Exam2 - Version 206 Exam 2 VANDEN BOUT(53585 1 This print-out should have 30 questions Multiple-choice questions may continue on the next column or

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 206 Exam 2 VANDEN BOUT (53585) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Balance the reaction MnO 4 + NO 2 MnO 2 + NO 3 in basic solution. What is the sum of the coefficients? 1. 23 2. 13 correct 3. 7 4. 15 5. 9 Explanation: The oxidation number of N changes from +3 to +5, so N is oxidized. The oxidation num- ber of Mn changes from +7 to +4, so Mn is reduced. We set up oxidation and reduction half reactions: Oxidation: NO 1 2 NO 1 3 Reduction: MnO 4 MnO 2 Mn and N atoms are balanced. In basic solu- tion we use H 2 O and OH to balance O and H atoms, adding the OH to the side needing oxygen: Oxid: 2 OH + NO 1 2 NO 1 3 + H 2 O Red: 2 H 2 O + MnO 4 MnO 2 + 4 OH Next we balance the total charge by adding electrons. In the reduction reaction thus far there is a total charge of- 1 in the left and- 4 on the right. Three electrons are added to the left. Oxid: 2 OH + NO 1 2 NO 1 3 + H 2 O + 2 e Red: 3 e + 2 H 2 O + MnO 4 MnO 2 + 4 OH The number of electrons gained by Mn must equal the number of electrons lost by N. We multiply the oxidation reaction by 3 and the reduction by 2 to balance the electrons: Oxid: 6 OH + 3 NO 1 2 3 NO 1 3 + 3 H 2 O + 6 e Red: 6 e + 4 H 2 O + 2 MnO 4 2 MnO 2 + 8 OH Adding the half-reactions gives 6 OH + 3 NO 1 2 + 4 H 2 O + 2 MnO 4 3 NO 1 3 + 3 H 2 O + 2 MnO 2 + 8 OH Canceling like terms gives the overall bal- anced equation 3 NO 1 2 + H 2 O + 2 MnO 4 3 NO 1 3 + 2 MnO 2 + 2 OH 002 10.0 points Arrange the agents I) Na + + e Na E red = +2 . 71 II) Ba 2+ + 2 e Ba E red =- 2 . 91 III) I 2 + 2 e 2 I E red = +0 . 54 IV) Au 3+ + 3 e Au E red = +1 . 40 V) Ti 3+ + e Ti 2+ E red =- . 37 in increasing order of reducing agent strength. 1. I, IV, III, V, II correct 2. IV, I, V, II, III 3. II, V, III, IV, I 4. I, II, III, IV, V 5. V, IV, III, II, I Explanation: 003 10.0 points NaHCO 3 , NaCl, and HBr are dissolved in water. How many equations are needed to describe this system? 1. 5 2. 8 correct 3. 4 4. 7 5. 6 Explanation: Version 206 Exam 2 VANDEN BOUT (53585) 2 The species Na + , H 2 CO 3 , HCO 3 , CO 2 3 , Cl , Br , H + , and OH will be present in the water. 004 10.0 points The overall reaction for the discharge of a nickel/cadmium cell is Cd + NiO 2 + 2 H 2 O Cd(OH) 2 + Ni(OH) 2 Which of the following statements is true? 1. It is impossible to recharge a nickel/cadmium cell....
View Full Document

This note was uploaded on 04/16/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

Page1 / 9

Exam2 - Version 206 Exam 2 VANDEN BOUT(53585 1 This print-out should have 30 questions Multiple-choice questions may continue on the next column or

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online