exam2dvb - Version 288 – Exam 2 – VANDEN BOUT –(53585...

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Unformatted text preview: Version 288 – 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. 9 3. 15 4. 13 correct 5. 7 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 What is the concentration of Pb 2+ and F − in a saturated solution of PbF 2 ? ( K sp = 3 . 7 × 10 − 8 ) [Pb 2+ ] [F − ] 1. 2 . 1 × 10 − 3 M 2 . 1 × 10 − 3 M 2. 1 . 9 × 10 − 4 M 3 . 8 × 10 − 4 M 3. 9 . 6 × 10 − 5 M 1 . 9 × 10 − 4 M 4. 1 . 05 × 10 − 3 M 2 . 1 × 10 − 3 M 5. 2 . 1 × 10 − 3 M 4 . 2 × 10 − 3 M correct 6. 3 . 3 × 10 − 3 M 6 . 6 × 10 − 3 M Explanation: 003 10.0 points Consider the voltaic cell In | In 3+ || Ru 3+ , Ru 2+ | Pt In 3+ + 3 e − → In E =- . 340 V Ru 3+ + 1 e − → Ru 2+ E =- . 080 V at 25 ◦ C. What is the equilibrium constant for the overall cell reaction? 1. 2 . × 10 21 2. 2 . 4 × 10 4 3. 1 . 3 × 10 1 4. 1 . 5 × 10 13 correct 5. 7 . 1 × 10 17 Version 288 – Exam 2 – VANDEN BOUT – (53585) 2 6. 1 . 2 × 10 7 Explanation: 004 10.0 points Write the charge balance equation for a dilute aqueous solution of HI. 1. [H 3 O + ] = [I − ] + [OH − ] correct 2. [HI] initial = [I − ] 3. [H 3 O + ] = [OH − ] 4. [I − ] = [OH − ] + [H 3 O + ] 5. [H 3 O + ] = [I − ] Explanation: 005 10.0 points To balance hydrogen and oxygen in the fol- lowing redox reaction in acidic solution, you would add ? to the left side and ? to the right side: ClO −...
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exam2dvb - Version 288 – Exam 2 – VANDEN BOUT –(53585...

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