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B. Basic Medium MnO, So, - so,?- + + MnO2 Balanced equation: Reducing Agent: Oxidizing Agent:

B. Basic Medium MnO, So, - so,?- + + MnO2 Balanced equation: Reducing Agent: Oxidizing Agent:

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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SO 1- Step 1: Write the "skeleton" half-equation based on the species undergoing oxidation and reduction. Oxidation : Reduction : k Step 2: Balance each half-equation 'atomically in the order a) atoms other than H and O SO D) O atoms by adding H,0 with the appropriate coefficient; H atoms by adding H* with the appropriate coefficient. S + 4 H;0 + SO + 8H- 2 Step 3: Examine the charge on both sides of the half-reaction. To the more positive side add the correct number of electrons to equalize the charge with the other side. s* • 4 H:0 2 + 2e +SOP • 8H • 8 e Step 4: Multiply each entire half-reaction by the lowest factor to equalize the number of electrons lost and gained. 1 [ S* + 4 H;0 4[ a • 2e- SO + 8H • 8e ·] → 21· ] Step 5: Obtain the net redox equation by combining the half-equations. Make all possible cancellations. 4H20 SO + 8H + 8e 4 + 8 e s* • 4k + 4 H;0 SO + 81.. 8 H- Step 6: Change from acidic to a basic medium by adding OH to both sides of the net equation; combining H* and OH to form H;0 and simplify. 8 OH • S • 4 12 + 4 H:0 SO • 8t + 8H 8 OH 8 H;0 8 OH + S*+ 4h + 44.0 The final balanced equation is: 8 OH + S + 4 l- SO + 81 + 4 H;0 Check: reactants products 80 8H 81 charge: -10 80 8 H 81 -10
Transcribed Image Text:SO 1- Step 1: Write the "skeleton" half-equation based on the species undergoing oxidation and reduction. Oxidation : Reduction : k Step 2: Balance each half-equation 'atomically in the order a) atoms other than H and O SO D) O atoms by adding H,0 with the appropriate coefficient; H atoms by adding H* with the appropriate coefficient. S + 4 H;0 + SO + 8H- 2 Step 3: Examine the charge on both sides of the half-reaction. To the more positive side add the correct number of electrons to equalize the charge with the other side. s* • 4 H:0 2 + 2e +SOP • 8H • 8 e Step 4: Multiply each entire half-reaction by the lowest factor to equalize the number of electrons lost and gained. 1 [ S* + 4 H;0 4[ a • 2e- SO + 8H • 8e ·] → 21· ] Step 5: Obtain the net redox equation by combining the half-equations. Make all possible cancellations. 4H20 SO + 8H + 8e 4 + 8 e s* • 4k + 4 H;0 SO + 81.. 8 H- Step 6: Change from acidic to a basic medium by adding OH to both sides of the net equation; combining H* and OH to form H;0 and simplify. 8 OH • S • 4 12 + 4 H:0 SO • 8t + 8H 8 OH 8 H;0 8 OH + S*+ 4h + 44.0 The final balanced equation is: 8 OH + S + 4 l- SO + 81 + 4 H;0 Check: reactants products 80 8H 81 charge: -10 80 8 H 81 -10
B. Basic Medium Mno4 So, - so,- + MnO2 Balanced equation: Reducing Agent: Oxidizing Agent:
Transcribed Image Text:B. Basic Medium Mno4 So, - so,- + MnO2 Balanced equation: Reducing Agent: Oxidizing Agent:
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