ph = pk a + log 10{[base]/[acid]}

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Mitchell Ross
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FORMULA SHEET (tear off) N A = 6.022 x 10 23 C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) + 32 1 amu = 1.661 x 10-27 kg C = K - 273.15 K = C + 273.15 1 atm = 760 torr = 760 mm Hg 1 atm = 1.

1 FORMULA SHEET (tear off) N A = x C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) amu = x kg C = K K = C atm = 760 torr = 760 mm Hg 1 atm = bar pv = nrt R = L atm/mol K 1 L atm = J R = J/mol K 1 J= 1 kg m 2 /s 2 F = C/mol (1 volt) (1 Coulomb) = 1 Joule p A = X A p A [B] = k p B p A = X Bp A T b = K b m B T f = K f m B = [B]RT H = U + pv G = H - TS G rxn = G rxn + RT ln Q ln K = - G rxn/rt K p = K C (RT) n If ax 2 + bx + c = 0, then x = ( - b [b 2-4ac] 1/2 ) 2a K a K b = K w = 1.0 x (at T = 25 C) ph = pk a + log 10{[base]/[acid]} G = - nfe cell E cell = E cell - (RT/nF) ln Q ln K = nfe cell/rt

2 GENERAL CHEMISTRY 2 FOURTH HOUR EXAM NOVEMBER 21, 2018 Name Panthersoft ID Signature Part 1 (16 points) Part 2 (30 points) Part 3 (34 points) TOTAL (80 points) Do all of the following problems. Show your work. Unless otherwise stated, you may assume T = 25.0 C in all problems.. 2

3 Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each] 1) The expression for K sp for lead II fluoride (PbF 2) is a) K sp = [Pb 2+ ] [F - ] b) K sp = [Pb 2+ ] [2 F - ] C c) K sp = [Pb 2+ ] [F - ] 2 d) K sp = [Pb 2+ ] [F - ] [PbF 2] e) K sp = [Pb 2+ ] [F - ] 2 [PbF 2] 2) Values of K sp are given below for four slightly soluble hydroxide compounds Cd(OH) 2 K sp = 5.3 x Co(OH) 2 K sp = 1.1 x Cu(OH) 2 K sp = 1.6 x Mg(OH) 2 K sp = 5.6 x An aqueous solution at an initial ph = 4.00 contains M of each of the following ions: Cd 2+, Co 2+, Cu 2+, and Mg 2+. The ph of the solution is slowly increased by the addition of a strong soluble base. Which of the following hydroxide compounds will be the first to precipitate out of solution? a) Cd(OH) 2 b) Cu(OH) 2 B c) Co(OH) 2 d) Mg(OH) 2 e) All of the hydroxide compounds will begin to precipitate at the same value for ph 3) Which of the following statements concerning galvanic cells is/are correct? a) In a galvanic cell, the oxidation reaction occurs at the anode b) In a galvanic cell, the reduction reaction occurs at the anode D c) In a galvanic cell, the cell reaction will be spontaneous if E cell > 0. V d) Both a and c e) Both b and c 4) For which of the following reduction reactions is the standard reduction potential defined as E red = 0.00 v? a) Ag + (aq) + e - Ag(s) b) Cu 2+ (aq) + 2 e - Cu(s) D c) Fe 3+ (aq) + 3 e- Fe(s) d) 2 H + (aq) + 2 e - H 2(g) e) I 2(s) + 2 e - 2 I - (aq) 3

4 Part 2. Short answer. 1) The solubility by mass of silver iodide (AgI, MW = g/mol) in water is 2.2 x 10-6 g/l. Based on this, find the numerical value for K sp for silver iodide. [8 points] Reaction is AgI(s) Ag + (aq) + I - (aq) K sp = [Ag + ] [I - ] The molar solubility is 2.2 x 10-6 g 1 mol = 9.4 x 10-9 mol/l L g For every mole of AgI that dissolves 1 mole of Ag + ions and 1 mole of I - ions are produced. Therefore K sp = (9.4 x 10-9 ) (9.4 x 10-9 ) = 8.8 x ) Half-cell reduction potentials are given below for four metal ions, and may be of use in answering the following questions. Cd 2+ (aq) + 2 e - Cd(s) E red = v Ni 2+ (aq) + 2 e - Ni(s) E red = v Co 2+ (aq) + 2 e - Co(s) E red = v Tl + (aq) + e - Tl(s) E red = v For each of the questions below circle the correct answer. There is one and only one correct answer per problem. a) Which of the following is most likely to be oxidized in an oxidation-reduction reaction (for standard conditions)? [4 points] Tl(s) Ni(s) Co(s) Cd(s) b) Which of the following is most likely to be reduced in an oxidation-reduction reaction (for standard conditions)? [4 points] Tl + (aq) Ni 2+ (aq) Co 2+ (aq) Cd 2+ (aq) 3) For a particular galvanic cell the cell potential is E cell = v. The number of moles of electrons transferred per mole of reaction is n = 4. What is the value for G rxn (in kj/mol)? [6 points] Grxn = - n F Ecell = - (4) ( C/mol) (0.161 v) = x 10 4 J/mol = kj/mol 4

5 4) Calcium metal (Ca, MW = 40.1 g/mol) can be produced in small amounts in the laboratory by electrolysis of calcium chloride (CaCl 2, MW = g/mol). The half-reaction for the electrolytic production of calcium metal is Ca e - Ca In a particular experiment, electrolysis is carried out on molten calcium chloride. A current i = 7.40 amps (1 amp = 1 C/s) is used. How many minutes will it take to produce 6.00 g of Ca metal in the experiment? [8 points] The number of Coulombs of electrons needed for the electrolysis is Coulombs = 6.00 g Ca 1 mol 2 mol e C = 2.89 x 10 4 C 40.1 g 1 mol Ca mol e - The amount of time it will take for this amount of charge to pass through the system is time = 2.89 x 10 4 C 1 s 1 min = 65.0 minutes 7.4 C 60 s Part 3. Problems. 1) As group 5 elements, arsenic (As) and nitrogen (N) can have a variety of different oxidation numbers. a) Assign the oxidation numbers for each of the atoms in the following substances [8 points total] As 2O 3 As +3 O -2 H 3AsO 4 H +1 As +5 O -2 NO 3 - N +5 O -2 HNO 2 H +1 N +3 O -2 b) Balance the unbalanced oxidation reduction reaction below for acid conditions [10 points] As 2O 3(s) + NO 3 - (aq) H 3AsO 4(aq) + HNO 2(aq) oxidation As 2O 3(s) + 5 H 2O(l) 2 H 3AsO 4(aq) + 4 e H + (aq) reduction ( NO - 3 (aq) + 2 e H + (aq) HNO 2(aq) + H 2O(l) ) x 2 net As 2O 3(s) + 2 NO - 3 (aq) + 3 H 2O(l) + 2 H + (aq) 2 H 3AsO 4(aq) + 2 HNO 2(aq) 5

6 2) The element indium (In) is used extensively in the semiconductor industry. Because of this, there is interest in the electrochemical properties of indium. Consider the following galvanic cell Zn(s) Zn 2+ (aq, 4.0 x 10-4 M) In 3+ (aq, 3.8 x 10-3 M) In(s) Half-cell reduction potentials relevant to this galvanic cell are given below, and may be of use in answering some of the problems. In 3+ (aq) + 3 e - In(s) E red = v Zn 2+ (aq) + 2 e - Zn(s) E red = v a) Give the half-cell oxidation reaction, the half-cell reduction reaction, and the net cell reaction for the above galvanic cell. [8 points] oxidation 3 Zn(s) 3 Zn 2+ (aq) + 6 e - E ox = v reduction 2 In3 + (aq) + 6 e - 2 In(s) E red = v net 3 Zn(s) + 2 In 3+ (aq) 3 Zn 2+ (aq) + 2 In(s) E cell = v b) Find the values for E cell and E cell for the above galvanic cell. [8 points] From above E cell = v For the above cell n = 6 Q = [Zn 2+ ] 3 = (4.0 x 10-4 ) 3 = 4.4 x 10-6 [In 3+ ] 2 (3.8 x 10-3 ) 2 So E cell = E cell - RT lnq = v - (8.314 J/mol K)(298.2 K) ln(4.4 x 10-6 ) nf (6)( C/mol) = v - ( v) = v 6

ph = pk a + log 10{[base]/[acid]}

ph = pk a + log 10{[base]/[acid]} FORMULA SHEET (tear off) N A = 6.022 x 10 23 C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) + 32 1 amu = 1.661 x 10-27 kg C = K - 273.15 K = C + 273.15 1 atm = 760 torr = 760 mm Hg 1 atm = 1.013 bar pv = nrt