Question

[Buddy]

Bài 3. Cân bằng các phản ứng oxi hóa – khử sau đây bằng Phương pháp giải thăng bằng electron

a) NaBr + Cl₂ → NaCl + Br₂

b) Fe₂O₃ + CO → Fe + CO₂

c) CO + I₂O₅ → CO₂ + I₂

d) Cr(OH)₃ + Br₂ + OH^- → CrO₄^2- + Br^- + H₂O

e) H⁺ + MnO₄^- + HCOOH → Mn²⁺ + H₂O + CO₂

Answer 1

[Kudo Shinichi]

a) 

\(2Br^{-1} \rightarrow Br_2^0+2e\)	`xx1`
\(Cl_2^0+2e\rightarrow2Cl^{-1}\)	`xx1`

`2NaBr + Cl_2 -> 2NaCl + Br_2`

b)

\(2Fe^{+3}+6e\rightarrow2Fe^0\)	`xx1`
\(C^{+2}\rightarrow C^{+4}+2e\)	`xx3`

\(Fe_2O_3+3CO\xrightarrow[]{t^o}2Fe+3CO_2\)

c) 

\(C^{+2}\rightarrow C^{+4}+2e\)	`xx5`
\(2I^{+5}+10e\rightarrow2I_2^0\)	`xx1`

d)

\(Cr^{+3}\rightarrow Cr^{+6}+3e\)	`xx2`
\(Br^0_2+2e\rightarrow2Br^{-1}\)	`xx3`

\(2Cr\left(OH\right)_3+3Br_2+10OH^-\rightarrow2CrO_4^{2-}+6Br^-+8H_2O\)

e) 

\(Mn^{+7}+5e\rightarrow Mn^{+2}\)	`xx2`
\(C^{+2}\rightarrow C^{+4}+2e\)	`xx5`

\(6H^++2MnO_4^-+5HCOOH\rightarrow2Mn^{2+}+8H_2O+5CO_2\uparrow\)