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Question Number 153050 by DELETED last updated on 04/Sep/21

Answered by DELETED last updated on 04/Sep/21

Mn^(+4) O_2 ^(−2) →Σbiloks=0   Mn+(−2×2)=0 →Mn−4=0  Mn=+4   Mn^(+4) →Mn^(+2)  disebut reduksi  Cl^(−1) →Cl_2 ^0  disebut oksidasi  kalau digabung disebut Redoks  sudah setara , jumlah atom masing  masing unsur sama sblm dan  sesudah reaksi.

$$\mathrm{M}\overset{+\mathrm{4}} {\mathrm{n}}\overset{−\mathrm{2}} {\mathrm{O}}_{\mathrm{2}} \rightarrow\Sigma\mathrm{biloks}=\mathrm{0}\: \\ $$$$\mathrm{Mn}+\left(−\mathrm{2}×\mathrm{2}\right)=\mathrm{0}\:\rightarrow\mathrm{Mn}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{Mn}=+\mathrm{4}\: \\ $$$$\mathrm{M}\overset{+\mathrm{4}} {\mathrm{n}}\rightarrow\mathrm{M}\overset{+\mathrm{2}} {\mathrm{n}}\:\mathrm{disebut}\:\mathrm{reduksi} \\ $$$$\mathrm{C}\overset{−\mathrm{1}} {\mathrm{l}}\rightarrow\mathrm{C}\overset{\mathrm{0}} {\mathrm{l}}_{\mathrm{2}} \:\mathrm{disebut}\:\mathrm{oksidasi} \\ $$$$\mathrm{kalau}\:\mathrm{digabung}\:\mathrm{disebut}\:\mathrm{Redoks} \\ $$$$\mathrm{sudah}\:\mathrm{setara}\:,\:\mathrm{jumlah}\:\mathrm{atom}\:\mathrm{masing} \\ $$$$\mathrm{masing}\:\mathrm{unsur}\:\mathrm{sama}\:\mathrm{sblm}\:\mathrm{dan} \\ $$$$\mathrm{sesudah}\:\mathrm{reaksi}. \\ $$

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