Quantitative Pratice Test

Quan
Chapter 15: Redox Titrations

1. (A) Which of the following is true of a redox titration?
(i) Since the redox reaction is spontaneous, the equilibrium constant does not have to be large for an effective titration.
(ii) The titration effectiveness is increased when the two half-reaction potentials are far apart.
(iii) Without an indicator the equivalence point cannot be detected.
(iv) For an effective redox titration, the reducing agent must always be in the buret.
(v) The equivalence point in a redox titration is found at the highest voltage reached during the titration.
(B) In some pretreatment situations, the analyte must be reduced prior to titration. Of the following, which does not show an analyte being converted to its reduced form(s)?
(i) CrO42- ( Cr2+
(ii)MnO4- ( Mn2+
(iii) Fe3+ ( Fe2+
(iv)Mo3+ ( MoO42-
(v) UO22+ ( U3+/U4+
(C) Which of the following, if any, provides a true statement about a redox Gran plot?
(i) The redox Gran plot is useful only if a constant ionic strength is maintained
(ii) The Y-axis data is obtained from VT x 10-nAEcell/0.05916V
(iii)The Y-axis intercept of the line in a proper redox Gran plot shows the volume of added titrant at the equivalence point.
(iv) In order to determine the volume of titrant needed to reach the equivalence point, several precise measurements near the equivalence point are required.
(v) none of the above are true statements

2. (A) Methylene blue is a common redox reaction indicator. A useful solution can be prepared by dissolving 0.10 gram in 100 mL of water. The Eo’ value for this indicator is 0.53V. (n = 2) What is the useful voltage range of this indicator?
(B) Starch indicators are useful when iodine titrations (organic oxidations) are performed. Why is it considered an advantage in these titrations to have iodine as the titrant instead of the analyte?

3. An environmental chemist is going to field test the chemical oxygen demand factor for a stream in Yellowstone Park.
(A) The dichromate that will be used for the redox titration of the stream was standardized before leaving on the hike. Based on the following data, what is the molarity of the potassium dichromate solution? 1.352 grams of K2Cr2O7 (294.19g/mol) are placed in enough water to form 250.0 mL of solution.
(B) A 25.0 mL of stream water was treated with 10.00 mL of acidified K2Cr2O7 from part “A” and warmed. After the reaction mixture was cooled, the excess K2Cr2O7 was titrated with 0.2505 M Fe2+. This back titration requires 35.67 mL. How many moles of K2Cr2O7 reacted with the original organic material in the stream? What is the COD for the stream sample? (Use 3/2 ratio for the reaction between oxygen and dichromate.)
(Overall titration reaction: 6 Fe2+ + Cr2O72- + 14H+ [pic] 6Fe3+ + 2Cr3+ + 7 H2O)

4. High nitrite ion (NO2-) levels in ground water can cause many health problems (nitrite can bond to hemoglobin). One method to determine nitrite levels can be titration with Ce4+.
Ce4+ (aq) + e- ( Ce3+ (aq) Eo = 1.72 V
NO3- (aq) + 2e- ( NO2- (aq) Eo = 0.940 V

(A) Balance the redox equation ___ NO2- + __ Ce4+ [pic] ___ NO3- + ___Ce3+
(B) How many mL of 0.00400 M Ce4+ solution would be required to titrate 100.0 mL of 0.00100 M NO2- solution to the equivalence point?
(C) At the equivalence point show the relationships indicated here by placing the proper coefficient in place of “?”: [? NO2-] = [Ce4+]; [ NO3-] = [?Ce3+]

(D) Assume that the titration is carried out with an apparatus similar to the one shown here using a S.C.E. reference electrode (+0.241V); What is the E cell after the addition of 5.00 mL of the Ce4+ solution?
(E) What is the Ecell voltage at the halfway point to the equivalence point?
(F) What is the Ecell at the equivalence point?
(G) The redox indicator Tris(5-nitro-1,10-phenanthroline)iron has been suggested to use with this particular titration. Justify, with calculations, either accepting or rejecting this choice. Tris(5-nitro-1,10-phenanthroline)iron (pale blue ( red violet) Eo = 1.25

5. Iron deficiency is one of the main world-wide nutritional problems. Studies are currently being done to add iron-containing supplements to cereal grains in an effort to increase iron in the diets of children in Africa. FeSO4 is commonly used. The iron in a bowl of cereal may be titrated with a standard solution of 0.00500 M Tl3+. (Tl = thallium)
The balanced redox titration reaction is: 2Fe2+(aq) +Tl3+(aq) [pic] 2Fe3+(aq) + Tl+ (aq)

(A) How many mL of the 0.00500 M Tl3+ titrant are required to oxidize the Fe2+ in 50.0 mL of cereal solution that is 0.00100 M Fe2+? (Essentially, what is Ve?) Tl3+ + 2e- [pic]Tl+ Eo = + 1.25 V Fe3+ + e- [pic]Fe2+ Eo = + 0.77 V
(B) What would be the Ecell at the halfway point of this titration?
(C) At the equivalence point, several concentration relationships are present: For example: [Fe3+] = ?[Tl+] What is the value of the “?”?
(D)What would be the Ecell at the equivalence point?
(E) What is the Ecell when you reach 2 x Ve?
(F) Which, if any, of the following would make the best indicator for this titration? (circle one choice) (i) Methylene Blue Eo = 0.53 V n = 2 (ii) Diphenylamine Eo = 0.76V n = 2 (iii) Ferroin Eo = 1.06 V n = 1 (iv) Tris(5-nitro-1, 10-phenanthroline) iron Eo = 1.25 V n =1 (v) none of the above

6. A dilute solution of hydrogen peroxide (H2O2) is commonly found as a household disinfectant. However, it can lose some of its potency over time. You could study the decomposition of aqueous H2O2 by performing a redox titration on a solution of H2O2 using a solution of KMnO4 as the titrant. (*Note H2O2 is not a pure liquid.)
O2 (g) + 2H+(aq) + 2e- [pic] H2O2 (aq) Eo = + 0.68 V MnO4-(aq) + 8H+(aq) + 5e- [pic] Mn2+(aq) + 4 H2O(l) Eo = +1.51V
The overall balanced titration reaction is: 2MnO4- + 6H+ + 5 H2O2 [pic] 2 Mn2+ + 5 O2 + 8 H2O
In the titration of 1.00 mL of 0.884 M H2O2 it was found that 17.65 mL of 0.0200 M KMnO4 was required to reach the equivalent point. (The [H+] in the solution was maintained at 1.00 M.)

(A) At the equivalence point the [Mn2+] would be equal to ___ [O2] (Fill in the blank using proper stoichiometry.)
(B) What was the cell voltage after the addition of 1.00 mL of KMnO4? ** The reference electrode used was the Ag/AgCl with a Eo = +0.222V.**

(C) What was the cell voltage at the halfway point in the titration?
(D) What was the cell voltage at the equivalence point?
(E) What was the cell voltage after adding twice the KMnO4 as used at the equivalence point?

7. Some analyses are best performed via the technique of “back titration.” For example, during the analysis of vitamin C with I3- a known excess of I3-,can be added to a sample. Then the un-reacted I3- excess can be back titrated with thiosulfate (S2O32-). The important analysis of glucose (C6H12O6) in nutritional and other biochemical processes can be performed in a similar I3- reaction. The stoichiometry is shown in the following balanced reaction of I3- and glucose: C6H12O6 + I3- + 3OH- ( C6H11O7- + 2H2O + 3 I-

To begin the analysis, 50.0 mL of 0.0526 M solution of I3- was added to a sample containing glucose. The excess (un-reacted) I3- required 23.7 mL of 0.0867 M S2O32- in a back titration. The stoichiometry of that is shown in the following balanced reaction of I3- reacting with S2O32-. 2S2O32- + I3- ( 3 I- + S4O62-

How many moles of glucose were in the sample?

8. Suppose a lipid (oil) is extracted from a sunflower seed and purified. This lipid sample has a mass of 12.56 g. The sample is treated with 20.00 mL of a 0.550 M Br2 solution. After excess KI is added to produce iodine, the excess iodine required 6.76 mL of 0.488 M Na2S2O3 to reach the end point.
(A) What is the iodine number of the extracted oil?
(B) Would another oil, with an iodine number of 12, have fewer or more carbon-carbon multiple bonds?

9. When hydrogen ion activity is maintained at 1.00, any [H+] components of the Nernst equation may be ignored. However, other H+ values must be utilized to determine the cell voltage. For example, suppose a permanganate titration of Fe2+ were carried out when the hydrogen ion activity was 0.010. What would be the resulting EInd value for the permanganate? Redox reaction: 5 Fe2+ + MnO4- + 8 H+ [pic] 5 Fe3+ + Mn2+ + 4 H2O
10. The permanganate and dichromate titrations of unknown Fe2+ samples are both H+ dependent reactions: 5Fe2+ + MnO4- + 8 H+ [pic] 5Fe3+ + Mn2+ + 4 H2O 6 Fe2+ + Cr2O72- + 14 H+ [pic] 6 Fe3+ + 2Cr3+ + 7 H2O
(A) If the molarities of MnO4- and Cr2O72- were the same, which would require the greater volume to titrate identical samples of Fe2+? Explain.
(B) Which titration would experience more of a change in the Ecell at the equivalence point (assume both were using the same reference cell) if the pH of the titrations was changed by one unit? Explain.
(C) At the equivalence point the calculation for EInd will have one key difference for the dichromate titration compared to the permanganate EInd. What is that difference? Why does it arise?

SOLUTIONS FOR TEST BANK CHAPTER 15

1. (A) (ii); (B) (iv); (C) (ii)

2. (A) Ecell = EoIn +/- (0.05916/n); 0.53 + (0.05916/2) = 0.56 V; to 0.53 – (0.05916/2) = 0.50 V
(B) As an analyte starch and iodine form a dark violet-purple color that is slow to dissociate. Therefore, the titrant may have difficulty reacting with the iodine.
3. (A) [pic]= 0.1839 M
(B) 0.2505 M Fe2+ x 0.03567 L x [pic]= 0.001489 moles Cr2O72- excess
0.01000L x 0.1839 mol/L = 0.001839 mol Cr2O72- added;
0.001839 – 0.001489 = 0.000350 mole Cr2O72- reacted
0.000350 mole Cr2O72- reacted x [pic]= 0.000525 mol O2
0.000525 mol O2 [pic]= 671.9mg/L = 672 mg/L

4. (A) ___ NO2- + 2 Ce4+ [pic] ___ NO3- + 2 Ce3+
(B) nTCTVT = nACAVA; 1(0.00400)VT = 2(100.0)(0.00100); VT = [pic] = 50.0mL
(C) [2 NO2-] = [Ce4+]; [NO3-] = [1/2 Ce3+]

(D) At 5.00 mL; EInd = ENO3- = Eo’NO3- - [pic] EInd = 0.940 – 0.02958 log[pic]= 0.9117; Ecell =0.9117 – 0.242 = 0.671V

(E) At halfway point; NO2- = NO3- so EInd = Eo’Ind; 0.940; Ecell = 0.940 – 0.242= 0.698 V
(F) At the equivalence point; EInd = [pic] EInd = [pic]= 1.20; Ecell = 1.20 - 0.242 = 0.958 = 0.96 V
(G) Ecell = EoInd +/- 0.05916/1 = 1.25 – 0.5916 = 1.19V – 0.242 = 0.95 V; 1.25 + 0.5916 = 1.31-.242 = 1.07V; Therefore, the color-changing range is from 0.95V to 1.07 V. This range covers the 0.96 V value for the equivalence point.

5. (A) nTCTVT = nACAVA; 1(0.00500)VT = 2(50.0)(0.00100); VT = [pic]= 20.0 mL
(B) Halfway to equivalence point; [Fe3+] = [Fe2+]; EInd = Eo’Ind = 0.77 – 0.222 = 0.548 = 0.55 V
(C) At equivalence point [Fe3+] = 2[Tl+]
(D) At equivalence point; EInd = [pic]= 1.09 V; Ecell = 1.09 - 0.222 = 0.868 = 0.87 V
(E) At 2 x Ve the [Tl3+] = [Tl+]; EInd = Eo’T = 1.25; Ecell = 1.25 – 0.222 = 1.028 = 1.03 V
(F) choice (v) none of the choices show a range that includes 0.87

6. (A) [Mn2+] = 2/5[O2]
(B) EInd = EH2O2 = Eo’H2O2- - [pic];
= 0.68 –[pic]= 0.644 ; Ecell = 0.644 – 0.222 = 0.42V
(C) At the halfpoint of the titration [H2O2] = [O2]; EInd = Eo’H2O2 = 0.68; Ecell = 0.68 - 0.222 = 0.46 V
(D) At the equivalence point EInd = [pic]= 1.273; Ecell = 1.273 - 0.222 = 1.05 V
(E) 5.00 mL + 17.65 = 22.65 mL titrant added; EInd = ET = Eo’MnO4- -[pic]
=1.51 - [pic]=1.50V
Ecell = 1.50 – 0.222 = 1.28 V

7. 0.0500 L x 0.0526 M = 0.00263 moles I3- were added.
0.0237 L x 0.0867M x [pic]= 0.00103 mole I3- in excess.
0.00263 – 0.00103 = 0.00160 mol I3- reacted with the glucose = 0.00163 mole glucose

8. mol Br2 added = 0.0200 L x 0.550 M = 0.0110 moles; mol S2O32- used = 0.00676 L x 0.488 M = 0.00330 moles mol iodine = 0.00330/2 = 0.00165 mol mol iodine reacted: 0.0110 – 0.00165 = 0.00935 mol; 100(0.00935)(253g/mol)/12.56= 18.8 = 19
(B) An oil with an iodine number of 12 has fewer multiple bonds than this oil.

9. EMnO4- = EoMnO4- - [pic]; = 1.51V - [pic]= 1.32 V

10. (A) The electron change for the Fe2+ sample in this reaction is one electron per mole. For the MnO4- there is a five-electron change. For Cr2O72- there is a six-electron change per mole. Therefore, each mole of dichromate will oxidize 6 moles of the iron. This means less dichromate will be required to oxidize all of the iron in the sample compared to the permanganate (where each mole oxidizes five moles of the iron in the sample). More permanganate will be required.
(B) The stoichiometry of the two reactions shows that the permanganate reaction involves 8 H+ per mole of Fe2+ present. The dichromate reaction shows a dependency of 14H+ per mole of Fe2+ present. When calculated in the Nernst equation, the exponent on the hydrogen activity for dichromate will be 14 compared to only 8 for the permanganate Nernst equation.
(C) The calculation for the EInd for dichromate will have a product component (0.05916 log [Cr3+] because the stoichiometry from Cr2O72- (reactant) to Cr3+ (product) is not 1:1.

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Ce4+

Titrant

NO2-