M NaCl solution. 0.500 M = x mol .175 L x mol = mol NaCl mol NaCl g NaCl = 5.11 g NaCl 1 mol NaCl 3/25/2017

Solution Concentration
Amounts of chemical present in solutions-expressed as concentration (amount of solute dissolved in given amount of solvent) Most common unit of concentration used for aqueous solutions-molarity # moles of solute per liter of solution 3/25/2017

Common Terms of Solution Concentration
Stock - routinely used solutions prepared in concentrated form Concentrated - relatively large ratio of solute to solvent (5.0 M NaCl) Dilute - relatively small ratio of solute to solvent (0.01 M NaCl) 3/25/2017

To calculate the concentration of each type of ion in a solution
Dissociate solid into its cations/anions This gives how many moles of ions are formed for each mole of solid Multiply number of ions by given molarity 3/25/2017

Calculate the molarity of all the ions in each of the following solutions:
Always write out the dissociation equation, so you have “ion-to-solute” mole ratio. 0.25 M Ca(OCl2) Ca(OCl2)(s)  Ca2+(aq) + 2OCl-(aq) Molarity of Ca2+ = molarity of Ca(OCl2) = M Molarity of 2OCl- = 2 x molarity of Ca(OCl2) = 0.50 M 2 M CrCl3 CrCl3 Cr3+(aq) + 2Cl-(aq) 3/25/2017

Standard solution-solution whose concentration is accurately known
a. Put a weighed amount of a substance (solute) into a volumetric flask and add a small quantity of water. c. Add more water (with gentle swirling) until the level of the solution just reaches the mark etched on the neck of the flask. Then mix solution thoroughly by inverting the flask several times. b. Dissolve the solid in the water by gently swirling the flask (with stopper in place). 3/25/2017

Example To prepare a 0.50 molar (0.50 M) solution of KCl, we first measure out 0.50 mole of solid KCl, that is 37.3 g When water is added up to this mark, the flask will contain 1.0 L of the KCl solution, and will contain 0.50 mole of KCl The number of moles of KCl in given amounts of the above solution is easy to find For instance, 0.10 L of 0.50 M KCl will contain 0.10 L soln x 0.50 mol KCl= mol KCl in 1 L soln 3/25/2017

Dilution Used to prepare less concentrated solution from more concentrated solution Adding more water to given amount of solution does not change number of moles of solute present in solution Moles of solute before dilution = moles of solute after dilution Since moles of solute equals solution volume (V) times molarity (M) we have molesi = molesf of MiVi = MfVf i/f stand for initial/final solution, respectively Note that Vf is always larger than Vi 3/25/2017

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Parts per million (mg/L)
An aqueous solution with a total volume of 750 mL contains mg of Cu2+. What is the concentration of Cu2+ in parts per million? 14.38 mg Cu2+ = 19.2 ppm Cu2+ 0.750 L soln 3/25/2017

Molarity to PPM A solution is 3 x 10-7 M in manganese(VII) ion. What is the Mn7+ 3 x 10-7 M concentration in ppm? 3 x 10-7 mol Mn g Mn mg Mn7+ = = 0.02 ppm Mn7+ L soln mol Mn g Mn7+ 3/25/2017

Homework: Read , pp Q pp , #11 c/d/i, 12, 15c, 16, 20, 23a, 28 3/25/2017

Solubility Rules Solubility of solute: amount dissolved in given quantity of solvent at given temperature Major ideas: Many salts dissociate into ions in aqueous solution If solid forms from combination of selected ions in solution Solid must contain anion part and cation part Net charge on solid must be zero Simple solubility rules used to predict products of reactions in aqueous solutions 3/25/2017

Any ionic compound can be broken apart into its cations and anion
Compounds containing 3/more different atoms break apart into appropriate polyatomic ion(s) Charges of all ions must add up to zero Subscripts on monatomic ions become coefficients for ions For polyatomic ions, only subscripts after parentheses become coefficients Whether or not ionic compound dissolves to appreciable extent in water depends on which cations/anions make up compound 3/25/2017

Soluble-good deal of solid visibly dissolves when added to water
Slightly soluble-only small amount of solid dissolves (Ionic compound can have low solubility and still be strong electrolyte) Insoluble-no solid dissolves (relative term-does not mean that no solute dissolves) 3/25/2017

All metallic oxides (O-2) NH4+, Group IA metals
NH4+/ Group IA None NO3-, ClO-, ClO4-, HCO3-, (slightly soluble) CrO42- All metallic oxides (O-2) NH4+, Group IA metals All metallic hydroxides (OH-) NH4+, Group IA/IIA from calcium down. Important-NaOH/KOH Ba(OH)2, Sr(OH)2, and Ca(OH)2 marginally soluble 3/25/2017

Types of Solution Reactions
Precipitation reactions AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) Acid-base reactions (Neutralization) NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) Oxidation-reduction reactions Fe2O3(s) + Al(s)  Fe(l) + Al2O3(s) 3/25/2017

Precipitation Reactions
2 soluble substances combined AX(aq) + BZ(aq)  AZ +BX Determine, using solubility rules, whether either will form solid (precipitate: insoluble solid that settles from solution) 3/25/2017

Chemical equations for precipitation reactions can be written in several ways:
Molecular equation: formulas of compounds are written as usual chemical formulas Pb(NO3)2(aq) + 2HCl(aq) PbCl2(s) + 2HNO3(aq) Precipitation reaction: more accurately represented by ionic equation which shows compounds as being dissociated in solution Pb2+(aq) + 2NO3(aq) + 2H+(aq) + 2Cl–(aq)  PbCl2(s) + 2H+(aq) + 2NO3 (aq) H+/NO3 ions not involved in formation of precipitate Spectator ions (identical species on both sides of equation can be omitted from equation because ions not involved in reaction) Net ionic equation shows only species that actually undergo a chemical change Pb2+(aq) + 2Cl–(aq)  PbCl2(s) 3/25/2017

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Homework: Read 4.5-4.6, pp. 148-top 156
Q pg. 182, #30, 34 a/c, 36 b/d, 38 3/25/2017

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What mass of Fe(OH)3 is produced when 35. mL of a 0
What mass of Fe(OH)3 is produced when 35. mL of a M Fe(NO3)3 solution is mixed with 55 mL of a M KOH solution? 0.250 M M x g Fe(NO3)3(aq) + 3KOH(aq)  Fe(OH)3(aq) + 3KNO3(aq) 0.250 M Fe(NO3)3 = x mol = mol/1 .035 L 0.180 M KOH = x mol = mol/3 = mol .055 L limiting reactant mol KOH 1 mol Fe(OH) g Fe(OH)3= .35 g Fe(OH)3 3 mol KOH mol Fe(OH)3 3/25/2017

Gravimetric Analysis- measurement of weight
An ore sample is to be analyzed for sulfur. As part of the procedure, the ore is dissolved and the sulfur is converted to sulfate ion, SO42-. Barium nitrate is added, which causes the sulfate to precipitate out as BaSO4. The original sample had a mass of g. The dried BaSO4 has a mass of g. What is the percent of sulfur in the original ore? (present in the g) 2.005 g BaSO4 1 mol BaSO mol S g S = g S g BaSO4 1 mol BaSO4 1 mol S % S = g S x 100% = % S in the ore 3.187 g in ore 3/25/2017

Homework: Read 4.7, pp Q pg. 182, #40, 42, 44 3/25/2017

Acid-Base Theories Attempt to explain what happens to molecules of acidic/basic substances in solution that gives rise to their characteristic properties.

Acids and Bases Solutions of acids Solutions of bases
Sour taste Change blue litmus to red (abr) Dissolve certain metals React w/carbonates to produce carbon dioxide gas Solutions of bases Bitter taste Feel slippery Change red litmus to blue (brb) Properties of acid neutralized by addition of base, and vice versa 3/25/2017

Svante Arrhenius proposed this acid-base theory
Acid: substance that produces hydrogen ions (H+, protons) in aqueous solution HNO3(aq)  H+(aq) + NO3–(aq) Base: substance that produces hydroxide ions (OH–) in aqueous solution Ca(OH)2(aq)  Ca2+(aq) + 2OH–(aq) All are electrolytes undergoing dissociation in aqueous solution Neutralization is combination of H+/OH– to form water H+(aq) + OH–(aq)  H2O(l) 3/25/2017

Strong acid-completely dissociates into its ions-strong electrolytes
HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4 Strong base-soluble ionic compounds containing hydroxide ion (OH-)-strong electrolytes Group IA (Li, Na, K, Rb, Cs)/heavy Group IIA (Ca, Sr, Ba) metal hydroxides, NH2+ Weak acid-dissociates (ionizes) only to a slight extent in aqueous solutions-weak electrolytes Acetic acid, HF Weak base-very few ions are formed in aqueous solutions-weak electrolyte NH3 3/25/2017

Acid/base both completely ionized in solution, ionic equation:
2H+(aq) + 2Cl–(aq) + Mg2+(aq) + 2OH–(aq)  2H2O(l) + Mg2+(aq) + 2Cl–(aq) Notice that Mg2+ ion and Cl– ion are spectator ions. Net ionic equation is: 2H+(aq) + 2OH–(aq)  2H2O(l) or H+(aq) + OH–(aq)  H2O(l) (coefficients can be cancelled) Neutralization equation for any strong acid and strong base is: H+(aq) + OH–(aq)  H2O(l) If we had carried out this reaction with two moles of HCl for each mole of base, there would be no left over acid or base, and the solution would be described as neutral. 3/25/2017

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Brønsted-Lowry Theory
More general theory of acid-base reactions than Arrhenius Acid: substance that donates proton Base: substance that accepts proton 3/25/2017

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All Arrhenius acids are also Brønsted acids
All Arrhenius bases are also Brønsted bases 3/25/2017

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Acid-Base reactions that form gases
2HCl + Na2S  H2S + 2NaCl 2H+ + S2-  H2S (net ionic equation) Carbonates/bicarbonates + acids First form carbonic acid which is unstable If carbonic acid is present in solution in sufficient concentrations, it decomposes to form CO2 gas and water 3/25/2017

Writing Equations for Acid-Base Neutralization Reactions
Neutralization reaction: between acid/base that produces water/salt

Metal ion from base/nonmetal ion from acid form salt
In neutralization reaction, H+ from acid/ OH- from base combine to form water Metal ion from base/nonmetal ion from acid form salt Neutralization reaction of hydrochloric acid and magnesium hydroxide 2HCl(aq) +Mg(OH)2(aq)  2H2O(l) +MgCl2(aq) Acid base water salt 3/25/2017

Neutralization List species present in solution before reaction
Write balanced net ionic reaction Find out number of moles of acid we need to neutralize Using stoichiometry of reaction, find out required moles of base to neutralize acid (limiting reactant if necessary) Determine volume of OH- needed to give that many moles Make sure moles of acid equals moles of base 3/25/2017

H+ (aq) + OH- (aq)  H2O(l) M1V1 = M2V2
How many mL of a M NaOH solution is needed to just neutralize mL of a M HCl solution? H+, Cl-, Na+, OH-, H2O H+ (aq) + OH- (aq)  H2O(l) M1V1 = M2V2 Since # H+ = OH-, we multiply both sides by 1 (van Hoft factor) (0.800 M)(x mL) = (0.600 m)(40.00 mL) = mL 3/25/2017

What volume of a 0. 100 M HCl solution is needed to neutralize 25
What volume of a M HCl solution is needed to neutralize 25.0 mL of a M NaOH? List species and decide what reaction will occur H+ Cl- Na+ OH- Na+(aq) + Cl-(aq)  NaCl(s)-soluble-spectator ions H+(aq) + OH-(aq)  H2O(l) Write balanced net ionic equation Determine limiting reactant Problem requires addition of just enough H+ ions to react exactly with OH- ions present, so not concerned with determining limiting reactant Calculate moles of reactant needed M1V1 = M2V2 (0.100 M)(x mL) = (0.350 M)(25.0 mL) 87.5 mL 3/25/2017

28. 0 mL of 0. 250 M HNO3 and 53. 0 mL of 0. 320 M KOH are mixed
28.0 mL of M HNO3 and 53.0 mL of M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH- ions in excess after the reaction goes to completion? H+, NO3-, K+, OH- for KNO3 which is soluble 28.0 mL HNO L mol H+ = x 10-3 mol H+ 1000 mL L HNO3 53.0 mL KOH L mol OH- = 1.70 x 10-2 mol OH- 1000 mL L KOH Limiting reactant is H+ so 7.00 x 10-3 mol H+ will react with = x 10-3 mol OH- to form 7.00 x 10-3 mol H2O. 1.70 x 10-2 mol OH x 10-3 mol OH- leaves 1.00 x 10-2 mol OH- Volume of combined solution is sum of individual volumes 28.0 mL mL = 81.0 mL = 8.1 x 10-2 L Molarity of OH- in excess is Mol OH x 10-2 mol OH- = M OH- L solution x 10-2 L 3/25/2017

Acid-Base Titrations Procedure for determining concentration of unknown acid (or base) solution (analyte) using known (standardized) concentration (titrant) of base (or acid) solution In titration of acid solution of unknown concentration, a known volume of a standardized base solution is added to acid until acid is just neutralized Point at which exactly enough base has been added to neutralize acid-equivalence or stoichiometric point Point often marked by indicator, a substance that changes color at (or very near) equivalence point Point where indicator actually changes-endpoint 3/25/2017

Use mole ratios from balanced neutralization reaction M1V1=M2V2
Goal is to choose indicator so endpoint (where indicator changes color) occurs exactly at equivalence point (where just enough titrant has been added to react with all the analyte) Because concentration (M) and required volume (V) of base are known, number of moles of base needed to neutralize acid can be calculated Use mole ratios from balanced neutralization reaction M1V1=M2V2 3/25/2017

Example What is the volume of 0.05-molar HCl that is required to neutralize 50 mL of a 0.10-molar Mg(OH)2 solution? Every mole of Mg(OH)2 dissociates to produce 2 moles of OH- ions, so a 0.10 M Mg(OH)2 solution will be a 0.20 M OH- solution Solution will be neutralized when # moles of H+ ions added is equal to # mole of OH- originally in the solution Moles = molarity x volume Moles of OH- = (0.20 M)(50 mL) = 10 millimoles = # moles H+ added Volume = moles/molarity Volume of HCl = 10 millimoles/0.05 M = 200 mL Another way to look at it: M1V1=M2V2 (0.05 M)(V) = (0.10 M)(50 mL)(2 moles) 3/25/2017

0.1704 M NaOH = x mol NaOH = 0.00725 mol NaOH 0.04255 L
You want to determine the molar mass of an acid. The acid contains one acidic hydrogen per molecule. You weigh out a g sample of the pure acid and dissolve it, along with 3 drops of phenolphthalein indicator, in distilled water. You titrate the sample with M NaOH. The pink endpoint is reached after addition of mL of the base. Calculate the molar mass of the acid. M NaOH = x mol NaOH = mol NaOH L 1 :1 ratio of OH- to H+ mol H+ = g acid = g/mol molar mass acid 3/25/2017

Homework: Read 4.8, pp Q pg. 183, #46 a/c, 48 b/c, 50 a/b, 52, 54 3/25/2017

The transfer of electrons
Redox Reactions The transfer of electrons

LEO goes GER In redox reactions electrons are transferred from one atom to another Oxidation Chemical change in which atom, ion, or molecule loses electrons Reduction Chemical change in which atom, ion, or molecule gains electrons Oxidation/reduction occur simultaneously 3/25/2017

Sodium is oxidized and chlorine is reduced.
In the reaction of metallic sodium with chlorine gas, an electron is transferred from the sodium atom to a chlorine atom. 2Na + Cl2  2Na+ + 2Cl– Sodium is oxidized and chlorine is reduced. In sodium-chlorine reaction: Sodium-reducing agent Chlorine-oxidizing agent Substance being oxidized is called reducing agent because it donates electrons which cause reduction in another substance. Substance being reduced is called oxidizing agent because it accepts electrons from another substance. 3/25/2017

Separated into two half-reactions
Oxidation half-reaction Reduction half-reaction For sodium-chlorine reaction they are oxidation 2Na  2Na+ + 2e– reduction 2e– + Cl2  2Cl– Summation of two half-reactions yields overall redox reaction shown above 3/25/2017

Oxidation number Oxidation state
Used to help keep account of electrons in reactions Designates positive or negative character Positive oxidation #: atom lost electrons compared to uncombined atom Negative oxidation #: atom gained electrons Oxidation numbers are useful in writing formulas, in recognizing redox reactions, and in balancing redox reactions Element oxidized in reaction whenever its oxidation number increases as result of reaction When oxidation number of element decreases in reaction, it is reduced 3/25/2017

(+1 when it is in a covalent compound, even if it is listed 2nd)
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Assign oxidation states for:
CaF2 C2H6 H2O ICl5 KMnO4 SO42- 3/25/2017

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Activity Series Different metals vary in ease of oxidation
Activity series gives order of decreasing ease of oxidation Predicts outcome of reactions between metals and either metal salts or acids Any metal on list can be oxidized by ions of elements below it 3/25/2017

Balancing Redox Reactions
Acid solutions Basic solutions

Oxidation number change method
Redox equation balanced by comparing increases and decreases in oxidation # Tip-off – If you are asked to balance an equation and if you are not told whether the reaction is a redox reaction or not, you can use the following procedure. 3/25/2017

Step 7: Balance the rest of the equation by inspection.
Step 1: Try to balance the atoms in the equation by inspection. If you successfully balance the atoms, go to Step 2. If you are unable to balance the atoms, go to Step 3. Step 2: Check to be sure that the net charge is the same on both sides of the equation. If it is, you can assume that the equation is correctly balanced. If the charge is not balanced, go to Step #3. Step 3: If you have trouble balancing the atoms and the charge by inspection, determine the oxidation numbers for the atoms in the formula, and go to Step 4. Step 4: Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced. Step 5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number (number of electrons lost = number of electrons gained). Step 6: Add coefficients to the formulas so as to obtain the correct ratio of the atoms whose oxidation numbers are changing. Step 7: Balance the rest of the equation by inspection. 3/25/2017

NO2(g) + H2(g)  NH3(g) + H2O(l)
The atoms in this equation can be balanced by inspection. (You might first place a 2 in front of the H2O to balance the O’s, then 7/2 in front of the H2 to balance the H’s, and then multiply all the coefficients by 2 to get rid of the fraction.) 2NO2(g) + 7H2(g)  2NH3(g) + 4H2O(l) We therefore proceed to Step #2. For the reaction between NO2 and H2, the net charge on both sides of the equation in Step #1 is zero. Because the charge and the atoms are balanced, the equation is correctly balanced. 3/25/2017

HNO3(aq) + H3AsO3(aq) ® NO(g) + H3AsO4(aq) + H2O(l)
Step #1: Try to balance the atoms by inspection. The H and O atoms are difficult to balance in this equation. Step #3: Is the reaction redox? The N atoms change from +5 to +2, so they are reduced. The As atoms, which change from +3 to +5, are oxidized. Step #4: Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced. As +3 to +5 Net Change = +2 N +5 to +2 Net Change = -3 Step #5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number. As atoms would yield a net increase in oxidation number of +6. (Six electrons would be lost by three arsenic atoms.) 2 N atoms would yield a net decrease of -6. (2 N gain 6 e-). Thus the ratio of As atoms to N atoms is 3:2. Step #6: To get the ratio identified in Step 5, add coefficients to the formulas which contain the elements whose oxidation number is changing. 2HNO3(aq) + 3H3AsO3(aq) ® NO(g) + H3AsO4(aq) + H2O(l) Step #7: Balance the rest of the equation by inspection. 2HNO3(aq) + 3H3AsO3(aq) ® 2NO(g) + 3H3AsO4(aq) + H2O(l) 3/25/2017

Cu(s) + HNO3(aq)  Cu(NO3)2(aq) + NO(g) + H2O(l)
Step #1: Try to balance the atoms by inspection. The N atoms and the O atoms are difficult to balance by inspection. Step #3: Is the reaction redox? The copper atoms are changing their oxidation number from 0 to +2, and some of the nitrogen atoms are changing from +5 to +2. Step #4: Determine the net increase in oxidation number for element oxidized and the net decrease in oxidation number for element reduced. Cu 0 to +2 Net Change = +2 Some N +5 to +2 Net Change = -3 Step #5: Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number. We need three Cu atoms (net change of +6) for every 2 nitrogen atoms that change (net change of -6). Because some of the nitrogen atoms are changing and some are not, we need to be careful to put the 2 in front of a formula in which all of the nitrogen atoms are changing or have changed. The 3 for the copper atoms can be placed in front of the Cu(s). Step #6: To get the ratio identified in Step 5, add coefficients to the formulas which contain the elements whose oxidation number is changing. 3Cu(s) + HNO3(aq)  Cu(NO3)2(aq) + 2NO(g) + H2O(l) Step #7: Balance the rest of the equation by inspection. 3Cu(s) + 8HNO3(aq) 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l) 3/25/2017

Some Examples: Al(s) + MnO2(s) ® Al2O3(s) + Mn(s)
SO2(g) + HNO2(aq) ® H2SO4(aq) + NO(g) HNO3(aq) + H2S(aq) ® NO(g) + S(s) + H2O(l) Al(s) + H2SO4(aq) ® Al2(SO4)3(aq) + H2(g) 3/25/2017

Half-reaction in acidic solutions
Redox Reactions Half-reaction in acidic solutions

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Cr2O72-(aq) + HNO2(aq)  Cr3+(aq) + NO3-(aq)
(acidic) Step #1: Write the skeletons of the oxidation and reduction half-reactions. Cr2O72-  Cr3+ HNO2  NO3- Step #2: Balance all elements other than H and O. Cr2O72-  2Cr3+ (need 2 in front of chromium) HNO2  NO3- Step #3: Balance the oxygen atoms by adding H2O molecules on the side of the arrow where O atoms are needed. The first half-reaction needs seven oxygen atoms on the right, so we add seven H2O molecules. Cr2O72-  2Cr3+ + 7H2O The second half-reaction needs one more oxygen atom on the left, so we add one H2O molecule. HNO2 + H2O  NO3- 3/25/2017

Step #5: Balance the charge by adding electrons.
Step #4: Balance H atoms by adding H+ ions on the side of the arrow where H atoms are needed. 1st half-reaction needs 14 H atoms on left to balance 14 hydrogen atoms in 7 H2O molecules-add 14 H+ ions to left. Cr2O72- + 14H+  2Cr3+ + 7H2O 2nd half-reaction needs 3 H atoms on right to balance 3 hydrogen atoms on left-add 3 H+ ions to the right. HNO2 + H2O  NO3- + 3H+ Step #5: Balance the charge by adding electrons. Sum of charges on left side of chromium half-reaction is +12 (-2 for the Cr2O72- plus +14 for the 14 H+). Sum of charges on right side of chromium half-reaction is +6 (for the 2 Cr3+). Add 6 electrons to left side-sum on each side becomes +6. 6e- + Cr2O72- + 14H+  2Cr3+ + 7H2O Sum on the left side of the nitrogen half‑reaction is zero. Sum on the right side of the nitrogen half-reaction is +2 (-1 for the nitrate plus +3 for the 3 H+). Add 2 e’s to right side-sum on each side becomes zero. HNO2 + H2O  NO3- + 3H+ + 2e- 3/25/2017

Step #7: Add the 2 half-reactions.
Step #6: If # e’s lost in oxidation half ≠ #e’s in reduction half, multiply one or both reactions by # making # e’s gained = # lost. For Cr half-reaction to gain 6 electrons, N half-reaction must lose 6 electrons-multiply coefficients in N half-reaction by 3. 6e- + Cr2O72- + 14H+  2Cr3+ + 7H2O 3(HNO2 + H2O  NO3- + 3H+ + 2e-) 3HNO2 + 3H2O  3NO3- + 9H+ + 6e- Step #7: Add the 2 half-reactions. 3 H2O in 2nd half-reaction cancel 3 of 7 H2O in 1st half-reaction to yield 4 H2O on the right of the final equation. 9 H+ on right of 2nd half-reaction cancel 9 of 14 H+ on left of 1st half-reaction leaving 5 H+ on left of final equation. Cr2O72- + 3HNO2 + 5H+  2Cr3+ + 3NO3- + 4H2O Step #8: Check to make sure that the atoms and the charge balance. The atoms in our example balance and the sum of the charges is +3 on each side, so our equation is correctly balanced. Cr2O72-(aq) + 3HNO2(aq) + 5H+(aq)  2Cr3+(aq) + 3NO3- (aq) + 4H2O(l) 3/25/2017

3/25/2017

Some Examples MnO4-(aq) + Br(aq)® MnO2(s) + BrO3-(aq)
I2(s) + OCl-(aq) ® IO3-(aq) + Cl-(aq) Cr2O72-(aq) + C2O42(aq)®Cr3+(aq) + CO2(g) Mn(s) + HNO3(aq) ® Mn2+(aq) + NO2(g) 3/25/2017

Half-reaction in basic solutions
Redox Reactions Half-reaction in basic solutions

Step 10: Cancel or combine the H2O molecules.
Steps 1-7: Begin by balancing the equation as if it were in acid solution. If you have H+ ions in your equation at the end of these steps, proceed to Step #8. Otherwise, skip to Step#11. Step 8: Add enough OH- ions to each side to cancel the H+ ions. (Be sure to add the OH- ions to both sides to keep the charge and atoms balanced.) Step 9: Combine the H+ ions and OH- ions that are on the same side of the equation to form water. Step 10: Cancel or combine the H2O molecules. Step 11: Check to make sure that the atoms and the charge balance. If they do balance, you are done. If they do not balance, re-check your work in Steps 1-10. 3/25/2017

Cr(OH)3(s) + ClO3-(aq)  CrO42-(aq) + Cl-(aq) (basic)
Step #1: Cr(OH)3  CrO42- ClO3-  Cl- Step #2: (Not necessary for this example) Step #3: Cr(OH)3 + H2O  CrO42- ClO3-  Cl- + 3H2O Step #4: Cr(OH)3 + H2O  CrO42- + 5H+ ClO3- + 6H+  Cl- + 3H2O Step #5: Cr(OH)3 + H2O  CrO42- + 5H+ + 3e- ClO3- + 6H+ + 6e-  Cl- + 3H2O Step #6: 2(Cr(OH)3 + H2O  CrO42- + 5H+ + 3e- ) 2Cr(OH)3 + 2H2O  2CrO42- + 10H+ + 6e- ClO3- + 6H+ + 6e-  Cl- + 3H2O Step #7: 2Cr(OH)3(s) + ClO3-(aq)  2CrO42-(aq) + Cl-(aq) + H2O(l) + 4H+(aq) 3/25/2017

Step #10: Cancel or combine the H2O molecules.
Step #8: Because there are 4 H+ on the right side of our equation above, we add 4 OH- to each side of the equation. 2Cr(OH)3 + ClO3- + 4OH-  2CrO42- + Cl- + H2O + 4H+ + 4OH- Step #9: Combine the 4 H+ ions and the 4 OH- ions on the right of the equation to form 4 H2O. 2Cr(OH)3 + ClO3- + 4OH-  2CrO42- + Cl- + H2O + 4H2O Step #10: Cancel or combine the H2O molecules. 2Cr(OH)3(s) + ClO3-(aq) + 4OH-(aq)  2CrO42-(aq) + Cl-(aq) + 5H2O(l) Step #11: The atoms in our equation balance, and the sum of the charges in each side is -5. Our equation is balanced correctly. 3/25/2017

Examples CrO42-(aq) + S2-(aq) ® Cr(OH)3(s) + S(s)
MnO4-(aq) + I-(aq) ® MnO2(s) + IO3-(aq) H2O2(aq) + ClO4-(aq) ® O2(g) + ClO2-(aq) S2-(aq) + I2(s) ® SO42-(aq) + I-(aq) 3/25/2017

Homework: Read , pp Q pp , #58 a/b/c/df/h/I, 60 b/c/g, 62 a/c/e, 64 a/c, 66 a/c Do one additional exercise and one challenge problem. Submit quizzes by to me: 3/25/2017

Solutions Labs Dry Lab 4 Oxidation-Reduction Equations

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Solutions Labs Dry Lab 4 Oxidation-Reduction Equations

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