OLDEXAM4_kookers - Version 173 – Exam 4 302 – sutcliffe...

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Unformatted text preview: Version 173 – Exam 4 302 – sutcliffe – (51060) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points While electroplating a spoon with silver (molar mass = 107.9 g/mol) from an Ag 2+ so- lution, 19 . 3 × 10 3 C of charge is used. By how much has the mass of the spoon increased? 1. 5.39 g 2. 107.9 g 3. 21.58 g 4. 10.79 g correct 5. 0 g 6. 53.9 g Explanation: This is a 2 electron process. g = Q(molar mass) nF = (19 . 3 × 10 3 C)(107 . 9 g / mol) (2)(9 . 65 × 10 4 C / mol) = 10 . 79 g 002 10.0 points Note: when this question says ”approxi- mately” that doesn’t mean you ignore those concentrations!! Consider the voltaic cell Pt | Sn 4+ (0.0010 M), Sn 2+ (0.10 M) || Ag + (0.010 M) | Ag Sn 4+ + 2 e − → Sn 2+ E = 0 . 15 V Ag + + 1 e − → Ag(s) E = 0 . 8 V The experimental cell potential for the cell is approximately 1. 0.65 V. 2. 0.68 V. 3. 0.72 V. 4. 0.62 V. 5. 0.59 V. correct Explanation: The species in contact with the electrode surfaces are Sn 2+ and Ag + . Note under standard conditions: Sn 2+ → Sn 4+ + 2 e − E anode =- . 15 V Ag + + 1 e − → Ag(s) E cathode = 0 . 8 V E cell = 0 . 65 V But these aren’t standard conditions, as the concentrations are not at 1 M. Use the Nernst equation to calculate E ’s at these concentra- tions: E = E- . 0592 V n log parenleftbigg [Red] y [Ox] x parenrightbigg For Sn 4+ + 2 e − → Sn 2+ , E = 0 . 15 V- . 0592 V 2 log parenleftbigg . 10 M . 0010 M parenrightbigg = 0 . 0908 V For Ag + + 1 e − → Ag(s), E = 0 . 8 V- . 0592 V 1 log parenleftbigg 1 . 010 parenrightbigg = 0 . 6816 V So, using the appropriate values of E , Sn 2+ → Sn 4+ + 2 e − E anode =- . 0908 V Ag + + 1 e − → Ag(s) E cathode = + 0 . 6816 V E cell = 0 . 5908 V 003 10.0 points Consider the voltaic cell Pt | H 2 (1 atm) | H + (? M) || Cl − (1 M) | AgCl(s) | Ag 2 H + + 2 e − → H 2 E = 0.00 V AgCl + 1 e − → Ag + Cl − E = 0.222 V If the measured cell potential for the cell is 0.430 volts, what is the pH of the solution? 1. 0.253 2. 3.75 Version 173 – Exam 4 302 – sutcliffe – (51060) 2 3. 4.00 4. less than 1.00 5. 3.52 correct Explanation: 004 10.0 points The equilibrium constant for the reaction 2 Hg( ℓ ) + 2 Cl − (aq) + Ni 2+ (aq) → Ni(s) + Hg 2 Cl 2 (s) is 5.6 × 10 − 20 at 25 ◦ C. Calculate the value of E ◦ for a cell utilizing this reaction. 1.- 1.14 V 2. + 0.57 V 3.- 0.57 V correct 4.- 0.25 V 5. + 1.14 V Explanation: 005 10.0 points If k = 2 . 7 × 10 − 6 M − 1 s − 1 for the reaction A → B which of the following is the correct rate law? 1. rate = k [A] 2. rate = k [A] [B] − 1 3. rate = k [A] 2 correct 4. rate = k [A] 1 5. rate = k [A] 2 [B] − 1 Explanation: 006 10.0 points The principle of inhibiting the corrosion of iron by using a sacrificial anode is to allow 1. a metal with a lower oxidation potential to be sacrificed....
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This note was uploaded on 03/26/2012 for the course CHEM 302 taught by Professor Mccord during the Fall '10 term at University of Texas.

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OLDEXAM4_kookers - Version 173 – Exam 4 302 – sutcliffe...

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