CHEM 1212L     PRINCIPLES OF CHEMISTRY II LAB

SPRING 2009

FINAL TEST

NAME:

DATE:
 

 

Freezing Point Depression

 

1.      The equation used to determine the freezing point depression is DTf = Kf x m. When you ran the experiment, which parameter was to be ultimately figured out at the end of the calculations?

a.       m, because DTf  and Kf were determined directly using related experimental data, and Kf is the only one that had to be determined using the above relation.

 

b.      Kf, because DTf  and m were determined directly using related experimental data, and Kf is the only one that had to be determined using the above relation.

 

c.       DTf, because m and Kf were determined using directly using related experimental data, and Kf is the only one that had to be determined using the above relation.

 

 

2.      How did you determine the freezing point depression (DTf) of cyclohexane?

a.       By measuring m and Kf of pure cyclohexane and a solution of para-dichlorobenzene in cyclohexane, then calculating DTf .

 

b.       By measuring the freezing temperatures of pure cyclohexane and a solution of cyclohexane in water, then calculating DTf as the difference between the two freezing temperatures.

 

c.       By measuring the freezing temperatures of pure cyclohexane and a solution of para-dichlorobenzene in cyclohexane, then calculating DTf as the difference between the two freezing temperatures.

 

 

Consider the following problem.

 

When 0.45 g of an unknown substance are dissolved in 10.0 ml of cyclohexane (d = 0.78 g/ml), the solution starts freezing at -3.0 deg C. The freezing point for pure cyclohexane is 7 deg C. The freezing point depression constant is 32.7 deg. C/m. What is the molar weight of the substance?

 

3.      What relation do you use to solve the problem? Are all parameters needed to solve the problem directly available? If not, which one should be determined next?

 

a.       MW = wt x #mol. The weight of the substance is known, but its number of moles is not known and should be determined next.

b.      MW = wt / #mol. The weight of the substance is known, but its number of moles is not known and should be determined next.

c.       MW = #mol / wt. The weight of the substance is known, but its number of moles is not known and should be determined next.

 

 

4.      How do you find the number of moles of solute? Are all parameters needed to solve the problem directly available? If so, what is the result?

a. #mol(solute) = m / kg’s (solvent). The number of kg’s of solvent can be determined by multiplying the grams of solvent by 1000, but the molality of the solution is not directly available and should be determined next.

 

b. #mol(solute) = m x kg’s (solvent). The number of kg’s of solvent can be determined by dividing the grams of solvent by 1000, but the molality of the solution is not directly available and should be determined next

 

c. #mol(solute) = M x V (solvent). The volume of solvent can be determined by dividing the grams of solvent by 1000, but the molarity of the solution is not directly available and should be determined next

 

5.      How do you determine the molality of the solution? Are all parameters needed to solve the problem directly available? Justify your answer.

a.      m = DTf / Kf = {Tf(solvent – Tf(solution)} / Kf = 10 deg.C / 32.7 (deg.C/m) = 0.306 mol / kg

 

b.      m = Kf / DTf = Kf / {Tf(solvent – Tf(solution)} = 32.7 deg.C / 10 (deg.C/m) = 3.27 mol / kg

 

c.        m = DTf x Kf = {Tf(solvent – Tf(solution)} x Kf = 10 deg.C x 32.7 (deg.C/m) =  3.49E3 mol/kg



6.      How should the correct answer to the previous question be used? Justify your answer.

a.       To calculate the freezing point depression of the solution. DTf = Kf x m = 32.7 x 0.306 =10.01 deg C.

 

b.      To calculate the volume of the solution. V = m x d. Density should be determined next.

 

c.       To calculate the number of moles of solute. #mol(solute) = m x kg’s(solvent) = 0.306 mol/kg x 10 mL x 0.78 g/mL / 1000 g/kg = 2.39E-3 mol

 

 

7.      How should the correct answer to the previous question be used? Justify your answer.

a.      To calculate the molar weight of the solute. MW = wt / #mol = 0.45 g / 2.39E-3 mol = 188.28 g/mol

 

b.      To calculate the molarity of the solute. M = #mol x V. V should be determined next.

 

c.       To calculate the number of molecules of the solute. #molecules = #mol x 6.022E23 molecules /mol = 2.39E-3 mol x 6.022E23 molecules/mol = 1.44E21 molecules.

 

 

 

8.      A 1.0 molal solution of NaCl freezes at -3.60 deg C. The freezing point depression constant is 1.86 deg C x kg/mol. The expected temperature based on the molality of NaCl as a molecule is -1.86 deg C. How do you explain the discrepancy?

a.       NaCl molecules lump up together in water solution, which increases the number of moles of solute and decreases the actual molality of the solution.

 

b.      NaCl molecules partially dfissociate into ions in water solution, which increases the number of dissolved particles and increases the actual molality of the solution.

 

c.       There must be an error. More NaCl molecules than is reported are dissolved in water.

 

 

Determination of Reaction Rates.

 

9.      In the Reaction Rate Determination experiment, how was x, the order of the reaction for S2O8(2-) determined?
a. By doubling the concentration of NaI and keeping the concentration of K2S2O8 constant.

 
b. By doubling the concentration of K2S2O8 and keeping the concentration of NaI constant.


c. By doubling both the concentration of K2S2O8 and the concentration of NaI.



10.  7. What was the goal of doubling the amount of NaI in the experiment using test tube 3?

 

a. To determine x, the order of the reaction with respect to S2O3(2-)
b. To determine y, the order of the reaction with respect to I(-)
c. To determine y, the order of the reaction with respect to S2O8(2-)

 

11.  The experiment mentioned in the previous question measures reaction rates, not x and y, the reaction orders. How are the values of x and y obtained?

 

a. Calculate  r2/r1 = 2^x and r3/r1 = 2^y , then find ln(r2/r1) = ln(2^x) = x / ln2 followed by x = ln(r2/r1) / ln2, and do likewise for y
b. Calculate  r2/r1 = 2^x and r3/r1 = 2^y , then find ln(r2/r1) = ln(2^x) = xln2 followed by x = ln(r2/r1) x ln2, and do likewise for y

c. Calculate  r2/r1 = 2^x and r3/r1 = 2^y , then find ln(r2/r1) = ln(2^x) = xln2 followed by x = ln(r2/r1) / ln2, and do likewise for y



12.  How do you determine of the rate constant k?

 

a. Use r = k[S2O8(2-)]^x [I(-)]^y, followed by  k = r + ([S2O8(2-)]^x [I(-)]^y)
b. Use r = k[S2O8(2-)]^x [I(-)]^y, followed by k = r x ([S2O8(2-)]^x [I(-)]^y)

c. Use r = k[S2O8(2-)]^x [I(-)]^y, followed by k = r / ([S2O8(2-)]^x [I(-)]^y)

 

 

13.  What is an example of the rate law mentioned in post-lab question 4? Explain your answer. 

a.       r = 3.00E-3 [0.036]^2 [0.12]^1, because r, [S2O8(2-)] and [I(-)] are the variables and k, x, and y are constant parameters.

 

b.       r = 3.00E-3 [S2O8(2-)]^2 [I(-)]^1, because r, [S2O8(2-)] and [I(-)] are the variables and k, x, and y are constant parameters.

 

c.        r = 3.00E-3 [S2O8(2-)]^2 [I(-)]^1, because r, [S2O8(2-)] and [I(-)] are the constant parameters and k, x, and y are variables.


14.  How could you figure out the energy of activation in the experiment you ran? Show the procedure.

a.      By using rate constants at different temperatures. Rearrange ln(k2/k1) = (Ea / R) x (1/T1 - 1/T2) to get Ea = R x ln(k2/k1) / (1/T1 - 1/T2)


b. By using rate constants at different concentrations. Rearrange ln(k2/k1) = (Ea / R) x (1/T2 - 1/T1) to get Ea = R x ln(k2/k1) / (1/T1 - 1/T2).

c. By using rate constants at different times. Rearrange ln(k2/k1) = (Ea / R) x (1/T1 - 1/T2) to get Ea = R / ln(k2/k1) / (1/T1 - 1/T2)

 

Factors that affect chemical equilibria

 

Equation 1: FeCl3 + 3NH4SCN à Fe(SCN)3 + 3NH4Cl

 

15.  How could the equilibrium in equation 1 be shifted toward the formation of more products?
a. By Adding extra FeCl3 or Fe(SCN)3
b. By Adding extra FeCl3 or NH4SCN
c. By Adding extra NH4Cl or NH4SCN


16.  How could the equilibrium in equation 1 be shifted toward the formation of more reactants without adding anything to the reaction?
a. By withdrawing FeCl3 or NH4SCN (aq)
b. By withdrawing Fe(SCN)3(aq) or NH4SCN
c. By withdrawing Fe(SCN)3(aq) or NH4Cl(aq)

 
Equation 10: Mg(2+)(aq) + 2OH(-)(aq) <---> Mg(OH)2(s)
Equation 11: NH3(aq) + H2O(aq) <---> NH4(+)(aq) + OH(-)(aq)


17.  What was the effect of adding NH4Cl to the equilibrium involving Mg(OH)2 in aqueous ammonia solution?
a. It triggered the production of NH4(+) and OH(-) in equation 11. The depleted H2O was replaced by dissolution of Mg(OH)2 (s) in equation 10.


b. It triggered the consumption of NH3 and H2O in equation 11. The depleted H2O was replaced by dissolution of Mg(OH)2 (s) in equation 10.


c. It triggered the consumption of NH4(+) and OH(-) in equation 11. The depleted OH(-) was replaced by dissolution of Mg(OH)2 (s) in equation 10.


 

Equation 12: NaHCO3(aq) + HCl(aq) --> NaCl(aq) + H2O(l) + CO2(g)


18.  When HCl is added to the NaHCO3 solution containing a pH color indicator, the color of the solution changes from red to yellow. However, when the solution is placed under vacuum, the color changes from yellow back to almost red. How do you explain that?
a. Applying vacuum to the solution removes CO2 from the equilibrium in equation 12, which triggers the consumption of NaCl and H2O, until CO2 is exhausted from the equilibrium, leaving a red solution of HCl
 
b. Applying vacuum to the solution adds CO2 to the equilibrium in equation 12, which triggers the consumption of HCl and NaHCO3, until HCl is exhausted from the equilibrium, leaving a red solution, of NaHCO3.

c. Applying vacuum to the solution removes CO2 from the equilibrium in equation 12, which triggers the consumption of HCl and NaHCO3, until HCl is exhausted from the equilibrium, leaving a red solution of NaHCO3.

 

 

19.  After the color change due to vacuum, is the solution basic or acidic? Justify your answer.

 

a. Basic, because that is the color caused by NaHCO3 which is a base

b. Acidic, because that is the color caused by NaHCO3 which is an acid.

c. Basic, because that is the color caused by CO2 which is a base



ACID BASE TITRATION

 

20.  What does the equation of the acid-base reaction of oxalic acid with NaOH tell about the relation between the amounts of acid and the amounts of base? Justify your answer.
 

a.      = #mol(NaOH) = 2 x #mol(OxAc) because, the mole ratio of NaOH to oxalic acid is 2/1

b.      = 2 x #mol(NaOH) = #mol(OxAc) because, the mole ratio of NaOH to oxalic acid is 2/1

c.       = #mol(NaOH) = #mol(OxAc) because, the mole ratio of NaOH to oxalic acid is 1/1


21.  How do you determine the molarity of NaOH solution?

 

a.   Using the formula M(NaOH) = V(NaOH) / #mol( NaOH) = V(NaOH) / 2 x #mol(Ox. Ac)

b.   Using the formula M(NaOH) = V(NaOH) x #mol(NaOH) = 2 x #mol(Ox. Ac) x V(NaOH)

c.   Using the formula M(NaOH) = #mol(NaOH) / V(NaOH) = 2 x #mol(Ox. Ac) / V(NaOH)

 

 

22.  In part B of the experiment what do you use the expression “#mol(NaOH) = #mol(Unk.Ac)” for? Justify your answer.

a. To determine the molarity of NaOH, because all that is needed is divide the #mol of NaOH by the volume.

 

b. To determine the #mol of the unknown acid, because the #mol of NaOH is can be determined using the molarity from part A of the experiment and the volume used for the titration in part B.

 

c. To determine the #mol of NaOH, because it is at the left side of the equal sign.

 

 

23.  How do you use the number of moles of unknown acid from the previous question? Explain.

a.      To calculate the molar weight of unknown acid. MW = wt / #mol. The weight was determined by weighing the unknown acid just before the titration.

 

b.      To calculate the molar weight of unknown acid. MW =#mol / wt. The weight was determined by weighing the unknown acid just before the titration.

 

c.       To calculate the molarity of  unknown acid. M = V / #mol. The volume is the same as the volume of NaOH used in the titration.

 

 

24.  What is the value of pKa of the unknown acid? Explain.
a. pH at the end of the reaction = pKa of the unknown acid, because at half-equivalence point, the expression pH = pKa + log([Conjugate base]/[Acid]) has log([Conjugate base]/[Acid]) = 1

b.  pH at equivalence point = pKa of the unknown acid, because at half-equivalence point, the expression pH = pKa + log([Conjugate base]/[Acid]) has log([Conjugate base]/[Acid]) = 10

 

c. pH at half-equivalence point = pKa of the unknown acid, because at half-equivalence point, the expression pH = pKa + log([Conjugate base]/[Acid]) has log([Conjugate base]/[Acid]) = 0



25.  Once you find the pKa, how do you find the value of the Ka of the unknown acid? Justify your answer.

a. Ka = 1E-pKa, because by definition, pKa = -log(Ka)

b. pKa = 1E-Ka, because by definition, Ka = -log(pKa)
c. Ka = 1E+pKa, because by definition, pKa = log(Ka).

 

Use the graphs below to answer the following questions .

26.  Use the graphics provided to determine graphically the Ka of the unknown acid that is titrated in the experiment. Show how you proceed both on the graphic and in your calculations. NO credit will be given if all you give is the final answer.

a.      Volume at half-equivalence point: 17.5 mL / 2 = 8.75 mL. Corresponding pH at half-equivalence point: 4.7 = pKa. Ka = 1E-pka = 1E-4.7 = 2.00E-5

 

b.      Volume at equivalence point: 17.5 mL. Corresponding pH at equivalence point: 10.83 = pKa. Ka = 1E-pka = 1E-10.83 = 1.48E-11

 

c.       Volume at equivalence point: 17.5 mL / 2 = 8.75 mL. Corresponding pH at equivalence point: 4.7 = pKa. Ka = 1E-pka = 1E-4.7 = 2.00E-5

 


BUFFERS

 

Phosphoric acid ionizes in water according to the following process:

Step 1: H3PO4 + H2O -->  H3O(+) + H2PO4(-)
Step 2: H2PO4(-) + H2O -->  H3O(+) + HPO4(2-)
Step 3: HPO4(2-) + H2O -->  H3O(+) + PO4(3-)
 

 

27.  Based on the table of components of the buffers you made in the lab, what is the expression of the pH of the buffers? Justify your answer.
a. pH = pKa1 + log([H2PO4(2-)]/[H3PO4]), because [H2PO4(-)] and [H3PO4] are the components of the buffer and are connected by Ka1


b. pH = pKa2 + log([HPO4(2-)]/[H2PO4(-)]), because [H2PO4(-)] and

[HPO4(2-)] are the components of the buffer and are connected by Ka2

 

c. pH = pKa3+ log([H2PO4(-)]/[HPO4(2-)]), because [H2PO4(-)] and [HPO4(2-)] are the components of the buffers and are connected by Ka3
 

 

28.  Three buffers are listed in the table on page 95 in your lab manual. Which one is expected to best resist pH changes due to additions of bases? Explain.

a.       The middle one, because the acid and conjugate base components of the buffer have the same concentration and therefore resist pH changes by added acids and bases equally.

 

b.       The first one, because its acid component is more concentrated than the conjugate base and better resists pH change by neutralizing more the added base.

 

c.        The third one, because its base component is more concentrated than the conjugate acid and better resists pH change by neutralizing more the added base.

29.  What is the pH of the correct buffer mentioned in the previous question?
a. pH = 7.21 + log(0.1/0.3) = 6.73
b. pH = 7.21 + log(0.2/0.2) = 7.21
c. pH = 7.21 + log(0.3/0.1) = 8.69

 

Ksp of Ca(OH)2

 

30.  On the report sheet of the Ksp experiment, you are asked to determine the number of moles of various species including oxalic acid. Should you just write in the one calculated on the previous page?

a.       Yes, because the #mol on the previous page is found using the relation #mol = wt / MW

b.      No, because the #mol on the previous page is found using the relation #mol = MW / wt

c.       No, because the #mol on the previous page is calculated in 100 mL, whereas the on in question is calculated in 10 mL of solution

 

 

31.  In the Ksp experiment you titrated oxalic acid in an Erlenmeyer flask by a solution of Ca(OH)2 in a burette. Would the numeric results of the experiment change if you titrated Ca(OH)2 in the flask by a solution of oxalic acid in the burette? Justify your answer.

a.       No, because even though the solution would start colored pink, it would turn colorless at the exact same end point it turned from colorless to pink in the actual experiment.

b.      Yes, because even though the solution would colored pink and turn colorless at end point.

c.       Yes, you would be titrating a base by an acid instead of an acid by a base.

 

 

 

32.  In the Ksp experiment all calculations refer to species related either to oxalic acid or to Ca(OH)2.  Why then bother to use phenolphthalein in the experiment?

a.       Because without phenolphthalein it would not be possible to see the end point of the reaction which was shown by change in color from pink to colorless.

b.      Because without phenolphthalein it would not be possible to see the end point of the reaction which was shown by change in color from colorless to pink.

c.       Actually it was not necessary to use phenolphthalein because its presence did not change the reaction between oxalic acid and Ca(OH)2.

 

 

 

THERMODYNAMICS

 

33.  How do you use the temperature change calculated in the “heat of Neutralization” experiment to determine DH, the heat of neutralization of HCl by NaOH? Explain.
a. DH (neutralization) = q(H2O) = (c x m x DT), because the heat released by the neutralization reaction is taken by the water solvent, which explains why the sign stays the same.


b. DH (neutralization) = -q(H2O) = -(c x m x DT), because the heat released by the neutralization reaction is taken by the water solvent, which explains the change of sign.


c. DH (neutralization) = -q(H2O) = -(c + m + DT), because the heat released by the neutralization reaction is taken by the water solvent, which explains the change of sign.


34.  How do you determine the heat of neutralization per mole of H(+) or OH(-) when the acid and the base used are HCl and NaOH? Which parameters needed for the determination are directly available, and which ones are not directly available? How can the latter accessed?

a. Divide the heat of neutralization by the number of moles of HCl or NaOH. The heat of neutralization is available from the previous question. The number of moles can be calculated using the formula: #mol = wt / MW. Both wt and MW are available

b. Multiply the heat of neutralization by the number of moles of HCl and NaOH. The heat of neutralization is available from the previous question. The number of moles can be calculated using the formula: #mol = M / V. Both M and V are available

c. Divide the heat of neutralization by the number of moles of HCl or NaOH. The heat of neutralization is available from the previous question. The number of moles can be calculated using the formula: #mol = M x V. Both M and V are available

 

35.  If the acid and base used were H3PO4 and Ba(OH)2, how would you determine the molar heat of neutralization in the previous question? Explain

a. The heat of neutralization which was equated to -q(H2O) would be divided:
* by 3 x the #mol of the acid because one mole of acid contains 3 moles of H(+) and
* by 2 x the #mol of the base, because one mole of the base contains 2 moles of OH(-).


b. There would no difference from the results in the previous question, because in both cases you would still have an acid and a base.

c. The heat of neutralization which was equated to -q(H2O)would be divided:
* by 2 x the #mol of the acid because one mole of acid contains 2 moles of H(+) and
* by 3 x the #mol of the base, because one mole of the base contains 3 moles of OH(-).

 


BALANCING REDOX EQUATIONS

 

36.  A redox reaction is made of two parts: the oxidation and the reduction. What happens to an atom whose oxidation number increases?

a.       It is being oxidized, because it is gaining electrons

b.      It is being reduced, because it is gaining electrons

c.       It is being oxidized, because it is losing electrons

 

  1. The principle to follow when balancing a redox reaction, is to balance the increase and the decrease in oxidation numbers of the atoms that are oxidized and reduced. What do you do when the the changes in oxidation numbers of the oxidized and reduced atoms are not equal?

 

    1. Find the balance by using the least common divider of increase and decrease in ON.

 

    1. Find the balance by using the greatest common multiple of increase and decrease in ON.

 

    1. Find the balance by using the least common multiple of increase and decrease in ON.

 

 

38.  What is done next in the process to balance a redox equation after finding the least common multiple of increase and decrease in ON?

g.      Find the oxidation coefficient used to multiply the species that is oxidized and  the reduction coefficient used to multiply the species that is reduced in the equation

h.      Find the oxidation coefficient used to multiply the species that is reduced and  the reduction coefficient used to multiply the species that is oxidized in the equation

i.        Balance the charges of the ionic species in the equation

 

39.  What are the next step in balancing redox equations?

j.        Balance the remaining atoms  and then balance the charges of the ionic species in the equation .

k.      Balance the charges of the ionic species in the equation and then balance the rest of the equation

l.        Balance the charges of the ionic species in the equation and then balance the oxidized  and reduced species in the equation

 

40.  Balance the following equation. Show how you proceed. NO credit will be given for the last answer alone.

MnO4-  +  S2-  +  H2O    →   MnO2  +  S8  +  OH-


Answer: _16__MnO4-  +  _24__S2-  +  _32__H2O    →   _16__MnO2  +  _3__S8  +  _64__OH-

 

 

ELECTROCHEMISTRY


The experiment on the titration of Fe(2+) by MnO4(-) is based on the following reactions

Equation 1: Reduction of Mn from Mn(7+) to Mn(2+):

8 H(+) (aq) + MnO4(-) (aq) + 5e(-) -> Mn(2+) (aq) + 4H2O

 

Equation 2: Oxidation of Fe from Fe(2+) to Fe(3+)

5 Fe(2+) (aq) -> 5 Fe(3+) (aq) + 5 e(-)

 

The equation of the overall redox reaction between MnO4(-) and Fe(2+0 is as follows:

Equation 3:

8 H(+) (aq) + MnO4(-) (aq) + 5 Fe(2+) (aq) -> Mn(2+) (aq) +  5 Fe(3+) (aq) + 4H2O

 

 

  1. In what circumstance would a titration operation use less KMnO4 than expected to titrate Fe(2+)? Justify your answer.

a. It would happen if some of the Fe(2+) ions are oxidized by another oxidizing agent than MnO4(-), such as O2 in the air.

 

b. It would happen if some of the Fe(3+) ions are oxidized by another oxidizing agent than MnO4(-), such as O2 in the air.

 

c. It would happen if some of the Fe(2+) ions are reduced by another oxidizing agent thant MnO4(-), such as O2 in the air.

 

 

  1. With the addition of H2SO4 to the sample, the reaction guaranteed to run in acidic conditions. Yet H3PO4 is added to the solution during the titration. What would happen if the H3PO4 was not added?

 

a. The end point of the titration would be harder to detect. In the absence of H3PO4 to keep the solution colorless as long as there is any Fe(2+) to oxidize to Fe(3+), the solution would take a deeper and deeper yellow color as the titration progresses. This would mask the light pink color that appears as soon as Fe(2+) is used up and the end point is reached.

 

b. The end point of the titration would be easier to detect. In the absence of H3PO4 to keep the solution yellow as long as there is any Fe(2+) to oxidize to Fe(3+), the solution would get more clear as the titration progresses. This would sharpen the light pink color that appears as soon as Fe(2+) is used up and the end point is reached.

 

c. The end point of the titration would be easier to detect. In the absence of H3PO4 to keep the solution yellow as long as there is any Fe(2+) to oxidize to Fe(3+), the solution would take a deeper and deeper pink color as the titration progresses. This would mask the light pink color that appears as soon as Fe(2+) is used up and the end point is reached.

 

 

  1. What is the following relation used for?

 

#mol(Fe(2+)) = 5 x #mol(KMnO4(-))

 

a. To determine the number of moles of Fe(2+) titrated by MnO4(-) in the sample, because the #mol of MnO4(-) is easily calculated from available information and each mole of Fe(2+) reacts with 5 moles of MnO4(-).

 

b. To determine the number of moles of Fe(2+) titrated by MnO4(-) in the sample, because the #mol of MnO4(-) is easily calculated from available information and each mole of MnO4(-) reacts with 5 moles of Fe(2+).

 

c. To determine the number of moles of MnO4(-)  titrated by Fe(2+) in the sample, because the #mol of Fe(2+) is easily calculated from available information and each mole of MnO4(-) reacts with 5 moles of Fe(2+).

 

 

  1. How do you determine the mass of Fe(2+) (MW = 55.9 g/mol) in a sample Fe(NH4)2(SO4).6H2O that is titrated by a given volume of KMnO4 of known molarity?

 

a. wt(Fe(2+)) = MW(Fe) x 1/5 x M(KMnO4) x V(L’s)

b. wt(Fe(2+)) = MW(Fe) / 5 x M(KMnO4) x V(L’s)

c. wt(Fe(2+)) = MW(Fe) x 5 x M(KMnO4) x V(L’s)

 

 

Suppose you want to build a galvanic cell and all the metals you have are aluminum foil and iron nails, together with Al(NO3)3 and FeCl3 salts. Based on that situation, and using data from the Standard Potentials table at the end of your textbook, answer the following questions:



45.  What are the reduction potentials of Al/Al3+ and Fe/Fe3+ couples? Why?
a. -1.66 V for Al/Al3+ and -0.036 V for Fe/Fe3+, because they are presented in the tables as corresponding to reactions in which the metals gain electrons, and therefore are reduced.


b. 1.66 V for Al/Al3+ and -0.036 V for Fe/Fe3+, because they are presented in the tables as corresponding to reactions in which the metals gain electrons, and therefore are reduced.


c. 1.66 V for Al/Al3+ and 0.036 V for Fe/Fe3+, because they are presented in the tables as corresponding to reactions in which the metals lose electrons, and therefore are oxidized.

 

 

46.  Write the equation of the reaction that occurs at the cathode of the  battery. Justify your choice.

a. Al --> Al(3+) + 3e, because it is a reduction reaction.

b. Fe(3+) + 3e --> Fe, because it is a reduction reaction

c. Fe(3+) + 3e --> Fe, because it is an oxidation reaction

 

 

47.  Write the equation of the reaction that occurs at the anode of the battery. Justify your choice.

a. Al(3+) + 3e --> Al , because it is a reduction reaction.

b. Al --> Al(3+) + 3e, because it is an oxidation reaction

c. Fe(3+) + 3e --> Fe, because it is a reduction reaction

 

 

 

48.  Write the equation of the balanced overall cell reaction. Justify your choice.

a. Al + Fe(3+) --> Al(3+) + Fe, both reactions use the exact same number of electrons

b. Al(3+) + Fe --> Al + Fe(3+), both reactions use the exact same number of electrons

c. 3Al(3+) + 3Fe --> 3Al + 3Fe(3+), each reaction uses 3 electrons and therefore the half reaction of Fe must be multiplied by 3 to be balanced.


49.  What redox couple combination should be used to build the best galvanic cell? Explain. What is the corresponding overal cell potential?
a. Fe/Fe3+ for the anode and Al/Al3+ for the cathode, because this combination provides the highest potential. E0cell = (1.66 – 0.036) V = 1.624 V

 

b. Al/Al3+ for the anode and Fe/Fe3+ for the cathode, because this combination provides the highest potential. E0cell = (-1.66 + 0.036) V = -1.624 V.


c. Al/Al3+ for the anode and Fe/Fe3+ for the cathode, because this combination provides the highest potential. E0cell = (1.66 – 0.036) V = 1.624 V.

 

 

 

ELECTROPLATING

 

50.  In the electrodeposition experiment that you ran in the lab, what could be a good explanation for collecting a smaller amount of Cu than expected?

a. A low number of electrons was transferred to Cu atoms during the reaction, therefore the current of the battery was low.

 

b. A low number of electrons was transferred to Cu(2+) ions during the reaction, therefore  the current of the battery was low.

 

c. A low number of electrons was transferred from Cu(2+) atoms, during the reaction, therefore  the potential of the battery was low.

 

 

d.      46. If the concentration of the Cu            SO4 solution was twice what you used in the lab, would you have deposited more Cu on the coin? Explain.

 

a. No, because the amount of material deposited depends only on the voltage of ther battery, according to Faraday’s law of electrolysis.

 

b. Yes, because the amount of material deposited depends only on the concentration of the solution, according to Faraday’s law of electrolysis.

 

c. No, because the amount of material deposited depends only on the amount of current used, according to Faraday’s law of electrolysis.

 

 

47. If the potential of the battery was half what your battery had, would you deposit half the amount of Cu on the coin? Explain.

a. No, because the amount of material deposited depends only on the voltage of ther battery, according to Faraday’s law of electrolysis.

 

b. Yes, because the amount of material deposited depends only on the concentration of the solution, according to Faraday’s law of electrolysis.

 

c. No, because the amount of material deposited depends only on the amount of current used, according to Faraday’s law of electrolysis.

 

 

  1. The copper you deposited in your experiment came from a solution of Cu(2+) ions. How does the experiment confirm the existence of electrons?

 

ANSWER: Cu(2+) ions are responsible for the blue color of the solution. Cu is brown. The only way for Cu(2+) ions to change to neutral Cu is to gain two negative charges. Those charges are provided by two electrons.

 

  1. The copper you deposited in your experiment came from a solution of Cu(2+) ions. Do the electrons travel to the Cu(2+) ions or is it the other way around? Justify your answer base on where the metallic copper is deposited

 

ANWER: The Cu(2+) ions travel to the electrons. If it were the electrons travelling to the ions, every time a Cu(2+) ion would receive two electrons to form neutral Cu, it would precipitate to the bottom of the reaction container. Electrons are collected at the negative pole of the power source. The Cu(2+) ions  are attracted by the negative charge of the pole where they go to pick up the electrons they need to achieve neutrality and form Cu atoms which adhere to the coin attached to the negative pole of the power source.

 

 

  1. In what conditions would no Cu would be deposited at all, no matter how much current is used? Explain using information on Cu/Cu(2+) standard potential

 

ANSWER: If the voltage of the power source is lower than the standard potential of the Cu(2+) + 2 e -> Cu (0.34 V), then the reaction would not take place. There would not be enough force to make Cu(2+) ins accept the electrons.

 

 

  1. If the copper plate was connected to the wire carrying the nickel and the nickel connected to the wire carrying the copper plate in the experiment, where would the copper be deposited? Justify your answer.

 

ANSWER: The Cu would be deposited on the Cu plate. Cu(2+) ions need electrons to become neutral Cu. Those electrons are collected at the negative pole of the power source. The Cu(2+) ions  are attracted by the negative charge of the pole where they go to pick up the electrons they need to achieve neutrality and form Cu atoms which adhere to the Cu plate attached to the negative pole of the power source.

 

 

Bonus Question a.

Solve the problem below. Show how you proceed. NO credit will be given if all you give is the final answer.

 

If it takes 25 mL of a solution of NaOH to titrate 0.126 g of oxalic acid dihydrate, what is the molarity of the NaOH solution?

 

 

Bonus Question b.

                                                    

Which experimental operation will it take you the longest time to forget? Justify your answer.