Questão 33 - 1ª Fase - IME 2027

i) ${\mathrm{Cr}}^{3+}$ em meio básico converte-se em ${{\mathrm{CrO}}_4}^{2-}$. ${\mathrm{Br}}_2$ reduz para ${\mathrm{Br}}^{-}$.

Por íon-elétron:

$4{\mathrm{H}}_2\mathrm{O}+\mathrm{CrC}{\mathcal{l}}_3 \to \mathrm{Cr}{{\mathrm{O}}_4}^{2-}+3\mathrm{C}{\mathcal{l}}^{-}+8{\mathrm{H}}^{+}+3{\mathrm{e}}^{-} (\times 2)$

$$\begin{gathered} \underline{2{\mathrm{e}}^{-}+{\mathrm{Br}}_2\to 2{\mathrm{Br}}^{-} (\times 3) } \\ 2\mathrm{CrC}{\mathcal{l}}_3+3{\mathrm{Br}}_2+8{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{Cr}{{\mathrm{O}}_4}^{2-}+6{\mathrm{Br}}^{-}+6\mathrm{C}{\mathcal{l}}^{-}+16{\mathrm{H}}^{+} \end{gathered}$$

Como o meio é básico, soma-se 16$O{H}^{-}$ aos dois lados, de modo que $16H^{+}+16O{H}^{-}=16H_{2}O$:

$2\mathrm{CrC}{\mathcal{l}}_3+3{\mathrm{Br}}_2+16{\mathrm{OH}}^{-}\to 2\mathrm{Cr}{{\mathrm{O}}_4}^{2-}+6{\mathrm{Br}}^{-}+6\mathrm{C}{\mathcal{l}}^{-}+8{\mathrm{H}}_2\mathrm{O}$

Adicionando espectadores:

$\boxed{2\mathrm{CrC}{\mathcal{l}}_3+3{\mathrm{Br}}_2+16\mathrm{KOH}\to 2{\mathrm{K}}_2{\mathrm{CrO}}_4+6\mathrm{KBr}+6\mathrm{KC}\mathcal{l}+8{\mathrm{H}}_2\mathrm{O}}$

 

ii) ${{\mathrm{NO}}_2}^{-}$ se oxida a ${{\mathrm{NO}}_3}^{-}$. ${{\mathrm{MnO}}_4}^{-}$ em meio ácido reduz a ${\mathrm{Mn}}^{2+}$.

Por íon-elétron:

$$\begin{gathered} {\mathrm{H}}_2\mathrm{O}+{{\mathrm{NO}}_2}^{-} \to { {\mathrm{NO}}_3}^{-}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-} (\times 5) \\ \underline{8{\mathrm{H}}^{+} + 5{\mathrm{e}}^{-} + {\mathrm{MnO}}_4^{-} \to {\mathrm{Mn}}^{2+} + 4{\mathrm{H}}_2\mathrm{O} (\times 2)} \\ 6{\mathrm{H}}^{+}+5{{\mathrm{NO}}_2}^{-}+2\mathrm{Mn}{{\mathrm{O}}_4}^{-} \to 5{{\mathrm{NO}}_3}^{-}+2{\mathrm{Mn}}^{2+}+3{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Adicionando espectadores:

$\boxed{3{\mathrm{H}}_2{\mathrm{SO}}_4+5{\mathrm{KNO}}_2+2{\mathrm{KMnO}}_4 \to 5{\mathrm{KNO}}_3+2{\mathrm{MnSO}}_4+3{\mathrm{H}}_2\mathrm{O}+{\mathrm{K}}_2{\mathrm{SO}}_4}$

 

iii) $\mathrm{C}{\mathcal{l}}_2$ em ${\mathrm{H}}_2\mathrm{O}$ a frio se converte em $\mathrm{HC}\mathcal{l}+HC\mathcal{l}O$, e os dois ácidos são neutralizados pelo $Ca{\left(OH\right)}_2$.

$$\begin{gathered} \mathrm{C}{\mathcal{l}}_2+{\mathrm{H}}_2\mathrm{O} \to \mathrm{HC}\mathcal{l}+\mathrm{HC}\mathcal{l}\mathrm{O} (\times 2) \\ 2\mathrm{HC}\mathcal{l}+\mathrm{Ca}{\left(\mathrm{OH}\right)}_{2 } \to \mathrm{CaC}{\mathcal{l}}_2+2{\mathrm{H}}_2\mathrm{O} \\ \underline{2\mathrm{HC}\mathcal{l}\mathrm{O}+\mathrm{Ca}{\left(\mathrm{OH}\right)}_{2 } \to \mathrm{Ca}{\left(\mathrm{C}\mathcal{l}\mathrm{O}\right)}_2+2{\mathrm{H}}_2\mathrm{O} } \\ \boxed{2\mathrm{C}{\mathcal{l}}_2+2\mathrm{Ca}{\left(\mathrm{OH}\right)}_2 \to \mathrm{Ca}{\left(\mathrm{C}\mathcal{l}\mathrm{O}\right)}_2+\mathrm{CaC}{\mathcal{l}}_2+2{\mathrm{H}}_2\mathrm{O}} \end{gathered}$$