Chemistry HW 10

Chemistry HW 10 - ekk86 hw10 Holcombe (53570) This...

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ekk86 – hw10 – Holcombe – (53570) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. You must use the Table oF Stan- dard Reduction Potentials on Dr. Mc- Cord’s course website For all prob- lems in which values are not given. http://courses.cm.utexas.edu/pmccord/ ch302/help/red-table.html Note: emF = electromotive Force, which is analogous to potential. 001 10.0 points A tin electrode in 0 . 414 M Sn(NO 3 ) 2 (aq) is connected to a hydrogen electrode in which the pressure oF H 2 is 1 bar. Sn | Sn 2+ || H + | H 2 | Pt IF the cell potential is 0 . 061 V at 25 , what is the pH oF the electrolyte at the hydrogen electrode? Correct answer: 1 . 52685. Explanation: The cell reaction is Sn(s) + 2 H + Sn 2+ +H 2 (g) . Using the Nernst equation, E = E - ± 0 . 025693 n ² ln ± [Sn 2+ ] P H 2 [H + ] 2 ² 0 . 061 = 0 . 14 - ± 0 . 025693 2 ² ln ³ (0 . 414)(1) [H + ] 2 ´ - 0 . 079 = - (0 . 0128465) ln ± 0 . 414 [H + ] 2 ² 6 . 14953 = ln ± 0 . 414 [H + ] 2 ² e 6 . 14953 = 0 . 414 [H + ] 2 [H + ] 2 = 0 . 414 e 6 . 14953 =0 . 000883672 [H + ]= 0 . 000883672 = 0 . 0297266 pH = - log(0 . 0297266) = 1 . 52685 002 10.0 points How much Au will be plated out (From a so- lution containing Au 3+ ) by the same amount oF current that plates 10.5 g oF Cu (From a solution oF Cu 2+ )? 1. 48.8 g 2. 0.110 g 3. 21.6 g correct 4. 10.5 g 5. 32.5 g Explanation: 003 10.0 points A concentration cell consists oF the same redox couples at the anode and the cathode, with di±erent concentrations oF the ions in the respective compartments. ²ind the unknown concentration For the Following cell. Pt(s) | ²e 3+ (aq , 0 . 1 M) , 2+ (aq , 1 M) || 3+ (aq , ?) , 2+ (aq , 0 . 001 M) | Pt(s) E . 1V Correct answer: 0 . 00490142 M. Explanation: E cell . M 1 . 1M M 2 = 1 M M 3 . 001 M F = 96485 C/mol RT F . 025693 V At the cathode, 3+ (aq ,x )+ e - 2+ (aq , 0 . 001 M) E . 77 V At the anode, 2+ (aq , 1 M) 3+ (aq , 0 . 1 M) + e - - E = - 0 . 77 V Equate the e - : 3+ (aq e - 2+ (aq , 0 . 001 M) At the anode, 2+ (aq , 1 M) 3+ (aq , 0 . 1 M) + e - The overall reaction is 3+ (aq ) + ²e 2+ (aq , 1 M) 2+ (aq , 0 . 001 M) + ²e 3+ (aq , 0 . 1 M) E cell . 77 V - 0 . 77 V = 0 V
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ekk86 – hw10 – Holcombe – (53570) 2 Using the Nernst equation, E cell = E cell - RT nF ln Q ln ± M 1 M 3 xM 2 ² = E cell - E cell ln ± M 1 M 3 2 ² = ( E cell - E cell ) M 1 M 3 2 = exp ³ ( E cell - E cell ) ´ 1 x = M 2 M 1 M 3 × exp ³± ² ( E cell - E cell ) ´ = 1M (0 . 1 M)(0 . 001 M) × exp ³ 1(0 V - 0 . 1 V) 0 . 025693 V ´ = 204 . 022 Thus x = [Fe 3+ ]= 1 204 . 022 =0 . 00490142 M . 004 10.0 points Consider the voltaic cell In | In 3+ (1 M) || Ru 3+ (1.0 M), Ru 2+ (0.010 M) | C In 3+ +3 e - In E 0 = - 0 . 34 V Ru 3+ +1 e - Ru 2+ E 0 = +0 . 25 V The experimental cell potential for the cell is approximately 1. +0.71 V. correct 2. +0.26 V.
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This note was uploaded on 11/01/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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Chemistry HW 10 - ekk86 hw10 Holcombe (53570) This...

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