Experimental Chemistry

Answer Practice Quiz 3 Copper-Silver Phase Diagram 1. At 700 ֯ C, what is maximum wt% solubility of Cu in Ag? Show line on above graph. If I draw a horizontal line at 700 ֯C (left red line going from 100 wt% Ag), and it meets the curve line at roughly 94 wt% Ag, hence solubility of Cu in Ag is 6 wt% Cu. 2. At 800 ֯ C, what is the maximum at% solubility of Ag in Cu? Show line on the graph. Draw horizontal line at 800 ֯C from 100 wt% Cu (green line), roughly 9 wt% Ag is soluble in Cu and 5 at% Ag is soluble in Cu. 3. Is Cu-Ag system isomorphous? Explain your answer Cu-Ag is not isomorphous because at room temperature there are two phases present (α and β), whereas isomorphous system has 100% solubility of components into each other at room temperature or presence of only one phase at room temperature. 4. How much wt% of Cu is present in Cu-Ag system at eutectic point? At eutectic point, there is 71.9 wt% of Ag and 28.1 wt% of Cu. Please draw the line in the exam. 5. At eutectic point, how much of α-phase is present if I have 5 kg of Cu-Ag system? Using the level rule at eutectic point to calculate α-phase W α = 91.2 − 71.9 91.2 − 8.0 W α = 0.2319 There is 5 Kg in the system, hence there is 0.2319 * 5 = 1.16 kg of α phase 6. At what wt% of Cu, there is eutectoid reaction? Show line on graph Eutectoid reaction is point or temperature at which a solid phase transforms into two solid phase. In above phase diagram, δ solid phase transforms into two solid phases γ and ε as shown by red line. This transformation is at 75 wt% Zn and 25 wt% Cu. 7. At what temperature solidification start for 60 wt% Cu in Cu-Zn system? Show on the graph Solidification starts at 900 ֯ C as shown by black line in the graph. 8. At what temperature solidification ends for 60 wt% Cu in Cu-Zn system? Show on the graph Solidification ends roughly at 875 C as shown by black line in the graph. 9. Calculate wt% of Zn in α-phase at 600 ֯C and 40 wt% Zn in Cu-Zn system? 37 wt% Zn is present in α-phase at 600 C with 40 wt% Zn in the system as shown with blue line. 10. Write one thing that you would like to see change in this course?

CHEM | 2640 | Experimental Chemistry | assessment | Answer Practice Quiz 3 Copper-

CH2640 Problem set I (Answers) 1. Write the balanced net reaction for a Sn (s) | SnCl 2 (aq) || AlBr 3 (aq) | Al (s) chemical cell. What is the standard potential for this cell? What is the cell potential if [Al 3+ ] = 17.0 mM and [Sn 2+ ] = 1.45 M? 3Sn(s) + 2Al 3+ (aq)  3Sn 2+ (aq) + 2Al(s) E cell 0 = E ox 0 + E red 0 = 0.13 V + (-1.66 V) = -1.53 V Q = [Sn 2+ ] 3 /[Al 3+ ] 2 = 1.45 3 /0.0170 2 = 10549 E cell = -1.53 V - 0.05916/6 log(10549) = -1.57 V 2. #14-7., page 323 in textbook a) 16 mol O 2 /day x 1 day/24 h x 1 h/3600 s x 4 mol e - /mol O 2 x 96485 C/mol e - = 71.5 A b) w = Eq = 1.1 V x (16 mol O 2 x 4 mol e - /mol O 2 x 96485 C/mol e - ) = 6.8 x 10 6 J 3. #14-12., page 323 in textbook a) 2H + (aq) + 2e -  H 2 (g) Hg 2 Cl 2 (s) + 2e -  2Hg(l) + 2Cl - (aq) E red = E o red – 0.05916/n log(P H2 /[H + ] 2 ) = 0.00 V – (0.05916/2) log(0.1/(10 -2.54 ) 2 ) E red = -0.120 V E red = E o red – 0.05916/n log([Cl - ] 2 ) = 0.268 V – (0.05916/2) log(0.200 2 ) = 0.309 V b) The hydrogen electrode has a negative reduction potential, so that reaction will be reversed and the oxidation of H 2 (g) will occur instead, using the Pt electrode as a surface to conduct electrons. This half-cell is the anode. c) E cell = 0.309 V – (-0.120 V) = 0.429 V 4. #14-14., page 323 a) Cu(s)  Cu + (aq) + e - E ox o = -0.518 V Cu 2+ (aq) + e -  Cu + (aq) E red o = 0.161 V E cell o = (-0.518 V + 0.161 V) = - 0.357 V K = 10 nE/0.05916 = 10 (1(-0.357)/0.05916) = 10 -6.03 = 9.33 x 10 -7 b) 2F 2 (g) + 4e -  4F - (aq) E red o = 2.890 V H 2 O(l) + 2F - (aq)  F 2 O(g) + 2H + (aq) + 4e - E ox o = -2.168 V E cell o = 2.890 V + (-2.168 V) = 0.722 V K = 10 (4(0.722)/0.05916) = 10 48.8 = 6.31 x 10 48 5. #14-21., page 324 in textbook a) Leave the first reaction as written, but reverse the second reaction and multiply by 5 to cancel the 10 total electrons being transferred. Result: 2IO 3 - (aq) + 16I - (aq) + 12H + (aq)  6I 3 - (aq) + 6H 2 O(l) b) E o cell = E o cathode – E o anode = 1.210 V – (0.535 V) = 0.675 V At equilibrium, Q = K and there is no net flow of electrons, giving a cell potential of 0 V. So, if E = E o – 0.05916/n log Q, then 0 = E o – 0.05916/n log K, and E o = 0.05916/n log K. Then, log K = nE o /0.05916 and K = 10^(nE o /0.05916). K = 10 (10(0.675)/0.05916) = 10 114 (strongly favors the products of this reaction!) c) E red = E o red – 0.05916/n log([I 3 - ]/([IO 3 - ] 2 [I - ][H + ] 12 ) = 1.210 V – (0.05916/10)log(1x10 -4 /[(0.01) 2 (0.01)(10 -6 ) 12 ]) = 0.772 V E ox = E o ox – 0.05916/n log([I 3 - ] 5 /[I - ] 15 ) = -0.535 V – (0.05916/10)log((1x10 -4 ) 5 /(0.01) 15 ) = -0.594 V E cell = E red + E ox = 0.178 V, spontaneous in the forward direction as written d) Reaction is at equilibrium when all given ion concentrations AND a new pH combine to make Q = K. K = 10 114 = [I 3 - ] 6 /([IO 3 - ] 2 [I - ] 16 [H + ] 12 ) log 10 114 = log( [I 3 - ] 6 /([IO 3 - ] 2 [I - ] 16 [H + ] 12 ) ) = log([I 3 - ] 6 /([IO 3 - ] 2 [I - ] 16 ) – 12 log[H + ] 114 = log( (1x10 -4 ) 6 /(0.01) 2 (0.01) 16 ) + 12 pH 114 = log (1x10 12 ) + 12 pH 114 = 12 + 12 pH pH = 8.5 6. What is the molar concentration of an aqueous HNO 3 solution with pH = 3.120? For strong acids, [H + ] in solution = initial [acid]. So, [H + ] = 10 -3.120 = 7.59 x 10 -4 M = [HNO 3 ] 7. What is the molar concentration of an aqueous solution of potassium cyanoacetate with a pH of 12.48 ? K a of cyanoacetic acid = 3.37 x 10 -3

CHEM | 2640 | Experimental Chemistry | question bank | CH2640 Problem set I (Answers)

CH2640_Practice_Exam_I_Answer_key.docx.docx-Label all pernent components (i.e. cathode,

divalent cations, but it will not show the presence of monovalent cations or any anions. To summarize, one technique detects fewer ions than you might want, and one detects too many. 12) In the context of chemical analysis, what is the difference between a primary standard and a secondary standard? Give examples of both types that have been used so far during experiments this term. A primary standard is something that can be reliably used as a known quantity in an experiment and is a point of comparison/equivalence to determine an unknown quantity. Substances used as primary standards are typically very stable solids (ionic or otherwise) that do not oxidize easily in air over the short term, are not hygroscopic or light sensitive, etc. An example of a primary standard is the MgSO 4 used to standardize the EDTA solution, or the H 2 SO 4 used to titrate the Ba(OH) 2 solution. A secondary standard is something that can be used in essentially all the same scenarios as a primary standard, but it must be standardized each time it is used in an experiment. Typically, the solid and/or solution of a secondary standard are susceptible to reaction with air/light/heat and its concentration is not stable over the short term. An example of this is the EDTA solution that had to be standardized with the MgSO 4 before titrating your water samples. 13) Find the wavelength and frequency of light whose energy is 1.50 x 10 -19 J/photon. In which range of the electromagnetic spectrum does this light fall? What effect does this light have when interacting with matter? E = hc/   = hc/E = (6.626 x 10 -34 J s/photon)(3.0 x 10 8 m/s)/(1.50 x 10 -19 J/photon)  = 13.252 x 10 -7 m = 1325 x 10 -9 m = 1325 nm E = h  (1.50 x 10 -19 J/photon) = (6.626 x 10 -34 J s/photon)   .226 x 10 15 1/s = 2.26 x 10 14 Hz This light falls in the IR range of the EM spectrum and will cause bonds within a molecule to vibrate. 14) A titration of a solution of RbOH is performed using 0.250 M sulfuric acid as the titrant. Identify three potential indicators that could be used to visualize the equivalence point of this titration. The titration is done ﬁve times, each with 75.00 mL of the base, and the following volumes to reach equivalence were recorded: 26.80 mL, 26.65 mL, 28.05 mL, 27.70 mL, and 27.45 mL. Calculate the mean concentration of the RbOH solution +/- two standard deviations. This is a strong acid/strong base reaction, so the equivalence point should be very close to pH = 7. A few possible indicators that change color around this pH are bromthymol blue, phenol red, and m -nitrophenol. 2RbOH(aq) + H 2 SO 4 (aq)  Rb 2 SO 4 (aq) + 2H 2 O(l) Eq. pt when mol RbOH = 2mol H2SO4 [RbOH]V RbOH = 2[H 2 SO 4 ]V H2SO4 so [RbOH] = 2[H 2 SO 4 ]V H2SO4 /V RbOH Using the given volumes, the five concentrations are: 0.178666 M, 0.177666 M, 0.187 M, 0.184666 M, and 0.183 M Mean value = 0.1821996 M St. dev. = 0.00396 To the correct # of sig figs, [RbOH] = 0.182 M +/- 0.004 Must match decimal place of standard deviation with rounded mean, since it is being added/subtracted.

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