Điện phân dung dịch X gồm CuSO4 và KCl...

A

- Hướng dẫn giải

Thứ tự oxh trên anot: Cl^- > H₂O; thứ tự khử trên catot: Cu²⁺>H₂O

\(\begin{gathered} TN1\left\{ \begin{gathered} BTE:\,\,2{n_{Cu}} + 2{n_{{H_2}}} = 2{n_{C{l_2}}} = \frac{{It}}{F} = 0,04 \hfill \\ {\overline M _{(C{l_2},\,\,{H_2})}} = \frac{{71{n_{C{l_2}}} + 2{n_{{H_2}}}}}{{{n_{C{l_2}}} + {n_{{H_2}}}}} = 48 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} {n_{C{l_2}}} = 0,02 \hfill \\ {n_{{H_2}}} = 0,01 \hfill \\ {n_{Cu}} = 0,01 \hfill \\ \end{gathered} \right. \Rightarrow X\,\,co\`u \,\,\left\{ \begin{gathered} CuS{O_4}:\,\,0,01\,\,mol \hfill \\ KCl:\,\,0,05\,\,mol \hfill \\ \end{gathered} \right\}\, \hfill \\ TH2\,\,\left\{ \begin{gathered} BTE:\,\,2\underbrace {{n_{Cu}}}_{0,01} + 2{n_{{H_2}}} = 2\underbrace {{n_{C{l_2}}}}_{0,025} + 4{n_{{O_2}}} \hfill \\ {m_{dd\,\,giam}} = 64\underbrace {{n_{Cu}}}_{0,01} + 2{n_{{H_2}}} = 71\underbrace {{n_{C{l_2}}}}_{0,025} + 32{n_{{O_2}}} = 2,715 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} {n_{{O_2}}} = 0,0075 \hfill \\ {n_{{H_2}}} = 0,03 \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow t = \frac{{96500.(0,01.2 + 0,03.2)}}{2} = \boxed{3860\,\,s} \hfill \\ \end{gathered} \)