{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW7solutions

# HW7solutions - Niu(qn269 Homework 7 owers(53885 This...

This preview shows pages 1–3. Sign up to view the full content.

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 -→ 0 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 -→ 0 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) .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 be driven by an external power supply, so it is an electrolytic cell. In this case the cathode would be made negative by the power sup- ply so that Na + cations would be attracted toward it and reduced. The anode would be made positive by the power supply so that Cl anions would be attracted toward it and oxidized.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern