Can you guys answer these questions?

[匿名]

2020-05-12 12:22 pm

1. For the unbalanced equation CH3OH(aq) + MnO4–(aq) --> CO32–(aq) + MnO42–(aq), what is the correct half-reaction for reduction in acidic solution?
Question options:a CH3OH(aq) + 2 H2O(l) → CO32–(aq) + 6 e– + 8OH–(aq) b. CH3OH(aq) + 2 H2O(l) → CO32–(aq) + 6 e– + 8 H+(aq)c. MnO4–(aq)→ MnO42–(aq) + e–d MnO4–(aq) + e– → MnO42–(aq)2. For the unbalanced reaction equation MnO4–(aq) + Cl–(aq) --> Mn2+(aq) + Cl2(g),Question options:a) chlorine is the oxidizing agentb) When charge is balanced, the total number of electrons transferred is 8c) 10 hydrogen ions must be added to balance the equationd) 8 water molecules must be added to balance the equation3. From the unbalanced equation above, what is the correct balanced half-reaction equation for the reduction half-reaction in an acidic medium?Cr2O72–(aq) + I–(aq) → Cr3+(aq) + IO3–(aq)Question options: a. I–(aq) → IO3–(aq) + 6 e– b I–(aq) + 3 H2O(l) → IO3–(aq) + 6 e– + 6H+(aq) c. Cr2O72–(aq + 6 e– + 14 H+(aq) → 2 Cr3+(aq) + 7 H2O(aq) d Cr2O72–(aq) + 3 e– → Cr3+(aq)

回答 (1)

fooks

2020-05-12 1:04 pm

1.
MnO₄⁻ undergoes reduction as the oxidation number of Mn decreases from +7 to +6.
Half-reaction for reduction of MnO₄⁻ in acidic solution is:
MnO₄⁻(aq) + e⁻ → MnO₄²⁻(aq)

The answer: d. MnO₄⁻(aq) + e⁻ → MnO₄²⁻(aq)

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2.
Oxidation half-equation: 10Cl⁻(aq) → 5Cl₂(g) + 10e⁻
Reduction half-equation: 2MnO₄⁻(aq) + 16H⁺(aq) + 10e⁻ → 2Mn²⁺(aq) + 8H₂O(ℓ)
Overall equation: 2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺(aq) →2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O(ℓ)

a. False. Chlorine is the reducing agent as it is oxidized.
b. False. It should be 10 instead.
c. False. It should be 16 instead.
d. True

The answer: d. 8 water molecules must be added to balance the equation

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3.
Cr₂O₇²⁻ undergoes reduction as the oxidation number of Cr decreases from +6 to +3.
Half-reaction for reduction of Cr₂O₇²⁻ in acidic solution is:
Cr₂O₇²⁻(aq) + 6e⁻ + 14H⁺(aq) → 2Cr³⁺(aq) + 7H₂O(ℓ)

The answer: c. Cr₂O₇²⁻(aq) + 6e⁻ + 14H⁺(aq) → 2Cr³⁺(aq) + 7H₂O(ℓ)

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