MnO4 2- and H2O2

2008-07-10 9:40 pm
大家都知有條咁既EQUATION

2MnO4-(aq) + 6H+(aq) + 5H2O2(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g)

呢條REDOX出現既問題,點解H2O2的電化序(electrochemical series)比

起MnO4 2-的VOLT數為正,個REACTION都行到?

仲有,從個SERIES去predict,H2O2的氧化力比ACIDIFIED既MnO42-

高,但係佢地出黎既reaction,係disproportionation?
更新1:

首先多謝你先,但呢條equation一定要heating 同埋要conc先行到?就咁加一定冇野睇? 同埋呢,我記得做lab,有d predict明明唔得,但又有野睇,咁可以點解釋?(冇 heating+ 1M),麻煩唒﹗

回答 (2)

2008-07-11 2:53 am
✔ 最佳答案
電化序係以溶液濃度為 1M 及温度為 25℃ 而排序的.


但實驗進行時, 啲 solutions 應該都幾濃, 斷估唔只 1M, 而且你會 warm 個 solution, right?


因為 reaction conditions (濃度和温度) 都改變了, 所以亦把行不到的 reaction 變為行到.
當考慮反應物那邊的 H2O2 個 O 時, 每個 O 的氧化數為 -1 (因為每個 H 為 +1 ); 但在生成物那邊的 H2O 個 O 的氧化數為 -2, O2 個 O 的氧化數為 0 (free element 的氧化數皆為 0). O 的氧化數同時由 -1 降到 -2 及由 -1 升到 0, 所以出來的 reaction係disproportionation.
2008-07-20 4:38 am
Here is the general rule: MnO4- oxidizes hydrogen peroxide. (no need to consider the stantard reduction potential)

If u use 1M of H2O2 and 1M of MnO4-, there is a possibility H2O2 oxidize
MnO4-. You may consider the reaction is a redox rather than
disproportionation, although it really looks like disproportionation.

water H2O comes from the acid, rather than from hydrogen peroxide.


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