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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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Tube correct answer is circled, I didn't get that answer, can someone explain where I went wrong? 

Balance the following redox reaction which occurs in acidic solution. When this equation is properly balanced, using the smallest whole-number coefficients, what is the coefficient on H₂O, and on which side of the equation does H₂O appear? Mn²+ (aq) + BiO3(aq) → Bi³(aq) + MnO4 (aq) b) a) 5, on the reactant side 15, on the product side c) 7, on the product side 1, on the reactant side d) e) 7, on the reactant side unbalanced
Transcribed Image Text:Balance the following redox reaction which occurs in acidic solution. When this equation is properly balanced, using the smallest whole-number coefficients, what is the coefficient on H₂O, and on which side of the equation does H₂O appear? Mn²+ (aq) + BiO3(aq) → Bi³(aq) + MnO4 (aq) b) a) 5, on the reactant side 15, on the product side c) 7, on the product side 1, on the reactant side d) e) 7, on the reactant side unbalanced
0 9 O O Mn 1 balance charges 4 H/₂0+ Mn₂² 2+ H 8 4 2+ Mn Bi 2é + 5+ Mn. tot 24 2+ 2+ → Bi → (aq) +7 → Mn² ²04 oxidation O 4 H₂O + Mn. (2)3 + Bioz →Bi 3+ (aq) 6+ -1 (2e + 6 H+ + Bi03 6 + -1 2+ S/MnO4 + 8H+ 8 H 04 Mn I 2+ - 3+ 2+ ↑ ↑ 3+ reduction ↑ → -1 -1 8+ → MnO4 + 8H+ + Sé -1 + 8+ 7 2+ 3+ Bi 3 + 3+ 8 +7 -2(4)-=-1 + Mn0a 3+ 641+ + Bio3 O + 3H2O) Bil 0:3 Hi 6 магт M₁²+ → MnO4 + 8H+ + 56 - ві 3+ Bil 0:3 Hi 6 + 3H₂O 5 22Ht 11 H₂O Jo é + 30+ + 5 BiO3 → 5 Bi³+ 5 BiD3 →→ 5 Bi³+ +15+₂0 3+ + 4+₂0 + 5e + 22H+ + Mn ²t + 5 BiD3 → Mn 04 + 5 Bist + 11 H ₂ O
Transcribed Image Text:0 9 O O Mn 1 balance charges 4 H/₂0+ Mn₂² 2+ H 8 4 2+ Mn Bi 2é + 5+ Mn. tot 24 2+ 2+ → Bi → (aq) +7 → Mn² ²04 oxidation O 4 H₂O + Mn. (2)3 + Bioz →Bi 3+ (aq) 6+ -1 (2e + 6 H+ + Bi03 6 + -1 2+ S/MnO4 + 8H+ 8 H 04 Mn I 2+ - 3+ 2+ ↑ ↑ 3+ reduction ↑ → -1 -1 8+ → MnO4 + 8H+ + Sé -1 + 8+ 7 2+ 3+ Bi 3 + 3+ 8 +7 -2(4)-=-1 + Mn0a 3+ 641+ + Bio3 O + 3H2O) Bil 0:3 Hi 6 магт M₁²+ → MnO4 + 8H+ + 56 - ві 3+ Bil 0:3 Hi 6 + 3H₂O 5 22Ht 11 H₂O Jo é + 30+ + 5 BiO3 → 5 Bi³+ 5 BiD3 →→ 5 Bi³+ +15+₂0 3+ + 4+₂0 + 5e + 22H+ + Mn ²t + 5 BiD3 → Mn 04 + 5 Bist + 11 H ₂ O
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