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HW13 Solutions

# HW13 Solutions - dsj294 hw10 Holcombe(53570 This print-out...

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dsj294 – hw10 – Holcombe – (53570) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find 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 . 601 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 . 44592. Explanation: The cell reaction is Sn(s) + 2 H + Sn 2+ + H 2 (g) . Using the Nernst equation, E = E - 0 . 025693 n ln [Sn 2+ ] P H 2 [H + ] 2 0 . 061 = 0 . 14 - 0 . 025693 2 ln (0 . 601)(1) [H + ] 2 - 0 . 079 = - (0 . 0128465) ln 0 . 601 [H + ] 2 6 . 14953 = ln 0 . 601 [H + ] 2 e 6 . 14953 = 0 . 601 [H + ] 2 [H + ] 2 = 0 . 601 e 6 . 14953 = 0 . 00128282 [H + ] = 0 . 00128282 = 0 . 0358165 pH = - log(0 . 0358165) = 1 . 44592 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. 32.5 g 2. 0.110 g 3. 10.5 g 4. 21.6 g correct 5. 48.8 g Explanation: 003 10.0 points A concentration cell consists of the same redox couples at the anode and the cathode, with di ff erent concentrations of the ions in the respective compartments. Find the unknown concentration for the following cell. Pt(s) | Fe 3+ (aq , 0 . 1 M) , Fe 2+ (aq , 1 M) || Fe 3+ (aq , ?) , Fe 2+ (aq , 0 . 001 M) | Pt(s) E = 0 . 1 V Correct answer: 0 . 00490142 M. Explanation: E cell = 0 . 1 V M 1 = 0 . 1 M M 2 = 1 M M 3 = 0 . 001 M F = 96485 C/mol R T F = 0 . 025693 V At the cathode, Fe 3+ (aq , x ) + e - Fe 2+ (aq , 0 . 001 M) E = 0 . 77 V At the anode, Fe 2+ (aq , 1 M) Fe 3+ (aq , 0 . 1 M) + e - - E = - 0 . 77 V Equate the e - : Fe 3+ (aq , x ) + e - Fe 2+ (aq , 0 . 001 M) At the anode, Fe 2+ (aq , 1 M) Fe 3+ (aq , 0 . 1 M) + e - The overall reaction is Fe 3+ (aq , x ) + Fe 2+ (aq , 1 M) Fe 2+ (aq , 0 . 001 M) + Fe 3+ (aq , 0 . 1 M) E cell = 0 . 77 V - 0 . 77 V = 0 V

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dsj294 – hw10 – Holcombe – (53570) 2 Using the Nernst equation, E cell = E cell - R T n F ln Q R T n F ln M 1 M 3 x M 2 = E cell - E cell ln M 1 M 3 x M 2 = n F ( E cell - E cell ) R T M 1 M 3 x M 2 = exp n F ( E cell - E cell ) R T 1 x = M 2 M 1 M 3 × exp n F R T ( E cell - E cell ) = 1 M (0 . 1 M)(0 . 001 M) × exp 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. +0.30 V.
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HW13 Solutions - dsj294 hw10 Holcombe(53570 This print-out...

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