HW7solutions - Niu(qn269 – Homework 7 – flowers –(53885 1 This print-out should have 28 questions Multiple-choice questions may continue on

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Unformatted text preview: Niu (qn269) – Homework 7 – flowers – (53885) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. HW 7 due 07 November 2008 at 11PM, Solutions to follow at 11:05 PM. 001 10.0 points Consider the oxidation-reduction reaction Na 2 SnO 2 + Bi(OH) 3-→ Bi + Na 2 SnO 3 + H 2 O What is the coefficient of Na 2 SnO 2 in the balanced equation? 1. 10 2. 2 3. 6 4. 3 correct 5. 12 6. None of these 7. 4 8. 8 9. 5 10. 1 Explanation: The unbalanced equation is Na 2 SnO 2 + Bi(OH) 3-→ Bi + Na 2 SnO 3 + H 2 O Balance the half reactions: Na 2 +2 SnO 2 +H 2 O-→ Na 2 +4 SnO 3 + 2 H + + 2 e − +3 Bi(OH) 3 + 3 H + + 3 e −-→ Bi + 3 H 2 O Equate the e − : 3 bracketleftBig Na 2 +2 SnO 2 + H 2 O-→ Na 2 +4 SnO 3 + 2 H + + 2 e − bracketrightBig 2 bracketleftBig +3 Bi(OH) 3 + 3 H + + 3 e −-→ Bi + 3 H 2 O bracketrightBig Add the balanced half reactions: 3 Na 2 SnO 2 + 3 H 2 O + 2 Bi(OH) 3-→ 3 Na 2 SnO 3 + 2 Bi + 6 H 2 O Neutralize the reamining charges with ex- isting counterions: 3 Na 2 SnO 2 + 2 Bi(OH) 3-→ 2 Bi + 3 Na 2 SnO 3 + 3 H 2 O 002 10.0 points Oxidation occurs 1. at the cathode. 2. at both anode and cathode. 3. at the anode. correct 4. in the electrolyte. 5. at either, depending on whether the cell is electrochemical or electrolytic. Explanation: 003 10.0 points In an electrochemical cell, the electrode at which reduction takes place is called the 1. anode. 2. It depends upon whether the cell is gal- vanic or electrolytic. 3. cathode. correct Explanation: Definition 004 10.0 points Consider the half-reactions and the bal- anced equation for the cell reaction repre- sented by the skeletal equation Mn(s) + Ti 2+ (aq) → Mn 2+ (aq) + Ti(s) . Niu (qn269) – Homework 7 – flowers – (53885) 2 What is the proper cell diagram for this reac- tion? 1. Mn(s) | Mn 2+ (aq) || Ti 2+ (aq) | Ti(s) cor- rect 2. Ti 2+ (aq) | Ti(s) || Mn(s) | Mn 2+ (aq) 3. Ti(s) | Ti 2+ (aq) || Mn 2+ (aq) | Mn(s) 4. Mn 2+ (aq) | Mn(s) || Ti(s) | Ti 2+ (aq) Explanation: The two half-reactions, written as reduc- tions, are Mn 2+ (aq) + 2 e − → Mn(s) Ti 2+ (aq) + 2 e − → Ti(s) Equate e − : Ti 2+ (aq) + 2 e − → Ti(s) Mn(s) → Mn 2+ (aq) + 2 e − Add the balanced half reactions: Mn(s) + Ti 2+ (aq) → Mn 2+ (aq) + Ti(s) The cell diagram is Mn(s) | Mn 2+ (aq) || Ti 2+ (aq) | Ti(s) 005 10.0 points The reaction 2 Na + + 2 Cl − → 2 Na + Cl 2 is indicative of ? cell; the sign of the anode is ? . 1. an electrolytic; negative 2. an electrolytic; positive correct 3. a voltaic; negative 4. a voltaic; positive Explanation: The relevant standard reduction potentials are: Na + + e − → Na E ◦ red =- 2 . 71 V Cl 2 + 2 e − → 2Cl − E ◦ red = +1 . 36 V In the reaction given the Cl reaction is an oxidation. Combining the standard reduction potentials gives: E ◦ cell = - 2.71V - (+ 1.36 V) = - 4.07 V This is a negative value so this is a NON- spontaneous reaction. It would have to bespontaneous reaction....
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This note was uploaded on 02/04/2009 for the course CH 53885 taught by Professor Flowers during the Fall '08 term at University of Texas at Austin.

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HW7solutions - Niu(qn269 – Homework 7 – flowers –(53885 1 This print-out should have 28 questions Multiple-choice questions may continue on

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