Cân bằng các phản ứng oxi hóa – khử sau đây bằng phương pháp thăng bằng electron

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Giải Hóa 10 Bài 13: Phản ứng oxi hóa – khử

Haylamdo biên soạn và sưu tầm lời giải Bài 3 trang 76 Hóa học 10 trong Bài 13: Phản ứng oxi hóa – khử, lời giải chi tiết sẽ giúp học sinh biết cách làm bài tập Hóa học 10.

Bài 3 trang 76 Hóa học 10: Cân bằng các phản ứng oxi hóa – khử sau đây bằng phương pháp 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

Lời giải:

a) NaBr + Cl₂ → NaCl + Br₂

$\stackrel{+1}{Na}\stackrel{-1}{Br}+{\stackrel{0}{Cl}}_2\to \stackrel{+1}{Na}\stackrel{-1}{Cl}+{\stackrel{0}{Br}}_2$

$\begin{array}{l}\stackrel{-1}{Br}\to {\stackrel{0}{Br}}_{}+1e\\ \stackrel{0}{Cl}+1e\to \stackrel{-1}{Cl}\end{array}$

$\begin{array}{c}1\times \\ 1\times \end{array}\left|\begin{array}{l}\stackrel{-1}{Br}\to {\stackrel{0}{Br}}_{}+1e\\ \stackrel{0}{Cl}+1e\to \stackrel{-1}{Cl}\end{array}\right.$

⇒ $\stackrel{-1}{Br}+\stackrel{0}{Cl}\to \stackrel{0}{Br}+\stackrel{-1}{Cl}$

2NaBr + Cl₂ → 2NaCl + Br₂

b) Fe₂O₃ + CO → Fe + CO₂

_{${\stackrel{+3}{Fe}}_2{\stackrel{-2}{O}}_3+\stackrel{+2}{C}\stackrel{-2}{O}\to \stackrel{0}{Fe}+\stackrel{+4}{C}{\stackrel{-2}{O}}_2$}

_{$\begin{array}{l}\stackrel{+3}{Fe}+3e\to \stackrel{0}{Fe}\\ \stackrel{+2}{C}\to \stackrel{+4}{C}+2e\end{array}$}

_{$\begin{array}{c}2\times \\ 3\times \end{array}\left|\begin{array}{l}\stackrel{+3}{Fe}+3e\to \stackrel{0}{Fe}\\ \stackrel{+2}{C}\to \stackrel{+4}{C}+2e\end{array}\right.$}

⇒ $2\stackrel{+3}{Fe}+3\stackrel{+2}{C}\to 2\stackrel{0}{Fe}+3\stackrel{+4}{C}$

Fe₂O₃ + 3CO → 2Fe + 3CO₂

c) CO + I₂O₅ → CO₂ + I₂

_{$\stackrel{+2}{C}\stackrel{-2}{O}+{\stackrel{+5}{I}}_2{\stackrel{-2}{O}}_5\to \stackrel{+4}{C}{\stackrel{-2}{O}}_2+{\stackrel{0}{I}}_2$}

_{$\begin{array}{l}\stackrel{+2}{C}\to \stackrel{+4}{C}+2e\\ \stackrel{+5}{I}+5e\to \stackrel{0}{I}\end{array}$}

_{$\begin{array}{c}5\times \\ 2\times \end{array}\left|\begin{array}{l}\stackrel{+2}{C}\to \stackrel{+4}{C}+2e\\ \stackrel{+5}{I}+5e\to \stackrel{0}{I}\end{array}\right.$}

⇒ $5\stackrel{+2}{C}+2\stackrel{+5}{I}\to 5\stackrel{+4}{C}+2\stackrel{0}{I}$

5CO + I₂O₅ → 5CO₂ + I₂

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

$\begin{array}{l}\stackrel{0}{Br}+1e\to B{r}^{-}\\ Cr{\left(OH\right)}_3+\text{5}O{H}^{-}\to Cr{O_4}^{2-}+\text{4}{H}_{2}O+3e\end{array}$

$\begin{array}{c}3\times \\ 1\times \end{array}\left|\begin{array}{l}\stackrel{0}{Br}+1e\to B{r}^{-}\\ Cr{\left(OH\right)}_3+\text{5}O{H}^{-}\to Cr{O_4}^{2-}+\text{4}{H}_{2}O+3e\end{array}\right.$

⇒ $Cr{\left(OH\right)}_3+\text{3}Br+\text{5}O{H}^{-}\to Cr{O_4}^{2-}+\text{3}B{r}^{-}+\text{4}{H}_{2}O$

2Cr(OH)₃ + 3Br₂ + 10OH^- → 2CrO₄^2- + 6Br^- + 8H₂O

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

_{$H^{+}+Mn{O_4}^{-}+\stackrel{+1}{H}\stackrel{+2}{C}\stackrel{-2}{O}\stackrel{-2}{O}\stackrel{+1}{H}\to M{n}^{2+}+H_{2}O+\stackrel{+4}{C}{\stackrel{-2}{O}}_2$}

_{$\begin{array}{l}\stackrel{+1}{H}\stackrel{+2}{C}\stackrel{-2}{O}\stackrel{-2}{O}\stackrel{+1}{H}\to \stackrel{+4}{C}{\stackrel{-2}{O}}_2+2H^{+}+2e\\ 8H^{+}+Mn{O_4}^{-}\text{+\hspace{0.17em}\hspace{0.17em}5e}\to M{n}^{2+}+\text{4}{H}_{2}O\end{array}$}

_{$\begin{array}{c}5\times \\ 2\times \end{array}\left|\begin{array}{l}\stackrel{+1}{H}\stackrel{+2}{C}\stackrel{-2}{O}\stackrel{-2}{O}\stackrel{+1}{H}\to \stackrel{+4}{C}{\stackrel{-2}{O}}_2+2H^{+}+2e\\ 8H^{+}+Mn{O_4}^{-}\text{+\hspace{0.17em}\hspace{0.17em}5e}\to M{n}^{2+}+\text{4}{H}_{2}O\end{array}\right.$}

⇒ $16H^{+}+\text{2}Mn{O_4}^{-}\text{+\hspace{0.17em}\hspace{0.17em}}\stackrel{+1}{5H}\stackrel{+2}{C}\stackrel{-2}{O}\stackrel{-2}{O}\stackrel{+1}{H}\to 2M{n}^{2+}+\text{8}{H}_{2}O+5\stackrel{+4}{C}{\stackrel{-2}{O}}_2+10H^{+}$

6H⁺ + 2MnO­₄^- + 5HCOOH → 2Mn²⁺ + 8H₂O + 5CO₂