Chapter 4_Chemical Reactions and Solution Stoichiometry

Chapter 4_Chemical Reactions and Solution Stoichiometry -...

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Unformatted text preview: Types of Chemical Reactions and Solution Stoichiometry Chapter 4 4.1 Water, the Common Solvent Water is not a linear molecule. It is bent at an angle of about 105. Water is a polar molecule because of its bent shape and because oxygen has a greater attraction for electrons than does hydrogen. 4.1 Polar vs. Nonpolar A polar molecule is a molecule that has a partially positive and a partially negative charge A nonpolar molecule there is no partial charges 4.1 Water, the Common Solvent "Like dissolves like" The following classes of molecules are miscible (dissolve in each other). Polar and Ionic Polar and Polar Nonpolar and Nonpolar The solubility of ionic substances in water varies greatly. The differences in the solubilities of ionic compounds in water typically depend on the relative attractions of the ions for each other. For example, sodium chloride is very soluble in water but silver chloride is only slightly soluble. 4.1 The Solution Process Hydration is the process that occurs when a solute is dissolved in a solvent The solvent molecules (H2O) attract the solute ions (Na+Cl) which causes the ions to break away from the crystal thus being dissolved. NaCl ( s ) Z+ H 2O ( l ) Na + (aq) + Cl - (aq) Exercies 4.1a Miscible or Immiscible Predict whether each pair of substances 1. NaNO3 and H2O miscible 2. C6H14 and H2O immiscible 3. I2 and C6H14 miscible 4. I2 and H2O immiscible are miscible or immiscible. Explain. Exercise 4.1b Dissociation Equations Complete each of the following dissociation equations: 1. CaCl2 ( s ) O H 2O ( l ) Ca (aq ) + 2Cl (aq ) 2+ - 1. Fe( NO3 )3 ( s ) O H 2O ( l ) Fe3+ (aq) + 3NO3- (aq) 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes A solution is a homogeneous mixture. A solute is a substance that is dissolved in a liquid to form a solution. A solvent is the dissolving medium in a solution. An aqueous solution means that water is the solvent. 4.2 Strong and Weak Electrolytes One useful property for characterizing a solution is its electrical conductivity, its ability to conduct electric current. Strong electrolytes are substances that are completely ionized when they are dissolved in water. They conduct an electric current very efficiently. Weak electrolytes are substances that exhibit a small degree of ionization in water. They conduct only a small current. Nonelectrolytes are substances that dissolve in water but do not produce any ions. They do not conduct an electric current. 4.2 Electrolytes vs. Nonelectrolytes 4.2 Strong and Weak Electrolytes Electrolyte Conductivity Degree of Dissociatio n strong high total weak non low to moderate none partial close to zero Examples strong acids (HCl), strong bases (NaOH), many salts (NaCl) weak organic acids (HC2H3O2), weak bases (NH3) sugar, AgCl, Fe2O3 4.2 Strong and Weak Acids and Bases The basis for the conductivity properties of solutions was first correctly identified by Svante Arrhenius (18591927). Arrhenius proposed that an acid is a substance that produces H+ ions when dissolved in water. Strong acids are acids in which virtually every molecule ionizes when placed in water. Weak acids only slightly ionize in water. Arrhenius' definition of a base is that they produce OH ions when dissolved in water. Strong bases are bases in which the cations and OH ions completely separate when dissolved in water. A weak base forms very few ions when dissolved in water. 4.2 Examples of Strong Electrolytes, Weak Electrolytes, and Nonelectrolytes Strong Inorganic Acids HCl, HBr, HI, HNO3, H2SO4, HClO4 NaOH, KOH Weak Heavy Metal Halides HgCl2, PbCl2 Inorganic Bases NH3, aniline Soluble Salts KCl, MgSO4, KClO3, CaCl2 Acetic Acid Some Bases Organic Acids Non Organic Compounds Sugars Exercise 4.2 Strong, Weak, or Nonelectrolyte 1. HClO4 strong 2. C6H12 non 3. LiOH strong 4. NH3 weak 5. CaCl2 strong 6. HC2H3O2 weak List whether each of the following is a strong, weak, or nonelectrolyte. 4.3 The Composition of Solutions Chemical reactions often take place when two solutions are mixed. To perform stoichiometric calculations we must know: 1. The nature of the reaction 2. The amounts of chemicals present in the solutions, usually expressed as concentration. Concentration is expressed in molarity (M) which is defined as moles of solute per liters of solution. 4.3 Molarity moles of solute M= liters of solution A solution that is 1.5 molar (written as 1.5 M) Keep in mind that moles millimoles micromoles = = liter milliliter microliter contains 1.5 moles of solute per liter of solution. Exercise 4.3a Calculating Molarity Calculate the molarity of a solution prepared by dissolving 11.85 g of solid KMnO4 in enough water to make 7.50 x 102 mL of solution. Solution: Remember that the units for molarity are moles per liter. Therefore, you must convert grams of KMnO4 to moles of KMnO4 and mL to L. Exercise 4.3a solution continued 1 mol KMnO4 11.85 g KMnO4 O = 0.07498 mol KMnO4 158.04 g KMnO4 1L 750 mL O = 0.750 L 1000 mL Divide moles by liters to get molarity. mol 0.07498 mol KMnO4 M= = = 0.100 M KMnO4 L 0.750 L solution Exercise 4.3b Mass from Molarity Calculate the mass of NaCl needed to prepare 175 mL of a 0.500 M NaCl solution. Solution: Use dimensional analysis to convert from volume to mass. Molarity can be used as a conversion factor. 1L 0.500 mol NaCl 58.44 g NaCl 175 mL O = 5.11 g NaCl 1000 mL 1L 1 mol NaCl 4.3 Concentration of Ions in Solution Remember, strong electrolytes completely dissociate in water. For example, KMnO4 ( s> ) H 2O ( l ) K + (aq) + MnO4 - (aq) This means that while it is generally acceptable to discuss your solution concentration as "molarity of KMnO4," it is more chemically correct to discuss "molarity of K+ ions and molarity of MnO4 ions." A solution that is 0.85 M in KMnO4 is really 0.85 M in K+ ion and 0.85 M in MnO4 ion, because KMnO4 completely dissociates. Exercise 4.3c Concentration of Ions Calculate the molarity of all of the ions in 1.5 M CrCl3. Solution: Hint: It is helpful to write out the dissociation equation. CrCl3 ( s > ) H 2O ( l ) Cr 3+ (aq ) + 3Cl - (aq ) molarity of Cr3+ = molarity of CrCl3 = 1.5 M molarity of Cl = three times the molarity of CrCl3 = 4.5 M Exercise 4.3d Concentration of Ions 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. Solution: 1. Calculate the solute concentration. 1 mol CuCl2 9.82 g CuCl2 O = 0.073038 mol CuCl2 134.45 g CuCl2 1L 600 mL O = 0.600 L 1000 mL mol 0.073038 mol CuCl2 M= = = 0.1217 M CuCl2 L 0.600 L Exercise 4.3d solution continued 1. Determine the molarity of Cl in solution using ion to solute ratio. CuCl2 ( s ) H 2O ( l ) Cu 2+ (aq ) + 2Cl - (aq ) There are two moles of Cl for every one mole of CuCl2. 0.1217 mol CuCl2 2 mol Cl - [Cl ] = = 0.243 M Cl - L 1 mol CuCl2 - 4.3 Dilution A standard solution is a solution whose concentration is accurately known. A stock solution is a concentrated solution that is purchased and stored in a laboratory's stock room. Water is added to a stock solution to achieve the molarity desired for a particular solution in a process called dilution. The key to a dilution calculation is to remember that the moles of solute after the dilution equals the moles of solute before dilution. To perform a dilution calculation we use the following formula: M1V1 = M2V2 M1 represents the initial molarity and V1 represents the initial volume. M2 and V2 represent the final molarity and volume after the dilution. Exercise 4.3f Preparation of a Dilute Solution What volume of 12 M hydrochloric acid must be used to prepare 600 mL of a 0.30 M HCl solution? Solution: M1 = 12 M HCl M2 = 0.30 M HCl V1 = ? V2 = 600 mL M 1V1 = M 2V2 (12 M )(V1 ) = (0.30 M )(600 mL) (0.30 M )(600 mL) V1 = = 15 mL (12 M ) 4.4 Types of Chemical Reactions Chemists group reactions into classes. There are three basic types of solution 1. 2. 3. reactions: We will define and illustrate each type in the following sections. Precipitation reactions Acidbase reactions Oxidationreduction reactions 4.5 Precipitation Reactions When two solutions are mixed, an insoluble substance sometimes forms; that is, a solid forms and separates from the solution. Such a reaction is called a precipitation reaction. The solid that forms and separates from the solution is called a precipitate. 4.5 Example of a Precipitation Reaction When an aqueous solution of potassium iodide reacts with an aqueous solution of lead (II) nitrate, a solid product (PbI2) is formed. KI, Pb(NO3)2, and KNO3 KI, are all soluble in water. PbI2 is insoluble in water and thus forms a precipitate in this reaction. 2KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2KNO3(aq) 4.5 Solubility Rules Solubility Rules for Ionic Compounds Compounds 1. Salts of alkali metals and ammonium Soluble 2. 3. 4. 5. Nitrate salts and chlorate salts Soluble Sulfate salts Soluble Chloride salts Carbonates, phosphates, chromates, sulfides, and hydroxides Most are Insoluble Compounds of Ag and some compounds of Hg and Pb Compounds of the alkali metals and of ammonium Compounds of Pb, Ag, Hg, Ba, Sr, and Ca Few exceptions Solubility Soluble Exceptions Some lithium compounds Exercise 4.5 Predicting Precipitates Complete and balance the following reactions, determining if a precipitate is formed. 1. KCl(aq) + Pb(NO3)2(aq) (aq) 2. AgNO3(aq) + MgBr2(aq) (aq) 3. Ca(OH)2(aq) + FeCl3(aq) (aq) 4. NaOH(aq) + HCl(aq) NaOH(aq) + HCl(aq) Exercise 4.5 solution 1. 2KCl(aq) + Pb(NO3)2(aq) 2KNO3(aq) + PbCl2(s) (aq) 2. 2AgNO3(aq) + MgBr2(aq) 2AgBr(s) + Mg(NO3)2(aq) (aq) 3. 3Ca(OH)2(aq) + 2FeCl3(aq) 3CaCl2(aq) + 2Fe(OH)3 (s) (aq) 4. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) NaOH(aq) + HCl(aq) 4.6 Describing Reactions in Solution There are three types of equations that are used to describe reactions in aqueous solutions. Molecular Equations which are the same as the balanced chemical equations. Complete Ionic Equations which show all substances that are strong electrolytes as ions. Net Ionic Equations which includes only those solution components directly involved in the formation of the precipitate. The other ions not involved in the chemical change (formation of the precipitate) are called spectator ions and are not included in the net ionic equation. 4.6 Molecular Equations The molecular equation gives the overall reaction stoichiometry but not necessarily the actual forms of the reactants and products in solution. Example: K 2CrO4 (aq ) + Ba( NO3 ) 2 (aq ) BaCrO4 ( s ) + 2 KNO3 ( aq) 4.6 Complete Ionic Equation The complete ionic equation represents as ions all reactants and products that are strong electrolytes. Example: 2 K + (aq ) + CrO4 2- (aq ) + Ba 2+ (aq) + NO3- (aq ) 2 BaCrO4 ( s ) + 2 K + (aq) + 2 NO3- (aq) 4.6 Net Ionic Equations The net ionic equation includes only those solution components undergoing a change. Spectator ions are ions not included in the change and are not included in the net ionic equation. Example: Ba (aq ) + CrO4 (aq ) 2+ 2- BaCrO4 ( s ) Chemists usually write the net ionic equation for a reaction in solution because it gives the actual forms of the reactants and products and includes only the species that undergo a change. Exercise 4.6 Molecular, Complete Ionic, and Net Ionic Equations Complete and write the molecular, complete ionic, and net ionic forms for each of the following: 1. Aqueous nickel (II) chloride reacts with aqueous sodium hydroxide 2. Aqueous sodium hydroxide reacts with aqueous phosphoric acid Exercise 4.6 solution 1 Write the balanced molecular equation. Write the complete ionic equation. Write the net ionic equation. Ni2+(aq) + 2OH(aq) Ni(OH)2(s) NiCl2(aq) + 2NaOH(aq) Ni(OH)2(s) + 2NaCl(aq) Ni2+(aq) + 2Cl(aq) + 2Na+(aq) + 2OH(aq) Ni(OH)2(s) + 2Na+(aq) + 2Cl(aq) Exercise 4.6 solution 2 Write the balanced molecular equation. Write the complete ionic equation. Write the net ionic equation. 3OH(aq) + 3H+(aq) 3H2O(l) 3NaOH(aq) + H3PO4(aq) 3H2O(l) + Na3PO4(aq) 3Na+(aq) + 3OH(aq) + 3H+(aq) + PO43(aq) 3H2O(l) + 3Na+(aq) + PO43(aq) 4.7 Stoichiometry of Precipitation Reactions 1. Identify the species present and determine what reaction occurs. Write the balanced chemical equation. 2. Calculate moles of reactants using molarity as the conversion factor from volume to moles. 3. If necessary, determine which reactant is limiting. 4. Calculate the moles of product. 5. Convert to grams or other units, as required. Steps for stoichiometric calculations for reactions in solution: Exercise 4.7 Stoichiometry of Precipitation Reactions What mass of precipitate is produced when 35 mL of a 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M KOH solution? Solution: Step 1: Determine what reaction occurs. Fe(NO3)3(aq) + 3KOH(aq) 3KNO3(aq) + Fe(OH)3(s) Exercise 4.7 solution continued Step 2: Calculate the moles of the reactants. 0.250 mol Fe( NO3 )3 1L 35 mL O = 0.00875 mol Fe( NO3 )3 1000 mL 1L 1L 0.180 mol KOH 55 mL O = 0.00990 mol KOH 1000 mL 1L Step 3: Determine limiting reactant. mol Fe( NO3 )3 1 = = 0.3333 mol KOH 3 mol Fe( NO3 )3 0.00875 = = 0.8838 mol KOH 0.00990 Since 0.8838 is greater than 0.3333 then KOH is the limiting reactant. Exercise 4.7 solution continued Steps 4 & 5: Convert to moles and to units wanted (grams). 1 mol Fe(OH )3 106.9 g Fe(OH )3 0.00990 mol KOH O = 0.35 g Fe(OH )3 3 mol KOH 1 mol Fe(OH )3 4.8 AcidBase Reactions Johannes N. Bronsted and Thomas M. Lowry defined acids and bases as follows: An acid is a proton donor. A base is a proton acceptor. This definition is more inclusive than Arrhenius's concept of acids and bases. A reaction in which an acid reacts with a base is called a neutralization reaction. The steps to an acidbase stoichiometric calculation are the same as for a precipitation reaction. Exercise 4.8a Neutralization of a Strong Acid How many mL of a 0.800M NaOH solution is needed to neutralize 40.00 mL of a 0.600 M HCl solution? Solution: Step 1: Write a balanced equation. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) Exercise 4.8a solution continued Step 2: There is no limiting reactant in this problem so we can use dimensional analysis to solve. 40.00 mL HCl x $ ' 1L 0.600 mol HCl 1 mol NaOH 1000 mL 1L 1 mol HCl 1L 1000 mL = 30.0 mL NaOH 0.800 mol NaOH 1L 4.8 Acidbase Titration Volumetric analysis is a technique for determining the amount of a certain substance by doing a titration. A titration involves delivery of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte). The point in the titration where enough titrant has been added to react exactly with the analyte is called the equivalence point. An indicator is a substance that is added at the beginning of the titration and changes color at the equivalence point. The point where the indicator changes color is called the endpoint of the titration. Exercise 4.8b Acidbase Titration You wish to determine the molarity of a solution of sodium hydroxide. To do this, you titrate a 25.00 mL aliquot of your sample, which has had 3 drops of phenolphthalein indicator added so that it is pink, with 0.1067 M HCl. The sample turns clear (indicating that the NaOH has been precisely neutralized by the HCl solution) after the addition of 42.95 mL of the HCl. Calculate the molarity of your NaOH solution. Exercise 4.8b solution Step 1: Write the balanced equation. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) Step 2: There is no limiting reactant in this problem so we will use dimensional analysis to determine the number of moles of NaOH that reacted with HCl. 1L 0.1067 mol HCl 1 mol NaOH 42.95 mL HCl O = 0.004583 mol NaOH 1000 mL 1L 1 mol HCl Step 3: Calculate the molarity of the NaOH sample. M= mol 0.004583 mol NaOH = = 0.1833 M NaOH L 0.02500 L 4.9 OxidationReduction Reactions A reaction in which one or more electrons are transferred is called an oxidationreduction reaction or redox reaction. The substance that is oxidized is the substance that loses electrons (increases in oxidation state). The substance that is reduced is the substance that gains electrons (decreases in oxidation state). LEO says GER 4.9 Oxidation States The use of oxidation states (oxidation numbers) allows us to keep track of electrons in redox reactions. In ionic compounds the oxidation state is the same as the charge of the ion. For covalent compounds, the oxidation states of atoms are obtained by arbitrarily assigning the electrons to particular atoms. The oxidation states of the atoms in a covalent compound is the imaginary charges the atoms would have if the shared electrons were divided equally between identical atoms bonded to each other or, for different atoms, were all assigned to the atom in each bond that has the greater attraction for electrons. 4.9 Rules for Assigning Oxidation Numbers 1. The oxidation of an atom in an element is 0. (Na, O2, 2. 3. 4. 5. 6. O3, Hg, etc.) The oxidation of a monatomic ion is the same as its charge. (Na+ = +1, Cl = 1) In its compounds, fluorine is always assigned an oxidation state of 1. Oxygen is assigned an oxidation state of 2 except for in peroxides, H2O2, (1) and OF2 (+2). Hydrogen is assigned an oxidation state of +1 except in metal hydrides, NaH, when it's 1. The sum of the oxidation states in an electrically neutral compound must equal zero. For a polyatomic ion, the sum of the oxidation states must equal the charge on the ion. Exercise 4.9a Assigning Oxidation States Assign oxidation states to each of the atoms in the following compounds: F = 1, Ca = +2 1. CaF2 2. 3. 4. 5. 6. C2H6 ICl5 H2SO3 KMnO4 SO42 H = +1, C = 3 H = +1, O = 2, S = +4 Cl = 1, I = +5 O = 2, K = +1, Mn = +7 O = 2, S = +6 4.9 Characteristics of Oxidation Reduction Reactions Oxidation is an increase in oxidation state. Reduction is a decrease in oxidation state. In a redox reaction, the reactant that accepts electrons from the other reactant is called the oxidizing agent. The reactant that donates electrons to the other reactant is called the reducing agent. Oxidizing agent = electron acceptor Reducing agent = electron donor Exercise 4.9b Redox Reactions For each of the following reactions, assign oxidation states to each of the atoms then identify which atoms undergo oxidation and reduction. Also, list the oxidizing and reducing agents. 1. 2H2(g) + O2(g) 2H2O(g) 0 0 +1 2 Oxidized: hydrogen (0 to +1) Reduced: oxygen (0 to 2) Oxidizing agent: O2 Reducing agent: H2 Exercise 4.9b continued +7 2 +2 1. 2MnO4(aq) + 16H+(aq) + 5C2O42(aq) 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Oxidized: carbon (+3 to +4) Reduced: manganese (+7 to +2) Oxidizing agent: MnO4 (permanganate ion) Reducing agent: C2O42 (oxalate ion) +4 2 +1 2 +1 +3 2 4.10 Balancing Oxidation Reduction Equations Oxidationreduction reactions are often complicated to balance. This section discusses a special technique, called the halfreaction method, that is used to balance these reactions. This process divides the reaction into two half reactions, one involving oxidation and the other reduction, balances the atoms and the charges for each half reaction and then combines the reactions to get the balanced equation. 4.10 Steps in the HalfReaction Method for Balancing Equations in an Acidic Solution 1. Write separate equations for the oxidation and 2. reduction halfreactions. For each half reaction, 3. If necessary, multiply one or both balanced half 4. 5. Balance all the elements except hydrogen and oxygen. Balance oxygen using H2O. Balance hydrogen using H+. Balance the charge using electrons. reactions by an integer to equalize the number of electrons transferred in the two halfreactions. Add the halfreactions, and cancel identical species. Check the equation to make sure it is balanced. Exercise 4.10a Balancing Redox Reactions in an Acidic Solution Balance the following equation in acidic solution using the halfreaction method. Cu(s) + HNO3(aq) Cu2+(aq) + NO(g) Solution: Step 1: Write the halfreactions. Cu Cu2+ (copper is oxidized) HNO3 NO (nitrogen is reduced) Exercise 4.10a solution continued Step 2: Balance each half reaction. Cu Cu Cu 2+ - Cu + 2e 2+ Cu is balanced so balance the charge by adding electrons. HNO3 O NO HNO3 + 3H + N is balanced, next balance O by adding H2O. Balance H by adding H+. Balance charge by adding e. HNO3 O NO + 2 H 2O HNO3 + 3H + + 3e - NO + 2 H 2O NO + 2 H 2O Exercise 4.10a solution continued Step 3: Equalize electron transfer. The same number of electrons gained by reduction has to be lost by oxidation. Step 4: Add the halfreactions and cancel. 3Cu 2 HNO3 + 6 H + + 6e - 3Cu 2++ 6e - 2 NO 4 H 2O + 3Cu 2+ +2 NO +4 H 2O +6e - 3Cu 3Cu 2++ 6e - + - 2 HNO3 + 6 H + 6e 2 NO + 4 H 2O 3Cu + 2 HNO3 + 6 H + +e - 6 Exercise 4.10a solution continued Canceling the electrons gives us the equation: 3Cu + 2 HNO3 +Y 6 H + 3Cu 2+ + 2 NO + 4 H 2O Step 5: Double check to make sure the equation is balanced. The equation is balanced! 4.10 Balancing Oxidation Reduction Reactions in a Basic Solution 1. 2. 3. To balance redox reactions in a basic solution, follow the steps 14 for an acidic solution then do the following: To both sides of the equation, add a number of OH ions that is equal to the number of H+ ions. This will eliminate H+ by forming H2O. Form H2O on the side containing H+ and OH ions, and eliminate the number of H2O molecules that appear on both sides of the equation. Check to make sure the equation is balanced. Exercise 4.10b Balancing Redox Reactions in a Basic Solution Balance the following equation in basic solution using the halfreaction method. Cr2O72(aq) + NO(g) Cr3+(aq) + NO3(aq) Solution: Step 1: Write the halfreactions. NO NO3 (nitrogen is oxidized) Cr2O72 Cr3+ (chromium is reduced) Exercise 4.10a solution continued NO Step 2: Balance each half reaction. NO3- NO + 2 H 2O N is balanced, next balance O by adding H2O. NO3- NO3- 4 H + + Balance H by adding H+. Balance charge by adding e. NO + 2 H 2O NO + 2 H 2O NO3- 4 H + 3e - Balanced! + + Balance Cr. Balance O by adding H2O. Balance H by adding H+. Cr2O7 2- Cr2O7 2- Cr 3+ Cr2O7 2- + 14 H + 2Cr 3+ 7 H 2O Balance charge by adding e. + Cr2O7 2- + 14 H + + 6e - 2Cr 3+ 7 H 2O + Balanced! Cr2O7 2- 2Cr 3+ 2Cr 3+ 7 H 2O + Exercise 4.10a solution continued Step 3: Equalize electron transfer. The same number of electrons gained by reduction has to be lost by oxidation. 2 NO + 4 H 2O 2 NO3- 8 H ++ 6e - + Cr2O7 2- + 14 H + + 6e - 2Cr 3+ 7 H 2O + Step 4: Add the halfreactions and cancel. 2 NO + 4 H 2O Cr2O7 2- + 14 H + + 6e - 2 NO3- 8H + 6e - + + 2Cr 3+ 7 H 2O + 2 NO3- 8H + 2Cr 3+ 3 H 2O 6e - + + + 7 + 6 2 NO + 4 H 2O + Cr2O7 2- + 14 H + + 6e - Exercise 4.10a solution continued Canceling gives us the equation: Cr2O7 2- + 2 NO +[6 H + Cr2O7 2- + 2 NO + 6 H + +OH - 6 2Cr 3+ + 2 NO3- + 3H 2O 2Cr 3+ +2 NO3- +3H 2O +6OH - Step 6: Because it's a basic solution, we need to add OH to get rid of the excess H+. Step 7: Combine H+ and OH to form water then cancel any water. Cr2O7 2- + 2 NO +[6 H 2O 2Cr 3+ + 2 NO3- + 3H 2O+ 6OH - 3 The balanced equation is: Cr2O7 2- ( aq ) + 2 NO ( g ) +[3H 2O (l ) 2Cr 3+ (aq )+ 2 NO3- (aq )+ 6OH - (aq ) ...
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This note was uploaded on 09/28/2009 for the course CHEM 102 taught by Professor Freeman during the Spring '08 term at South Carolina.

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