Reactions in Aqueous Solutions

Document Sample
Reactions in Aqueous Solutions Powered By Docstoc
					        Reactions in Aqueous Solutions
1.    Water as a universal solvent
2.    Aqueous solutions - solutions with water as solvent
3.    Solubility of ionic compounds in water
4.    Solubility of molecular compounds in water
5.    Strong, weak, and non-electrolyte solutions
6.    Solution compositions - concentrations
7.    Types of reactions in aqueous solutions
8.    Chemical equations for reactions in aqueous solutions
9.    Stoichiometry for reaction in aqueous solution
10.   Redox reactions in aqueous solutions.
         Water as a Universal Solvent
• Water molecule is very polar;
• Water interacts strongly with ionic and polar molecules;
• Strong interactions enable water to dissolve many solutes –
  ionic and nonionic;
• Solubility of ionic compounds depends on the relative
  strength of ion-dipole interactions between ions and water
  molecules and ionic bonds within the compounds;
• Many ionic compounds dissolve in water because of strong
  ion-dipole interactions;
• Polar nonionic compounds dissolve in water due to strong
  dipole-dipole interactions or hydrogen bonding;
      Electrolytes and Nonelectrolytes
• Electrolytes – solutions capable of conducting electric
  current - contain ions that move freely;
• Nonelectrolytes – solutions not capable of conducting
  electric current - contains neutral molecules only;
• Strong electrolytes – ionic compounds, strong acids and
  strong bases; they dissociate completely when dissolved in
  water, producing a lot of free ions;
• Weak electrolytes – weak acids or weak bases; they only
  dissociate (ionize) partially when dissolved in water –
  solutions contain mostly neutral molecules and very little
  free ions.
          Strong and Weak Electrolytes
• Examples of strong electrolytes – they ionize completely:
   – NaCl(aq)  Na+(aq) + Cl-(aq)
   – H2SO4(aq)  H+(aq) + HSO4-(aq)
   – Ca(NO3)2(aq)  Ca2+(aq) + 2NO3-(aq);

• Examples of weak electrolytes – they do not ionize
  completely:
   – HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
   – NH4OH(aq)  NH4+(aq) + OH-(aq);
   – Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq)
                   Nonelectrolytes
• Substances that do not ionize in aqueous solution are
  nonelectrolytes;
• Most organic compounds are nonelectrolytes;
• Solutions containing such substances cannot conduct
  electricity, because they do not have freely moving ions.
• Examples of nonelectrolytes:
   – C6H12O6, C12H22O11, CH3OH, C2H5OH, C3H7OH,
     HOC2H4OH, etc.
                Solution Concentrations

• The concentration of a solution may be expressed in:
   – Percent by mass, percent by volume, or molarity;
   Percent (by mass) = (Mass of solute/Mass of solution) x 100%
   Percent (by volume) = (Vol. of solute/Vol. of solution) x 100%


   Molarity = (Mol of solute/Liter of solution)
   Mol of solute = Liters of solution x Molarity
                    Percent by Mass

• Example:
    – A sugar solution contains 25.0 g of sugar dissolved in
      100.0 g of water. What is the mass percent of sugar in
      solution?
    – Percent sugar = {25.0 g/(25.0 g + 100.0 g)} x 100%
•                   = 20.0% (by mass)
      Calculation of Mass from Percent
• Example:
  – Seawater contains 3.5% (by mass) of NaCl. How many
    grams of sodium chloride can be obtained from 5.00
    gallons of seawater? (1 gall. = 3.785 L; assume density
    of seawater = 1.00 g/mL)

  – Mass of seawater =
        5.00 gall x (3785 mL/gall.) x (1.00 g/mL) = 18925 g;

  – Mass of NaCl = 18925 g sw x (3.5/100) = 662 g
                 Percent by Volume

• Example:
    – A solution is prepared by mixing 150. mL of methanol,
      100. mL of acetone, and 250. mL of water. What is the
      volume percent of methanol and acetone in solution?

    – Percent methanol = (150. mL/500. mL) x 100%
•                      = 30.0% (by volume)
    – Percent acetone = (100. mL/500. mL) x 100%
    –                 = 20.0% (by volume)
                Molar Concentration

• Example:
    – 4.0 g of sodium hydroxide, NaOH, is dissolved in
      enough water to make a 100.-mL of solution. Calculate
      the molarity of NaOH.

    – Mole of NaOH = 4.0 g NaOH x (1 mole/40.0 g)
•                    = 0.10 mole
    – Molarity of NaOH = 0.10 mol/0.100 L = 1.0 M
    Calculation of Solute Mass in Solution

• Example:
   – How many grams of NaOH are present in 35.0 mL of
     6.0 M NaOH solution?

• Mole of NaOH = (6.0 mol/L) x 35.0 mL x (1 L/1000 mL)
•              = 0.21 mol
• Mass of NaOH = 0.21 mol x (40.0 g/mol)
•              = 8.4 g NaOH
      Preparing Solutions from Pure Solids
1.   From the volume (in liters) and molarity of solution,
     calculate the mole and mass of solute needed;
2.   Weigh the mass of pure solute accurately;
3.   Transfer solute into a volumetric flask of appropriate
     size;
4.   Add deionized water to the volumetric flask, well below
     the narrow neck, and shake well to dissolve the solute.
5.   When completely dissolved, add more distilled water to
     fill the flask to the mark and mix the solution well.
            Preparing Solution from Solid
• Example:
   – Explain how you would prepare 1.00 L of 0.500 M NaCl solution.

• Calculate mass of NaCl needed:
   – Mole of NaCl = 1.00 L x (0.500 mol/L) = 0.500 mol
   – Mass of NaCl = 0.500 mol x (58.44 g/mol) = 29.2 g
• Preparing the solution:
   – Weigh 29.2 g of NaCl accurately and transfer into 1-liter
     volumetric flask. Fill the flask half way with distilled water, shake
     well until all solid has dissolved. Fill the flask to the 1-liter mark
     with more distilled water and mix the solution well by inverting
     the flask back and forth several times.
          Preparing Solution from Stock
1.   Calculate volume of stock solution needed using the
     formula: MiVi = MfVf (i = initial; f = final)
2.   Measure accurately the volume of stock solution and
     carefully transfer to a volumetric flask of appropriate
     size;
3.   Dilute stock solution with distilled water to the required
     volume (or to the “mark” on volumetric flask)
4.   Mix solution well.
     (Note: if diluting concentrated acid, place some distilled
     water in the flask, add the concentrated acid, and then
     add more distilled water to the required volume.)
           Preparing Solution from Stock
• Example:
   – Explain how you would prepare 1.0 L of 3.0 M H2SO4 solution
     from concentrated H2SO4, which is 18 M.

• Calculate volume of concentrated H2SO4 needed:
   – Vol. of conc. H2SO4 = (1.0 L x 3.0 M/18 M) = 0.17 L = 170 mL
• Preparing the solution:
   – Place some distilled water in the 1-liter volumetric flask (that
     would fill the flask to about a quarter full). Measure accurately 170
     mL of conc. H2SO4 and transfer carefully to the volumetric flask
     that already contains some distilled water. Then fill the flask to the
     1-liter mark with more distilled water and mix the solution well by
     inverting the flask back and forth several times.
         Reactions in Aqueous Solution
1. Double-Displacement Reactions
  1. Precipitation reactions
  2. Neutralization (or Acid-Base) reactions
2. Oxidation-Reduction (Redox) Reactions
  1.   Combination reactions
  2.   Decomposition reactions
  3.   Combustion reactions
  4.   Single-Replacement reactions
  5.   Reactions involving strong oxidizing reagents
                Precipitation Reactions
• Reactions that produce insoluble products (or precipitates)
  when two aqueous solutions are mixed.
• Examples:
   – 1) AgNO3(aq) + KBr(aq)  AgBr(s) + KNO3(aq)

   – 2) Pb(NO3)2(aq) + K2CrO4  PbCrO4(s) + 2KNO3(aq)

   – 3) BaCl2(aq) + H2SO4(aq)  BaSO4(s) + 2NaCl(aq)

   – 4) 3Hg(NO3)2 (aq) + 2Na3PO4 (aq)  Hg3(PO4)2(s) + 6NaNO3(aq)
Solubility Rules for Predicting Solid Products
• Soluble salts:
   – All compounds of alkali metals and NH4+
   – All compounds containing nitrate, NO3-, and acetate. C2H3O2-,
     except silver acetate, which is sparingly soluble;
   – Most chlorides, bromides, and iodides, except AgX, Hg2X2, PbX2,
     and HgI2; where X = Cl-, Br-, or I-.
   – Most sulfates, except CaSO4, SrSO4, BaSO4, PbSO4 and Hg2SO4.

• Insoluble or slightly soluble salts:
   – Most hydroxides (OH-), sulfides (S2-), carbonates (CO32-),
     chromates (CrO42-), and phosphate (PO43-), except those associated
     with the Group 1A metals or NH4+.
       Predicting Precipitation Reactions
• Complete and balance the following reactions in aqueous
  solution and identify the precipitate, if formed.

• a) CaCl2(aq) + Na2CO3(aq)  ?

• b) NH4NO3(aq) + MgCl2(aq)  ?

• c) Pb(NO3)2(aq) + KI(aq)  ?

• d) AgNO3(aq) +    Na3PO4(aq)  ?
     Equations for Precipitation Reactions
• Molecular equation:
   – Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)

• Total Ionic equation:
   – Pb2+ + 2NO3- + 2K+ + CrO42-  PbCrO4(s) + 2K+ + 2NO3-
   – (K+ and NO3- are spectator ions)

• Net ionic equation:
   – Pb2+(aq) + CrO42-(aq)  PbCrO4(s)
     Acid-Base (Neutralization) Reactions
• Acid – a compound that produces hydrogen ions (H+)
  when dissolved in aqueous solution;

• Base – a compound that produces hydroxide ions (OH-) in
  aqueous solutions.

• Some examples of acids and strong bases:
• Acids: HCl, HClO4, HNO3, H2SO4, H3PO4, and HC2H3O2;
• Bases: NaOH, KOH, Ba(OH)2, and NH3.
                 Acid-Base Reactions
• Some example of acid-base reactions:

   – HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)

   – H2SO4(aq) + KOH(aq)  H2O(l) + K2SO4(aq)
   – HC2H3O2(aq) + NaOH(aq)  H2O(l) + NaC2H3O2(aq)

   – 2HClO4(aq) + Ba(OH)2(aq)  2 H2O(l) + Ba(ClO4)2(aq)
        Equations for Acid-Base Reactions
• An example of strong acid and strong base reaction:
• Molecular equation:
   – HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

• Total ionic equation:
   H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq) + H2O(l)

• Net ionic equation:
   – H+(aq) + OH-(aq)  H2O(l)
       Equations for Acid-Base Reactions
• An example of week acid and strong base reaction:
• Molecular equation:
   – HC2H3O2(aq) + NaOH(aq)  NaC2H3O2(aq) + H2O(l)

• Total ionic equation:
   HC2H3O2(aq) + Na+(aq) + OH-(aq)  Na+(aq) + C2H3O2-(aq) + H2O(l)

• Net ionic equation:
   – HC2H3O2(aq) + OH-(aq)  C2H3O2-(aq) + H2O(l)
       Equations for Acid-Base Reactions
• An example of strong acid and weak base reaction:
• Molecular equation:
   – HCl(aq) + NH3(aq)  NH4Cl(aq)

• Total ionic equation:
   – H+(aq) + Cl-(aq) + NH3(aq)  NH4+(aq) + Cl-(aq)

• Net ionic equation:
   – H+(aq) + NH3(aq)  NH4+(aq)
       Stoichiometry in Aqueous Solution

• Example-1: Precipitation reaction
   – How many grams of BaSO4 will be formed when 100.0 mL of
     0.100 M BaCl2 is reacted with 150.0 mL of 0.100 M Na2SO4?
   – Reaction: BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)

• Solution-1:
   – Find the limiting reactant by calculating the mole of each reactant,
     where the moles can be calculated by multiplying the volume (in
     liters) of the solution with the molarity.

   – Mole of BaCl2 = 0.1000 L x 0.100 mol/L = 0.0100 mole;
   – Mole of Na2SO4 = 0.1500 L x 0.100 mol/L = 0.0150 mole;
       Stoichiometry in Aqueous Solution

• Solution-1 (continued)
   – BaCl2 is the limiting reactant;
   – Therefore, mole of BaSO4 expected = 0.0100 mole;
   – Mass of BaSO4 formed = 0.0100 mol x 233.39 g/mol
   –                          = 2.33 g

   – If 2.05 g of barium sulfate was actually obtained, what is the
     percent yield?

   – Percent yield = (2.05 g/2.33 g) x 100% = 88.0%
Acid-Base Titration
                    Acid-Base Titration

• Example-1:
   – In a titration experiment, 32.40 mL of 0.2560 M NaOH solution
     was required to neutralize acetic acid in a 10.00-mL sample of
     vinegar. Calculate the molarity of acetic acid in the vinegar. If the
     density of vinegar is 1.0 g/mL, calculate the mass percent of acetic
     acid in the vinegar.
• Reaction: HC2H3O2(aq) + NaOH(aq)  H2O(l) + NaC2H3O2(aq)

• Solution-1:
   – Calculate mole of NaOH using its volume and molarity;
   – According to the equation, mole of acetic acid = mole of NaOH
                  Acid-Base Titration

• Solution-1 (continued)
   – Mole of NaOH used = 0.03240 L x 0.2560 mol/L
   –                    = 0.008294 mol;
   – Mole of HC2H3O2 reacted = 0.008294 mole;
   – Molarity of HC2H3O2 in vinegar = 0.008294 mol/0.01000 L
   –                                = 0.8294 M

   – Mole of HC2H3O2 in 100.0 mL   = 0.1000 L x 0.8294 mol/L
   –                               = 0.08294 mol
   – Mass of HC2H3O2 in 100.0 mL   = 0.08294 mol x 60.05 g/mol
   –                               = 4.981 g
                    Acid-Base Titration

• Solution-1 (continued)

    – Mass of 100.0 mL vinegar = 100.0 mL x 1.0 g/mL
    –                              = 1.0 x 102 g
    – Percent of acetic acid in vinegar = (4.981 g/ 1.0 x 102 g) x 100%
•                                     = 5.0% (by mass)
                 Acid-Base Reactions

• Example-2:
   – How many milliliters of 0.2765 M NaOH solution will be required to
     neutralize 20.00 mL of 0.1500 M H2SO4?
   Reaction: H2SO4(aq) + 2NaOH(aq)  2H2O(l) + Na2SO4(aq)


• Solution-2:
   – Mole of H2SO4 present = 0.02000 L x 0.1500 mol/L = 0.003000 mol;
   – Mole of NaOH needed = 0.003000 mol x 2 = 0.006000 mol;
   – Volume of 0.2765 M NaOH needed to neutralize the acid
   –       = (0.006000 mol)/(0.2765 mol/L) = 0.02170 L
   –                                         = 21.70 mL
                 Acid-Base Reactions

• Example-3:
   – A 4.00-mL sample of sulfuric acid is diluted to 100.0 mL. 20.00
     mL of the dilute acid is then titrated with 0.2750 M NaOH
     solution. If 35.60 mL of the base were required to neutralize the
     acid, calculate the molarity of the original sulfuric acid solution.
   Reaction: H2SO4(aq) + 2NaOH(aq)  2H2O(l) + Na2SO4(aq)

   Solution-3:
     Mole of NaOH used = 0.03560 L x (0.2750 mol/L) = 0.009790 mole;
     Mole of H2SO4 titrated = ½ (0.009790 mol) = 0.004895 mole
                 Acid-Base Reactions

• Solution-3 (continued):
   – Molarity of dilute acid = (0.004895 mol)/(0.02000 L) = 0.24475 M
   – Molarity of undiluted acid = 0.24475 M x 100.0 mL/4.00 mL
   –                            = 6.12 M
           Reactions That Produce Gas
• Reactions producing CO2 gas:
   – CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g);

   – NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g);

• Reaction that produces SO2 gas:
      Na2SO3(s) + 2HCl(aq)  2NaCl(aq) + H2O(l) + SO2(g);

• Reaction that produces H2S gas:
      Na2S(aq) + 2HCl(aq)  2NaCl(aq) + H2S(g);
           Oxidation-Reduction Reactions

• Oxidation  loss of electrons and increase in oxidation number;
• Reduction  gain of electrons and decrease in oxidation number;

• Oxidation-reduction (or Redox) reaction  one that involves transfer
  of electrons from one reactant to the other;

• Oxidizing agent  the reactant that gains electrons and got reduced;
• Reducing agent  the reactant that loses electrons and got oxidized.
            Types of Redox Reactions

• Reactions between metals and nonmetals;
• Combustion reactions (reactions with molecular oxygen);
• Single replacement reactions;
• Decomposition reactions that form free elements;
• Reactions in aqueous solution involving oxidizing and
  reducing agents;
• Disproportionation reactions.
             Types of Redox Reactions

• Reactions between metals and nonmetals:
   – 4Al(s) + 3 O2(g)  2Al2O3(s);
   – 3Mg(s) + N2(g)  Mg3N2(s);


• Combustion reactions:
   – CH4(g) + 2 O2(g)  CO2(g) + 2H2O(g);
   – 2C8H18(l) + 25 O2(g)  16CO2(g) + 18H2O(g);
   – C2H5OH(l) + 3 O2(g)  2CO2(g) + 3H2O(g);
             Types of Redox Reactions

• Single-Replacement Reactions:
   – Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g);
   – Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s);
   – Cl2(aq) + 2KBr(aq)  2KCl(aq) + Br2(aq);


• Decomposition Reactions:
   – 2HgO(s)  2Hg(l) + O2(g);
   – (NH4)2Cr2O7(s)  Cr2O3(s) + N2(g) + 4H2O(g);
               Types of Redox Reactions

• Reactions in aqueous solutions that involve strong oxidizing
  reagents:
    – MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + H2O(l);

    – Cr2O72-(aq) + 3H2O2(aq) + 8H+(aq)  2Cr3+(aq) + 7H2O(l) + 3 O2(g);

    – 2Cr(OH)4-(aq) + 3H2O2(aq) + 2 OH-(aq)  2CrO42-(aq) + 8H2O(l);

• Disproportionation reaction:
    – Cl2(g) + 2NaOH(aq)  NaOCl(aq) + NaCl(aq) + H2O(l);

    – 3Br2(aq) + 6NaOH(aq)  NaBrO3(aq) + NaBr(aq) + H2O(l);
Guidelines for Determining Oxidation Numbers of Elements


  1.   Atoms in the free elemental form are assigned oxidation number zero (0);
  2.   The sum of oxidation number in neutral molecules or formula units is 0;
       the sum of oxidation number (o.n.) of atoms in a polyatomic ion is equal to the
       net charge of the ion (in magnitude and sign);
  3.   In their compounds, each Group IA metal is assigned an o.n. of +1; each Group
       IIA metal an o.n. of +2; boron and aluminum each an o.n. of +3, and fluorine an
       o.n. of –1;
  4.   Hydrogen is assigned an o.n. of +1 in compounds or polyatomic ions with
       nonmetals, and an o.n. of –1 in metal hydrides;
  5.   In compounds and polyatomic ions, oxygen is assigned an o.n. of -2, except in
       peroxides, in which its o.n. is –1;
  6.   In binary compounds with metals, chlorine, bromine, and iodine each has an o.n.
       of -1; sulfur, selenium, and tellurium each has an o.n. of -2.
            Oxidation-Reduction Reactions

• In the following equations, identify all reactions that are redox
  reactions:
•   1.   2KMnO4(aq) + 16HCl(aq)  2MnCl2(aq) + 2KCl(aq) + 5Cl2(aq) + 8H2O(l);

• 2.     2KClO3(s)  2KCl(s) + 3 O2(g);

• 3.     CaCO3(s)  CaO(s) + CO2(g);

• 4.     Mg(OH)2(s)  MgO(s) + H2O(g);

• 5.     Mg(s) + ZnSO4(aq)  MgSO4(aq) + Zn(s);
•   6.   Cr2O72-(aq) + 3C2H5OH(l) + 2H+(aq)  2Cr3+(aq) + 3CH3COOH(aq) + 4H2O(l)
    Balancing Redox Reactions by Half-Equation Method


• Example-1: Balance the following oxidation-reduction reaction in aqueous
   solution:
    –          MnO4-(aq) + Fe2+(aq)  Mn2+(aq) + Fe3+(aq)

• Solution-1:
    – Note: the above equation is both not balanced and not complete. It only shows the
      components (reactants) that undergoes changes is oxidation numbers;
    – Redox reactions in acidic solution means that you need to add H+ ion in the
      equation, which produces water as one of the products.
    Balancing Redox Reactions by Half-Equation Method


• Solution-1 (continued)
   – Balancing the equation – first step, break up the equation into two half
     equations: oxidation and reduction half-equations:

   –       MnO4-(aq)              Mn2+(aq)
   – In this case, all the four oxygen in MnO4- will become water in acidic solution. So,
     add enough H+ ion to form H2O with the four oxide ions in in MnO4-. The half-
     equation with all atoms balanced will look like this:

   –       MnO4-(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l)
    Balancing Redox Reactions by Half-Equation Method


• Solution-1 (continued)
   – Next step, check the total charges on both side of the half-equation; note
     that they are not equal. Add enough electrons on the side that has more
     positive charges (or less negative charges), so that the total charges on
     both sides become equal (in magnitude and sign).

   –    MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l);

   – Now we have a balanced reduction half-equation.
   – (How do you know it is “reduction” half-equation?)
    Balancing Redox Reactions by Half-Equation Method


• Solution-1 (continued)
   – Next, write the other half-equation and balance it:
   –       Fe2+(aq)  Fe3+(aq) + e-;         (this is oxidation half-equation)

   – To obtain the overall equation, we add the two balanced half-equations, but make
     sure the number of electrons on both half-equations are equal, so that they cancel
     out. The overall equation should not contain any electrons. In this case, we multiply
     the above oxidation half-equation by 5 and obtain:
   –       5Fe2+(aq)  5Fe3+(aq) + 5e-;
   – Now, adding the two half-equations yields the following balanced net ionic
     equation:

   – MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l);
    Balancing Redox Reactions by Half-Equation Method


• Example-2: Balance the following oxidation-reduction reaction in
   acidic solution:
    –     Cr2O72-(aq) + H2O2(aq)  Cr3+(aq) + O2(g) + H2O(l);

• Solution-2: Write the two half-equations and balance them.
    – Reduction half-equation:
    – Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l);

    – Oxidation half-equation:
    –      H2O2(aq)  O2(g) + 2H+(aq) + 2e-;
    Balancing Redox Reactions by Half-Equation Method


• Solution-2 (continued)
   – Multiply the oxidation half-equation by 3 to make the number of electrons
     equal with that of the reduction-half equation.
   –       3H2O2(aq)  3 O2(g) + 6H+(aq) + 6e-;

   – Then add the two half-equations, canceling all the electrons on both side
     of the equation, all the H+ ions on the right-hand side, and the same
     number of H+ ions on the left-hand side. The final (balanced) net ionic
     equation will be as follows:

   – Cr2O72-(aq) + 3H2O2(aq) + 8H+(aq)  2Cr3+(aq) + 3O2(g) + 7H2O(l);
    Balancing Redox Reactions by Half-Equation Method


• Example-3: Balance the following oxidation-reduction reaction in basic
   solution:
    – Cr(OH)4-(aq) + H2O2(aq)  CrO42-(aq) + H2O(l);

• Solution-3:
    – Write the two half-equations and balance them, adding electrons on either side as
      needed to balance the charges.
    – Note that this reaction is in basic solution, which means the overall equation should
      contains OH- ion instead of H+ ions.
    Balancing Redox Reactions by Half-Equation Method


• Solution-3 (continued):
   – Oxidation half-equation:
   –    Cr(OH)4-(aq) + 4OH-(aq)  CrO42-(aq) + 4H2O(l) + 3e-;

   – Reduction half-equation:
   –      H2O2(aq) + 2e-  2OH-;

   – Multiply the oxidation half-equation by 2 and the reduction half-equation
     by 3 to make the number of electrons in both half-equations equal.
   –    2Cr(OH)4-(aq) + 8OH-(aq)  2CrO42-(aq) + 8H2O(l) + 6e-;
   –           3H2O2(aq) + 6e-  6OH-;
    Balancing Redox Reactions by Half-Equation Method


• Solution-3 (continued):
   – Now add the two half-equations, canceling all the electrons on both sides
     of the equation, all the 6OH- ions on the right-hand side, and the same
     number of OH- ions on the left-hand side.
   – The overall balanced net ionic equation will appear as follows:


   – 2Cr(OH)4-(aq) + 3H2O2(aq) + 2OH-(aq)  2CrO42-(aq) + 8H2O(l);
   Uses of Reactions in Aqueous Solutions

• 1. Dissolving Insoluble Compounds
   – Fe2O3(s) + 6HNO3(aq)  2Fe(NO3)3(aq) + 3H2O(l);
   – Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)

• 2. Syntheses of Inorganic Compounds
   – AgNO3(aq) + NaBr(aq)  AgBr(s) + NaNO3(aq);
   – Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)

• 3. Extraction of Metals form Solution
   – Mg2+(aq) + Ca(OH)2(aq)  Mg(OH)2(s) + Ca2+(aq);

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:195
posted:5/25/2012
language:English
pages:52