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Answers to Study Guide for CHEM 102 Test 3, Fall 2007

  1. For the reactions below find the equilibrium concentrations of all products and reactants. Assume the initial concentration of all reactants is 2.90 M and the initial concentration of all products is zero (0.00 M).

    1. Mn2(CO)10 + CH3I <------> CH3Mn(CO)5 + IMn(CO)5

      Kc = 1.88 x 10-6

      Kc = {[CH3Mn(CO)5][IMn(CO)5]}/{[Mn2(CO)10][CH3I]}

      Kc is small and the amount of products will be small. The amount of each of the reactants before equilibrium is reached will be nearly identical to the starting concentration. Therefore the following is a good approximation:

      Kc = x2/(2.90)2
      Kc(2.90)2 = x2
      (1.88 x 10-6)(2.90)2 = 1.581 x 10-5 = x2
      (x2) = (1.581 x 10-5)
      x = 3.98 x 10-3 M = [CH3Mn(CO)5] = [IMn(CO)5]
      Equilibrium concentration of [Mn2(CO)10] = [CH3I] = 2.90 M -(3.98 x 10-3 M) = 2.90 M

    2. Cl3CCOOH(aq) <------> H+(aq) + Cl3CCOO-(aq)

      Ka = {[H+][Cl3CCOO-]/[Cl3CCOOH]

      Ka = 1.29 x 10-1

      Ka is large enough that the quadratic equation should be used. Approximating the reaction without the quadratic gives the following: Ka = x2/[2.90]

      Ka[2.90] = x2

      (1.29 x 10-1)(2.90) = 0.3741 = x2

      (0.3741) = (x2) = x = 0.612 M = [H+] = [Cl3CCOO-]

      [Cl3CCOOH] = 2.90 - 0.612 = 2.29 M

      Using the quadratic equation:

      Ka = x2/[2.90 - x]

      Ka[2.90-x] = x2

      (2.90)Ka-Kax = x2 x2 + Kax -(2.90)Ka = 0

      x2 + (1.29 x 10-1)x - 0.3741 = 0

      x ={-b + (b2 -4ac)}/2a= {-0.129 + [(0.129)2-4(1)(-0.3741)]}/2(1)

      x = {-0.129 + (1.513041)}/2(1)

      x = {-0.129 + 1.230057316}/2 = 0.0.551 M = [H+] = [Cl3CCOO-]

      [Cl3CCOOH] = 2.90 - 0.551 = 2.35 M

  2. Provide the equilibrium expressions for the reactions below:

    1. HCl(aq) + Rh(CO){P{C6H5}3Cl(aq) < -----> HRh(CO){P(C6H5}3Cl2(aq)

      Kc = [HRh(CO){P(C6H5)3Cl2]/{[HCl][Rh(CO){P(C6H5)3}Cl]}

    2. 2 SO2(g) + O2(g) <-----> 2 SO3(g)

      Kc = [SO3]2/{[SO2]2[O2]}

    3. HC2H3O2(aq) <-----> H+(aq) + C2H3O2-(aq)

      Kc = {[H+][C2H3O2-]}/[HC2H3O2]

    4. N2O4(g) <-----> 2 NO2(g)

      Kc = [NO2]2/[N2O4]

    5. PCl5(g) <-----> PCl3(g) + Cl2(g)

      Kc = {[PCl3][Cl2]}/[PCl5]

  3. What is the Kc or Kp for the reactions below if the concentrations are those at equilibrium for the reaction?

    1. 2 CO(g) + O2(g) <-----> 2 CO2(g)

      [CO] = 753 torr, [O2] = 325 torr, [CO2] = 1200 torr

      KP = (pCO2)2/{(pCO)2(pO2)} = 7.81 x 10-3

    2. HCl(aq) + RhCl3(aq) <------> HRhCl4(aq)

      [HCl] = 6.25 M, [RhCl3] = 1.33 M, [HRhCl4] = 0.0551 M

      Kc = [HRhCl4]/{[HCl][RhCl3]} = 6.63 x 10-3

    3. H2(g) + I2(g) <-----> 2 HI(g)

      [H2] = 0.00926 atm, [I2] = 0.00516 atm, [HI] = 0.00510 atm

      KP = (pHI)2/{(pH2)(pI2) = 0.544

  4. What is the equilibrium concentration of the reactant(s) and product(s) in the reactions below given the equilibrium constants and starting concentrations of the reactants?

    1. HCl(aq) + Rh(CO){P(C6H5)3}2Cl(aq) <-----> HRh(CO){P(C6H5)3}2Cl2(aq)

      [HCl] = 1.000 M, [Rh(CO){P(C6H5)3}2Cl] = 1.000 M, Kc = 150.00

      Kc = [HRh(CO){P(C6H5)}2Cl2]/{[HCl][Rh(CO){P(C6H5)3}2Cl]}

      Kc = x/{[1-x][1-x]} = x/(1 - 2x + x2)

      Kc(1 - 2x + x2) = x

      Kc - 2Kcx + Kcx2 = x

      150 -300x + 150x2 = x

      150x2 -301x + 150 = 0

      Use the quadratic equation

      [301 + {(-301)2 -4(150)(150)}]/2(150) =

      {301 + (601)}/300 = {301 + 245}/300

      Adding 301 + 245 = 546 and 546/300 = 1.82 M. This answer is obviously not the real root, since the maximum we can expect to see is 1.00 M in product if the reaction went to completion. Subtracting 301-245 = 56 and 56/300 = 0.187 M = product concentration. Reactant concentrations must be 1.00 M -0.187 M = 0.813 M. The quadratic must be used because of the high value of Kc.

    2. HC2H3O2(aq) -----> H+(aq) + C2H3O2-(aq)

      [HC2H3O2] = 1.977 M, KA = 1.800 x 10-5

      In this case KA is small enough that we will not need the quadratic equation.

      KA = {[H+][C2H3O2-]}/[HC2H3O2] = x2/[HC2H3O2]

      KA[HC2H3O2] = x2

      (1.8 x 10-5)(1.977) = 3.56 x 10-5 = x2

      (3.56 x 10-5) = x = 5.97 x 10-3 = [H+] = [C2H3O2-]

      Equilibrium conc. of [HC2H3O2] = 1.977-(5.97 x 10-3) = 1.971 M

    3. CO(g) + Cl2(g)<:-----> COCl2(g)

      [CO] = 455 torr, [Cl2] = 455 torr, Kp = 1.67 x 10-3

      Method is identical to part a of this problem above.

      pCOCl2 = 139 torr
      pCO = pCl2 = 455 - 139 = 316 torr

  5. Describe the effects of changing the conditions that chemical a reaction is run under upon the equilibrium of a reaction. Apply LeChatelier's principle using the example given below:

    2 NH3(g) <-------> N2H4(g) + H2(g) the reaction being run in a constant volume vessel and the reaction is endothermic.

    Which direction would the equilibrium shift if:

    1. the temperature is increased - shifts to right since reaction is endothermic
    2. more NH3 is supplied - shifts to right
    3. more H2(g) is supplied - shifts to left
    4. the N2H4(g) is removed from the system as fast as it is produced - shifts to right
    5. the volume of the reaction vessel is doubled - no effect
    6. the volume of the reaction vessel is halved - no effect

    Increasing temperature usually increases reaction rate. The rule of thumb is that the reaction rate increases by two fold for every 10oC of temperature increase. But both the forward & reverse reaction rates approximately double. To succesfully determine the effect of temperature on a reaction, you must know whether the reaction is exothermic or endothermic. In endothermic reactions heat is treated like a reactant. In exothermic reactions, heat is treated like a product. Applying LeChatelier's principle to the reaction, one must consider concentratons of reactants and products, volume, number of particles of product compared to reactant, endothermic or exothermic, etc. Equilibrium constants are determined for each temperature.

    1. For the reactions below find the equilibrium concentrations of all products and reactants. Assume the initial concentration of all reactants is 2.90 M and the initial concentration of all products is zero (0.00 M).

      1. Mn2(CO)10 + CH3I <------> CH3Mn(CO)5 + IMn(CO)5

        Kc = 1.88 x 10-6

        Kc = {[CH3Mn(CO)5][IMn(CO)5]}/{[Mn2(CO)10][CH3I]}

        Kc is small and the amount of products will be small. The amount of each of the reactants before equilibrium is reached will be nearly identical to the starting concentration. Therefore the following is a good approximation:

        Kc = x2/(2.90)2
        Kc(2.90)2 = x2
        (1.88 x 10-6)(2.90)2 = 1.581 x 10-5 = x2
        (x2) = (1.581 x 10-5)
        x = 3.98 x 10-3 M = [CH3Mn(CO)5] = [IMn(CO)5]
        Equilibrium concentration of [Mn2(CO)10] = [CH3I] = 2.90 M -(3.98 x 10-3 M) = 2.90 M

      2. Cl3CCOOH(aq) <------> H+(aq) + Cl3CCOO-(aq)

        Ka = {[H+][Cl3CCOO-]/[Cl3CCOOH]

        Ka = 1.29 x 10-1

        Ka is large enough that the quadratic equation should be used. Approximating the reaction without the quadratic gives the following: Ka = x2/[2.90]

        Ka[2.90] = x2

        (1.29 x 10-1)(2.90) = 0.3741 = x2

        (0.3741) = (x2) = x = 0.612 M = [H+] = [Cl3CCOO-]

        [Cl3CCOOH] = 2.90 - 0.612 = 2.29 M

        Using the quadratic equation:

        Ka = x2/[2.90 - x]

        Ka[2.90-x] = x2

        (2.90)Ka-Kax = x2 x2 + Kax -(2.90)Ka = 0

        x2 + (1.29 x 10-1)x - 0.3741 = 0

        x ={-b + (b2 -4ac)}/2a= {-0.129 + [(0.129)2-4(1)(-0.3741)]}/2(1)

        x = {-0.129 + (1.513041)}/2(1)

        x = {-0.129 + 1.230057316}/2 = 0.0.551 M = [H+] = [Cl3CCOO-]

        [Cl3CCOOH] = 2.90 - 0.551 = 2.35 M

    2. Provide the equilibrium expressions for the reactions below:

      1. HCl(aq) + Rh(CO){P{C6H5}3Cl(aq) < -----> HRh(CO){P(C6H5}3Cl2(aq)

        Kc = [HRh(CO){P(C6H5)3Cl2]/{[HCl][Rh(CO){P(C6H5)3}Cl]}

      2. 2 SO2(g) + O2(g) <-----> 2 SO3(g)

        Kc = [SO3]2/{[SO2]2[O2]}

      3. HC2H3O2(aq) <-----> H+(aq) + C2H3O2-(aq)

        Kc = {[H+][C2H3O2-]}/[HC2H3O2]

      4. N2O4(g) <-----> 2 NO2(g)

        Kc = [NO2]2/[N2O4]

      5. PCl5(g) <-----> PCl3(g) + Cl2(g)

        Kc = {[PCl3][Cl2]}/[PCl5]

    3. What is the Kc or Kp for the reactions below if the concentrations are those at equilibrium for the reaction?

      1. 2 CO(g) + O2(g) <-----> 2 CO2(g)

        [CO] = 753 torr, [O2] = 325 torr, [CO2] = 1200 torr

        KP = (pCO2)2/{(pCO)2(pO2)} = 7.81 x 10-3

      2. HCl(aq) + RhCl3(aq) <------> HRhCl4(aq)

        [HCl] = 6.25 M, [RhCl3] = 1.33 M, [HRhCl4] = 0.0551 M

        Kc = [HRhCl4]/{[HCl][RhCl3]} = 6.63 x 10-3

      3. H2(g) + I2(g) <-----> 2 HI(g)

        [H2] = 0.00926 atm, [I2] = 0.00516 atm, [HI] = 0.00510 atm

        KP = (pHI)2/{(pH2)(pI2) = 0.544

    4. What is the equilibrium concentration of the reactant(s) and product(s) in the reactions below given the equilibrium constants and starting concentrations of the reactants?

      1. HCl(aq) + Rh(CO){P(C6H5)3}2Cl(aq) <-----> HRh(CO){P(C6H5)3}2Cl2(aq)

        [HCl] = 1.000 M, [Rh(CO){P(C6H5)3}2Cl] = 1.000 M, Kc = 150.00

        Kc = [HRh(CO){P(C6H5)}2Cl2]/{[HCl][Rh(CO){P(C6H5)3}2Cl]}

        Kc = x/{[1-x][1-x]} = x/(1 - 2x + x2)

        Kc(1 - 2x + x2) = x

        Kc - 2Kcx + Kcx2 = x

        150 -300x + 150x2 = x

        150x2 -301x + 150 = 0

        Use the quadratic equation

        [301 + {(-301)2 -4(150)(150)}]/2(150) =

        {301 + (601)}/300 = {301 + 245}/300

        Adding 301 + 245 = 546 and 546/300 = 1.82 M. This answer is obviously not the real root, since the maximum we can expect to see is 1.00 M in product if the reaction went to completion. Subtracting 301-245 = 56 and 56/300 = 0.187 M = product concentration. Reactant concentrations must be 1.00 M -0.187 M = 0.813 M. The quadratic must be used because of the high value of Kc.

      2. HC2H3O2(aq) -----> H+(aq) + C2H3O2-(aq)

        [HC2H3O2] = 1.977 M, KA = 1.800 x 10-5

        In this case KA is small enough that we will not need the quadratic equation.

        KA = {[H+][C2H3O2-]}/[HC2H3O2] = x2/[HC2H3O2]

        KA[HC2H3O2] = x2

        (1.8 x 10-5)(1.977) = 3.56 x 10-5 = x2

        (3.56 x 10-5) = x = 5.97 x 10-3 = [H+] = [C2H3O2-]

        Equilibrium conc. of [HC2H3O2] = 1.977-(5.97 x 10-3) = 1.971 M

      3. CO(g) + Cl2(g)<:-----> COCl2(g)

        [CO] = 455 torr, [Cl2] = 455 torr, Kp = 1.67 x 10-3

        Method is identical to part a of this problem above.

        pCOCl2 = 139 torr
        pCO = pCl2 = 455 - 139 = 316 torr

    5. Describe the effects of changing the conditions that chemical a reaction is run under upon the equilibrium of a reaction. Apply LeChatelier's principle using the example given below:

      2 NH3(g) <-------> N2H4(g) + H2(g) the reaction being run in a constant volume vessel and the reaction is endothermic.

      Which direction would the equilibrium shift if:

      1. the temperature is increased - shifts to right since reaction is endothermic
      2. more NH3 is supplied - shifts to right
      3. more H2(g) is supplied - shifts to left
      4. the N2H4(g) is removed from the system as fast as it is produced - shifts to right
      5. the volume of the reaction vessel is doubled - no effect
      6. the volume of the reaction vessel is halved - no effect

      Increasing temperature usually increases reaction rate. The rule of thumb is that the reaction rate increases by two fold for every 10oC of temperature increase. But both the forward & reverse reaction rates approximately double. To succesfully determine the effect of temperature on a reaction, you must know whether the reaction is exothermic or endothermic. In endothermic reactions heat is treated like a reactant. In exothermic reactions, heat is treated like a product. Applying LeChatelier's principle to the reaction, one must consider concentratons of reactants and products, volume, number of particles of product compared to reactant, endothermic or exothermic, etc. Equilibrium constants are determined for each temperature.

      1. For the reactions below find the equilibrium concentrations of all products and reactants. Assume the initial concentration of all reactants is 2.90 M and the initial concentration of all products is zero (0.00 M).

        1. Mn2(CO)10 + CH3I <------> CH3Mn(CO)5 + IMn(CO)5

          Kc = 1.88 x 10-6

          Kc = {[CH3Mn(CO)5][IMn(CO)5]}/{[Mn2(CO)10][CH3I]}

          Kc is small and the amount of products will be small. The amount of each of the reactants before equilibrium is reached will be nearly identical to the starting concentration. Therefore the following is a good approximation:

          Kc = x2/(2.90)2
          Kc(2.90)2 = x2
          (1.88 x 10-6)(2.90)2 = 1.581 x 10-5 = x2
          (x2) = (1.581 x 10-5)
          x = 3.98 x 10-3 M = [CH3Mn(CO)5] = [IMn(CO)5]
          Equilibrium concentration of [Mn2(CO)10] = [CH3I] = 2.90 M -(3.98 x 10-3 M) = 2.90 M

        2. Cl3CCOOH(aq) <------> H+(aq) + Cl3CCOO-(aq)

          Ka = {[H+][Cl3CCOO-]/[Cl3CCOOH]

          Ka = 1.29 x 10-1

          Ka is large enough that the quadratic equation should be used. Approximating the reaction without the quadratic gives the following: Ka = x2/[2.90]

          Ka[2.90] = x2

          (1.29 x 10-1)(2.90) = 0.3741 = x2

          (0.3741) = (x2) = x = 0.612 M = [H+] = [Cl3CCOO-]

          [Cl3CCOOH] = 2.90 - 0.612 = 2.29 M

          Using the quadratic equation:

          Ka = x2/[2.90 - x]

          Ka[2.90-x] = x2

          (2.90)Ka-Kax = x2 x2 + Kax -(2.90)Ka = 0

          x2 + (1.29 x 10-1)x - 0.3741 = 0

          x ={-b + (b2 -4ac)}/2a= {-0.129 + [(0.129)2-4(1)(-0.3741)]}/2(1)

          x = {-0.129 + (1.513041)}/2(1)

          x = {-0.129 + 1.230057316}/2 = 0.0.551 M = [H+] = [Cl3CCOO-]

          [Cl3CCOOH] = 2.90 - 0.551 = 2.35 M

      2. Provide the equilibrium expressions for the reactions below:

        1. HCl(aq) + Rh(CO){P{C6H5}3Cl(aq) < -----> HRh(CO){P(C6H5}3Cl2(aq)

          Kc = [HRh(CO){P(C6H5)3Cl2]/{[HCl][Rh(CO){P(C6H5)3}Cl]}

        2. 2 SO2(g) + O2(g) <-----> 2 SO3(g)

          Kc = [SO3]2/{[SO2]2[O2]}

        3. HC2H3O2(aq) <-----> H+(aq) + C2H3O2-(aq)

          Kc = {[H+][C2H3O2-]}/[HC2H3O2]

        4. N2O4(g) <-----> 2 NO2(g)

          Kc = [NO2]2/[N2O4]

        5. PCl5(g) <-----> PCl3(g) + Cl2(g)

          Kc = {[PCl3][Cl2]}/[PCl5]

      3. What is the Kc or Kp for the reactions below if the concentrations are those at equilibrium for the reaction?

        1. 2 CO(g) + O2(g) <-----> 2 CO2(g)

          [CO] = 753 torr, [O2] = 325 torr, [CO2] = 1200 torr

          KP = (pCO2)2/{(pCO)2(pO2)} = 7.81 x 10-3

        2. HCl(aq) + RhCl3(aq) <------> HRhCl4(aq)

          [HCl] = 6.25 M, [RhCl3] = 1.33 M, [HRhCl4] = 0.0551 M

          Kc = [HRhCl4]/{[HCl][RhCl3]} = 6.63 x 10-3

        3. H2(g) + I2(g) <-----> 2 HI(g)

          [H2] = 0.00926 atm, [I2] = 0.00516 atm, [HI] = 0.00510 atm

          KP = (pHI)2/{(pH2)(pI2) = 0.544

      4. What is the equilibrium concentration of the reactant(s) and product(s) in the reactions below given the equilibrium constants and starting concentrations of the reactants?

        1. HCl(aq) + Rh(CO){P(C6H5)3}2Cl(aq) <-----> HRh(CO){P(C6H5)3}2Cl2(aq)

          [HCl] = 1.000 M, [Rh(CO){P(C6H5)3}2Cl] = 1.000 M, Kc = 150.00

          Kc = [HRh(CO){P(C6H5)}2Cl2]/{[HCl][Rh(CO){P(C6H5)3}2Cl]}

          Kc = x/{[1-x][1-x]} = x/(1 - 2x + x2)

          Kc(1 - 2x + x2) = x

          Kc - 2Kcx + Kcx2 = x

          150 -300x + 150x2 = x

          150x2 -301x + 150 = 0

          Use the quadratic equation

          [301 + {(-301)2 -4(150)(150)}]/2(150) =

          {301 + (601)}/300 = {301 + 245}/300

          Adding 301 + 245 = 546 and 546/300 = 1.82 M. This answer is obviously not the real root, since the maximum we can expect to see is 1.00 M in product if the reaction went to completion. Subtracting 301-245 = 56 and 56/300 = 0.187 M = product concentration. Reactant concentrations must be 1.00 M -0.187 M = 0.813 M. The quadratic must be used because of the high value of Kc.

        2. HC2H3O2(aq) -----> H+(aq) + C2H3O2-(aq)

          [HC2H3O2] = 1.977 M, KA = 1.800 x 10-5

          In this case KA is small enough that we will not need the quadratic equation.

          KA = {[H+][C2H3O2-]}/[HC2H3O2] = x2/[HC2H3O2]

          KA[HC2H3O2] = x2

          (1.8 x 10-5)(1.977) = 3.56 x 10-5 = x2

          (3.56 x 10-5) = x = 5.97 x 10-3 = [H+] = [C2H3O2-]

          Equilibrium conc. of [HC2H3O2] = 1.977-(5.97 x 10-3) = 1.971 M

        3. CO(g) + Cl2(g)<:-----> COCl2(g)

          [CO] = 455 torr, [Cl2] = 455 torr, Kp = 1.67 x 10-3

          Method is identical to part a of this problem above.

          pCOCl2 = 139 torr
          pCO = pCl2 = 455 - 139 = 316 torr

      5. Describe the effects of changing the conditions that chemical a reaction is run under upon the equilibrium of a reaction. Apply LeChatelier's principle using the example given below:

        2 NH3(g) <-------> N2H4(g) + H2(g) the reaction being run in a constant volume vessel and the reaction is endothermic.

        Which direction would the equilibrium shift if:

        1. the temperature is increased - shifts to right since reaction is endothermic
        2. more NH3 is supplied - shifts to right
        3. more H2(g) is supplied - shifts to left
        4. the N2H4(g) is removed from the system as fast as it is produced - shifts to right
        5. the volume of the reaction vessel is doubled - no effect
        6. the volume of the reaction vessel is halved - no effect

        Increasing temperature usually increases reaction rate. The rule of thumb is that the reaction rate increases by two fold for every 10oC of temperature increase. But both the forward & reverse reaction rates approximately double. To succesfully determine the effect of temperature on a reaction, you must know whether the reaction is exothermic or endothermic. In endothermic reactions heat is treated like a reactant. In exothermic reactions, heat is treated like a product. Applying LeChatelier's principle to the reaction, one must consider concentratons of reactants and products, volume, number of particles of product compared to reactant, endothermic or exothermic, etc. Equilibrium constants are determined for each temperature.

      6. Calculate Eo for the cells below at standard conditions. Are the cells galvanic (voltaic) or electrolytic?

        1. Zn(s)|Zn2+|sat'd KCl|Fe2+|Fe(s)

          Zn -----> Zn2+ + 2 e- Eo = 0.76 V
          2 e- + Fe2+ -----> Fe Eo = -0.44 V
          Eocell = 0.32 V, galvanic (voltaic)

        2. Ni(s)|Ni2+|sat'd KCl|Cd2+|Cd(s)
          Ni -----> Ni2+ + 2 e- Eo = 0.25 V
          Cd2+ + 2 e- -----> Cd Eo = -0.40 V
          Eocell = -0.15 V, electrolytic

        3. Pb(s)|Pb2+|sat'd KCl|Au3+|Au(s), 1.63 V, galvanic
        4. Au(s)|Au3+|sat'd KCl|Ag+|Ag(s), -0.70 V, electrolytic
        5. Cr(s)|Cr3+|sat'd KCl|Hg22+|Hg(l), 1.59 V, galvanic
        6. Al(s)|Al3+|sat'd KCl|Mg2+|Mg(s), -0.71 V, electrolytic
        7. Sn(s)|Sn2+|sat'd KCl|Be2+|Be(s), -1.71 V, electrolytic
        8. Ba(s)|Ba2+|sat'd KCl|Cr3+|Cr(s), 2.16 V, galvanic
        9. Mn(s)|Mn2+|sat'd KCl|Pb2+|Pb(s), 1.05 V, galvanic
        10. Cd(s)|Cd2+|sat'd KCl|Zn2+|Zn(s), -0.36 V, electrolytic

      7. Calculate Ecell for the cells below at the non-standard conditions given and 298 K. Are the cells galvanic (voltaic) or electrolytic?

        1. Sn(s)|Sn2+(1.953 M)|sat'd KCl|Zn2+(0.3689 M)|Zn(s)

          Balance the equation & find Eocell:

          Sn -----> Sn2+ + 2 e- Eo = 0.14 V
          2 e- + Zn2+ -----> Zn Eo = -0.76 V
          Sn + Zn2+ -----> Sn2+ + Zn, Eocell = -0.62 V, n = 2

          Ecell = Eocell - (0.0591/2)logQ,

          Q = [Sn2+]/[Zn2+] = 1.953/0.3689 = 5.294

          Ecell = -0.62 -[(0.0591/2)log5.294]
          Ecell = -0.62 - [(0.0591)(0.7238)/2 = -0.62 V - 0.02139
          Ecell = -0.64138 = -0.64 V, electrolytic

        2. Hg(l)|Hg22+(1.564 M)|sat'd KCl|Au3+(3.596 M)|Au(s)

          Balance the equation & find Eocell:

          6 Hg -----> 3 Hg22+ + 6 e- Eo = -0.85 V
          6 e- + 2 Au3+ -----> 2 Au Eo = 1.50 V
          6 Hg + 2 Au3+ ----> 3 Hg22+ + 2 Au, Eocell = 0.65 V

          n = 6

          Ecell = Eocell - (0.0591/n)logQ,

          Q = [Hg22+]3/[Au3+]2 = (1.564)3/(3.596)2 = 0.2959

          Ecell = 0.65 - (0.0591/6)log(0.2959)
          Ecell = 0.65 - [(0.0591)(-0.5288)/6]
          Ecell = 0.65 + [(0.0591)(0.5288)/6] = 0.65521 = 0.66 V, galvanic

        3. Au(s)|Au3+(1.112 M)|sat'd KCl|Ag+(0.7888 M)|Ag(s), -0.71 V, electolytic
        4. Cu(s)|Cu2+(3.500 M)|sat'd KCl|Ag+(4.000 M)|Ag(s), 0.44 V, galvanic
        5. Mn(s)|Mn2+(0.7542 M)|sat'd KCl|Sr2+(1.556 M)|Sr(s), -1.70 V, electrolytic
        6. Sn(s)|Sn2+(4.9934 M)|sat'd KCl|Pb2+(0.7848 M)|Pb(s), -0.01 V, electrolytic
        7. Mn(s)|Mn2+(3.000 M)|sat'd KCl|Zn2+(4.000 M)|Zn(s), 0.42 V, galvanic
        8. Mg(s)|Mg2+(3.984 M)|sat'd KCl|Ni2+(0.8347 M)|Ni(s), 2.10 V, galvanic
        9. Fe(s)|Fe2+(0.3908 M)|sat'd KCl|Hg22+(2.387 M)|Hg(l), 1.31 V, galvanic
        10. Ni(s)|Ni2+(3.478 M)|sat'd KCl|Ag+(2.408 M|Ag(s), 1.06 V, galvanic

      8. What is the Eo for a battery made from aluminum and iodine under standard conditions according to the cell diagram below:

        Al(s)|Al3+(1.00 M)|sat'd KCl|(I2(s)/Pt)| I-(aq)(1.00 M)

        2 Al -----> 2 Al3+ + 6 e- Eo = 1.66 V
        3 I2 + 6 e- -----> 6 I- Eo = 0.53 V
        2 Al + 3 I2 -----> 2 Al3+ + 6 I- , Eocell = 2.19 V

        What would be the voltage of this cell if the Al3+ concentration is 0.5612 M and the I- concentration is 6.00 M?

        Ecell = Eocell - (0.0591/n)logQ

        Q = [Al3+]2[I-]6 = (.5612)2(6.00)6 = 14,694
        log(14,694) = 4.167

        Ecell = 2.19 V - (0.0591)(4.167)/6 = 2.15 V

        The problem with this cell is that the reduced species, I- is always increasing in concentration as the cell produces electricity, lessening the voltage of the cell.

      9. Describe the major differences between a "wet cell" (eg., lead-acid battery) and a "dry" cell (eg.,carbon-zinc or alkaline battery).

        A wet cell is a cell in which the electrolyte is a liquid. In dry cells the electrolyte is in the form of a paste or polymeric material.

      10. What is the differences between a battery and a fuel cell. What are the advantages and disadvantages of each?

        A battery stores electrical energy and may be either the wet cell type or dry cell type. A fuel cell generates electrical energy from a chemical reaction using fuel supplied to the cell. Theoretically a fuel cell should last forever as long as their are reactants for the chemical reaction. However, a solvent system is needed for the fuel cell. In applications where the solvent is a disadvantage a dry cell might be a better choice.

      11. Define:

        1. anode-electrode where oxication takes place in an electrochemical cell
        2. cathode-electrode where reduction takes place in an electrochemical cell
        3. oxidizing agent-a chemical which oxidizes another chemical and is itself reduced
        4. reducing agent-a chemical which reduces another chemical and is itself oxidized
        5. battery-a device for the storage of electrical energy
        6. fuel cell-a device which produces electricity from a chemical reaction for which fuel is supplied to the cell
        7. corrosion-the deterioration of metal by an electrochemical process
        8. galvanic (voltaic)-an electrochemical cell which is spotaneous and produces voltage
        9. electrolytic-an electrochemical cell which is non-spontaneous and must have a voltage supplied to proceed
        10. cathodic protection-the electrochemical protection of a metal by attaching to it a more electropositive (less electronegative) metal which will reduce the oxidized metal on the piece to be protected
        11. sacrificial anode or electrode-the piece of more electropositive metal used to protect the cathode in cathodic protection

      12. You have a choice of using Mg, Al or Zn as a sacrificial anode for cathodic protection of a piece of iron. You are limited to a volume of 50.00 mL for your sacrificial anode. Which material will you choose and why? (Consider potentials, densities and grams of metal per mole of electrons produced in your answer.) dMg = 1.738 g/mL, dAl = 2.700 g/mL, dZn = 7.140 g/mL

        g of Mg = (50.00 mL)(1.738 g/mL) = 86.90 g of Mg
        86.90 g of Mg = 86.90 g Mg/(24.31 g/mol) = 3.575 moles of Mg
        Mg -----> Mg2+ + 2 e- so 3.576 mol Mg will supply 7.150 mol of e-

        g of Al = (50.00 mL)(2.700 g/mL) = 135.0 g of Al
        135.0 g of Al = 135.0 g/(26.98 g/mol) = 5.004 mol of Al
        Al -----> Al3+ + 3 e- so 5.004 mol Al will supply 15.012 mole of e-

        g of Zn = (50.00 mL)(7.140 g/mL) = 357.0 g of Zn
        357.0 g of Zn = 357.0 g/(65.38 g/mol) = 5.460 mol of Zn
        Zn -----> Zn2+ + 2 e- so 5.460 mol Zn will supply 10.920 mol e-

        The best choice is the Al, since it will supply electrons for a longer time. All three metals are electropositive enough so that they will work as sacrificial electrodes.

      13. How long must a current of 0.600 A be passed through a cell containing molten BeCl2 to produce 50.00 g of Be metal?

        Be2+ + 2 e- -----> Be

        50.00 g of Be/(9.012 g/mol) = 5.548 mol of Be
        (5.548 mo l)(2 e- required for reduction) = 11.10 mol of e- or F

        1 F = 96,500 C so (11.10 F)(96,500 C/F) = 1.071 x 106 C
        Coulombs = Amperes x seconds, thus seconds = Coulombs/amperes so

        seconds = 1.071 x 106 C/0.600 A = 1.785 x 106 s
        (1.785 x 106 s)/(3600 s/h) = 496 h

      14. Calculate the EMF for the voltaic cells below:

        1. Al(s)|Al3+(1 M, aq)|KCl(sat'd)|Co2+(1 M, aq)|Co(s)

          Al(s) -----> Al3+(aq) + 3 e- Eo = 1.66 V
          Co2+(aq) + 2 e- -----> Co(s) Eo = -0.28 V
          Eocell = 1.38 V

        2. Cd(s)|Cd2+(0.500 M,aq)|KCl (sat'd)|Sn2+(1.50 M,aq)|Sn(s)

          Cd -----> Cd2+(aq) + 2 e- Eo = 0.40 V
          Sn2+(aq) + 2 e- -----> Sn Eo = -0.14 V
          Cd + Sn2+ ----> Cd2+ + Sn Eocell = 0.26 V

          Ecell = Eocell - (0.0591/2)logQ

          n = 2, Q = [Cd2+]/[Sn2+] = 0.500/1.500 = 0.333
          Ecell = 0.26 V - (0.0591/2)log(0.333) = 0.26 + 0.0141 = 0.2741 = 0.27 V

        3. Cu(s)|Cu2+(1 M, aq)|KCl (sat'd)|Au3+(1 M, aq)|Au(s) Cu -----> Cu2+ + 2 e- Eo = -0.34 V3+ + 3 e- -----> Au Eo = 1.50 V
          Eocell = 1.16 V


      15. Are the cells below electrolytic or galvanic at standard conditions?

        1. Zn|Zn2+|sat'd KCl|Mg2+|Mg

          Zn -----> Zn2+ + 2 e- Eo = 0.76 V
          Mg2+ + 2 e- -----> Mg Eo = -2.37 V
          Electrolytic: Eocell = -1.61 V

        2. Cd|Cd2+|sat'd KCl|Ag+|Ag

          Cd -----> Cd2+ + 2 e- Eo = 0.40 V
          Ag+ + e- -----> Ag Eo = 0.80 V
          Voltaic (galvanic): Eocell = 1.20 V

      16. How much silver will be plated out on an electrode if 5.00 amperes are passed through a silver nitrate solution for 37.0 minutes? Assume that there is plenty of silver nitrate available in the solution.

        1 Coulomb = (1 Ampere)(1 second) or 1 C = (1 A)(1 s)
        and
        96,500 C = 1 Faraday = 1 mole of electrons

        (37.00 min)(60 s/min) = 2,220 s
        (2,220 s)(5.00 A) = 11,100 C
        (11,100 C)/(96,500 C/F) = 0.1150 F (or mole of e-)

        Ag+ + e- -----> Ag
        so 0.1150 F will reduce 0.1150 mol of Ag+ to Ag
        g Ag = (0.1150 mol)(107.9 g/mol) = 12.4 g of Ag will have been plated out

      17. Balance the redox equations below in acid:

        1. MnO4- + Al -----> Mn2+ + Al3+

          Half-reactions:
          5 e- + 8 H+ + MnO4- -------> Mn2+ + 4 H2O
          Al -----> Al3+
          so 15 e- are needed so

          15 e- + 24 H+ + 3 MnO4- -----> 3 Mn2+ + 12 H2O
          5 Al -----> 5 Al3+ + 15 e-
          24 H+ + 3 MnO4- + 5 Al -----> 3 Mn2+ + 5 Al3+ + 12 H2O

        2. BrO3- -----> Br- + BrO4-

          3 BrO3- + 3 H2O -----> 3 BrO4- + 6 H+ + 6 e-
          6e- + 6 H+ + BrO3- -----> Br- + 3 H2O

          4 BrO3- -----> 3 BrO4- + Br-

        3. Cr2O72- + Sn2+ -----> Cr3+ + Sn4+

          6 e- + 14 H+ + Cr2O72- -----> 2 Cr3+ + 7 H2O
          3 Sn2+ -----> 3 Sn4+ + 6 e-

          14 H+ + 3 Sn2+ + Cr2O72- -----> 3 Sn4+ + 2 Cr3+ + 7 H2O

        4. IO4- + Mn -----> I2 + Mn2+

          14 e- + 16 H+ + 2 IO4- -----> I2 + 8 H2O
          7 Mn -----> 7 Mn2+ + 14 e-

          16 H+ + 7 Mn + 2 IO4- -----> I2 + 7 Mn2+ + 8 H2O

        5. Cr6+ + Br2 -----> Cr3+ + BrO3-

          18 H2O + 3 Br2 -----> 6 BrO3- + 36 H+ + 30 e-
          30 e- + 10 Cr6+ -----> 10 Cr3+

          18 H2O + 3 Br2 + 10 Cr6+ -----> 6 BrO3- + 36 H+ + 10 Cr3+

      18. Balance the redox equations below in base:

        1. Pb + Ag2O -----> Ag + PbO2

          4 OH- + Pb -----> PbO2 + 2 H2O + 4 e-
          4 e- + 2 H2O + 2 Ag2O -----> 4 Ag + 4 OH-
          Pb + 2 Ag2O -----> 4 Ag + PbO2

        2. BrO- -----> Br- + BrO3-

          4 e- + 2 H2O + 2 BrO- -----> 2 Br- + 4 OH-
          4 OH- + BrO- -----> BrO3- + 2 H2O + 4 e-

          3 BrO3- -----> 2 Br- + BrO3-

        3. Zn + MnO4- -----> MnO2 + Zn2+

          3 Zn -----> 3 Zn2+ + 6 e-
          6 e- + 4 H2O + 2 MnO4- -----> 2 MnO2 + 8 OH-

          3 Zn + 4 H2O + 2 MnO4- -----> 3 Zn2+ + 2 MnO2 + 8 OH-

        4. Li + Zn(OH)2 -----> LiOH + Zn

          2 OH- + 2 Li -----> 2 LiOH + 2 e-
          2 e- + Zn(OH)2 -----> Zn + 2 OH-

          2 Li + Zn(OH)2 -----> 2 LiOH + Zn

        5. Au3+ + MnO2 -----> Au + MnO4-

          Au3+ + 3 e- -----> Au
          MnO2 + 4 OH- -----> MnO4- + 2 H2O + 3 e-

          Au3+ + MnO2 + 4 OH- -----> MnO4- + 2 H2O + Au

      19. What is the molar concentration of a solution of Na2Cr2O7, if 25.00 mL of this solution is titrated with 36.89 mL of 2.668 M SnCl2 solution in acid? The approriate unbalanced half reactions are shown below:
        Cr2O72- ---------> Cr+3

        Sn2+ ---------> Sn4+

        Balanced half-reactions:

        6 e- + 14 H+ + Cr2O72- ---------> 2 Cr+3 + 7 H2O

        Sn2+ ---------> Sn4+ + 2 e-

        NRA (Sn2+) = 2 eq/mol x 2.668 mol/L = 5.336 eq/L = 5.336 N

        NRAVRA = NOAVOA

        NOA = NRAVRA/VOA

        NOA = (5.336 N)(36.89 mL)/(25.00 mL) = 7.874 N Cr2O72-

        M Cr2O72- = (7.874 eq/L)/(6 eq/mol) = 1.312 mol/L = 1.312 M


      20. Define:

        1. equilibrium - When the forward and reverse reaction rates are equal and concentrations no longer change over time.

        2. equilibrium expression for the reaction a A + b B <---------> c C + d D, the equilibrium expression is

          K = {[C]c[D]d}/{[A]a[B]b}

        3. equilibrium constant KC - equilibrium constant for concentration,concentration is expressed in molarity

        4. equilibrium constant, KP - equilibrium constant for pressure, concentations are expressed in pressure units

        5. LeChatelier's principle - When a stress is placed on a system at equilibrium, the equilibrium will shift to relieve that stress.

        .
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