Post-lab 8 Bleach Redox-solutions

Post-lab 8 Bleach Redox-solutions - choudhury (fac489) –...

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Unformatted text preview: choudhury (fac489) – Post-lab 8 Bleach Redox – lyon – (51155) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Balance the redox reaction Fe 2+ (aq) + MnO- 4 (aq) → Fe 3+ (aq) + Mn 2+ (aq) that occurs in an acidic solution. What is the coefficient of Fe 2+ ? 1. 1 2. 2 3. 4 4. 5 correct 5. 3 Explanation: The oxidation number of Fe changes from +2 to +3, so Fe is oxidized. The oxidation number of Mn changes from +7 to +2, so Mn is reduced. We set up oxidation and reduction half-reactions: Red: MnO- 4 → Mn 2+ Oxid: Fe 2+ → Fe 3+ Mn and Fe atoms are balanced. Since this is an acidic solution, we use H 2 O and H + to balance O and H atoms, adding the H 2 O to the side needing oxygen: Red: 8 H + + MnO- 4 → Mn 2+ + 4 H 2 O We balance the total charge in each half- reaction by adding electrons. In the preceding reduction reaction there is a total charge of +7 on the left and +2 on the right. Five electrons are added to the left: Red: 5 e- + 8 H + + MnO- 4 → Mn 2+ + 4 H 2 O Oxid: Fe 2+ → Fe 3+ + 1 e- The number of electrons gained by Mn must equal the number of electrons lost by Fe. We multiply the oxidation reaction by 5 to bal- ance the electrons: Red: 5 e- + 8 H + + MnO- 4 → Mn 2+ + 4 H 2 O Oxid: 5 Fe 2+ → 5 Fe 3+ + 5e- Adding the half-reactions gives the balanced equation 8 H + + MnO- 4 + 5 Fe 2+ → Mn 2+ + 4 H 2 O + 5 Fe 3+ 002 10.0 points Using the smallest possible integer coefficients to balance the redox equation MnO- 4 + C 2 O- 2 4 → Mn +2 + CO 2 (acidic solution), the coefficient for C 2 O 2- 4 is 1. 2. 2. 7. 3. 5. correct 4. 4. 5. The correct coefficient is not given. Explanation: The oxidation number of C changes from +3 to +4, so C is oxidized. The oxidation num- ber of Mn changes from +7 to +2, so Mn is reduced. We set up oxidation and reduction half-reactions: Red: MnO- 4 → Mn 2+ Oxid: C 2 O 2- 4 → CO 2 Mn atoms are balanced. We need 2 CO 2 molecules to balance C: Oxid: C 2 O 2- 4 → 2 CO 2 Since this is an acidic solution, we use H 2 O and H + to balance O and H atoms, adding the H 2 O to the side needing oxygen: Red: 8 H + + MnO- 4 → Mn 2+ + 4 H 2 O Oxid: C 2 O 2- 4 → 2 CO 2 We balance the total charge in each half- reaction by adding electrons. In the preceding reduction reaction there is a total charge of +7 on the left and +2 on the right. Five electrons are added to the left: Red: 5 e- + 8 H + + MnO- 4 → Mn 2+ + 4 H 2 O Oxid: C 2 O 2- 4 → 2 CO 2 + 2 e- The number of electrons gained by Mn must equal the number of electrons lost by C. We choudhury (fac489) – Post-lab 8 Bleach Redox – lyon – (51155) 2 multiply the reduction reaction by 2 and the oxidation reaction by 5 to balance the elec- trons: Red: 10 e- + 16 H + + 2 MnO- 4 → 2 Mn 2+ + 8 H 2 O Oxid: 5 C 2 O 2- 4 → 10 CO 2 + 10 e- Adding the half-reactions gives the overall balanced equation: 5 C 2 O 2- 4 + 16 H + + 2 MnO- 4...
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This note was uploaded on 12/13/2011 for the course CH 204 taught by Professor Leytner during the Fall '08 term at University of Texas at Austin.

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Post-lab 8 Bleach Redox-solutions - choudhury (fac489) –...

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