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Question

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

Kudo Shinichi · Accepted Answer

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`$6H^++2MnO_4^-+5HCOOH\rightarrow2Mn^{2+}+8H_2O+5CO_2\uparrow$Câu trả lời: 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`$6H^++2MnO_4^-+5HCOOH\rightarrow2Mn^{2+}+8H_2O+5CO_2\uparrow$

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)

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

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

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

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

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