Balance the following equation in acidic solution:
Cr2O72-(aq) + NO(g) -> Cr3+(aq) NO3-(aq)
This is what I have so far:
14H+(aq) + Cr2O72-(aq) -> 2Cr3+(aq) + 7H2O(l) + 4e-
2H2O(l) + NO(g) -> NO3-(aq) + 4H+(aq)
I'm confused on what to do next as the equation 2H2O(l) + NO(g) -> NO3-(aq) + 4H+(aq) has charges 0 on both sides. What am I doing wrong?
Chemistry balancing redox reactions: Balance the following equation in acidic solution?
2016-06-21 2:15 pm
回答 (1)
2016-06-21 2:23 pm
✔ 最佳答案
Reduction half equation : Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l) ...... [1]Oxidation half equation : NO(g) + 2H₂O(l) → NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ ...... [2]
[1] + [2]×2, and cancel 6e⁻, 8H⁺(aq) and 4H₂O(l) on the both sides of the equation. The overall equation is :
Cr₂O₇²⁻(aq) + 2NO(g) + 6H⁺(aq) → 2Cr³⁺(aq) + 2NO₃⁻(aq) + 3H₂O(l)
收錄日期: 2021-04-18 15:08:20
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