Using the half-reaction method to complete and balance the following redox reaction. Include the net ionic equation in its simplified form. Remember to dissociate or ionize where possible.
I. Cu (s) + HNO3 (aq) --> Cu(NO3)2 (aq) + NO (g) + h2o (l)
II. H2SO3 (aq) + HIO3 (aq) --> H2SO4 (aq) + HI (aq)
II. MnO4 - (aq) + SO3 2+ (aq) --> Mn 2+ (aq) + SO4 2- (aq)
hello! chemistry help please ??? ?
2020-07-06 4:04 pm
回答 (1)
2020-07-06 4:29 pm
✔ 最佳答案
I.Oxidation half equation: Cu(s) → Cu²⁺(aq) + 2e⁻
Reduction half equation: NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(ℓ)
(Oxidation half equation)×3 + (Reduction half equation)×2, and cancel 6e⁻ on each side.
The overall net ionic equation is:
3Cu(s) + 2NO₃⁻(aq) + 4H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 4H₂O(ℓ)
====
II.
Oxidation half equation: SO₃²⁻(aq) + H₂O(ℓ) → SO₄²⁻(aq) + 2H⁺(aq) + 2e⁻
Reduction half equation: IO₃⁻(aq) + 6H⁺(aq) + 6e⁻ → I⁻(aq) + 3H₂O(ℓ)
(Oxidation half equation)×3 + (Reduction half equation)×1, and cancel 6e⁻, 6H⁺(aq) and 3H₂O(ℓ) on each side.
The overall net ionic equation is:
3SO₃²⁻(aq) + IO₃⁻(aq) → 3SO₄²⁻(aq) + I⁻(aq)
====
III.
Oxidation half equation: SO₃²⁻(aq) + H₂O(ℓ) → SO₄²⁻(aq) + 2H⁺(aq) + 2e⁻
Reduction half equation: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(ℓ)
(Oxidation half equation)×5 + (Reduction half equation)×2, and cancel 10e⁻, 6H⁺(aq) and 5H₂O(ℓ) on each side.
The overall net ionic equation is:
2MnO₄⁻(aq) + 5SO₃²⁻(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 5SO₄²⁻(aq) + 3H₂O(ℓ)
收錄日期: 2021-04-18 18:33:21
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