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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 +...

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`

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

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