Điện phân dung dịch chứa đồng thời NaCl và CuSO4 (điện cực trơ, màng

* Hướng dẫn giải

Đáp án B.

Khi $t_1=1930:$

$$\begin{gathered} {\text{Cu}}^{2+}+2\text{e}\to \text{Cu }2{\text{Cl}}^{-}\to {\text{Cl}}_2+2\text{e} \\ {n}_{\text{Cu}}=\frac{m}{64}\to n_{\text{e}}=\frac{2m}{64}\to n_{{\text{Cl}}_2}=\frac{m}{64} \\ {m}_{\text{dd}\downarrow }=m_{\text{Cu}}+m_{{\text{Cl}}_2}=2,7 \\ \to m+\frac{m}{64}.71=2,7\to m=1,28 \\ \to n_{\text{e}}=\frac{2m}{64}=0,04 \end{gathered}$$         

Khi ${\text{t}}_2=7720$: Khối lượng catot vẫn tăng chứng tỏ  vẫn chưa điện phân hết.

$$\begin{gathered} n_{\text{Cu}}=\frac{4m}{64}=0,08\to n_{\text{e}}=0,16 \\ {\text{Cu}}^{2+}+2\text{e}\to \text{Cu }\begin{array}{l}2{\text{Cl}}^{-}\to {\text{Cl}}_2+2\text{e}\\ x2x\end{array} \\ \begin{array}{l}2{\text{H}}_2\text{O}\to 4{\text{H}}^{+}+{\text{O}}_2+4\text{e}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4y\end{array} \\ \to \left\{\begin{array}{l}2x+4y=0,16\left(n_{\uparrow }\right)\\ 71x+32y+4.1,28=9,15\left(m_{\text{dd}\downarrow }\right)\end{array}\to \left\{\begin{array}{l}x=0,05\\ y=0,015\end{array}\right.\right. \end{gathered}$$         

$\to n_{\text{NaCl}}=n_{{\text{Cl}}^{-}}=0,05.2=0,1$ lúc này Fe( điện phân hết)

Khi thời gian ${\text{t}}_3$: Có thể ${\text{CuSO}}_4$ đã điện phân hết và ${\text{H}}_2\text{O}$  ở catot đã điện phân.