MCQ
The passage of current liberates ${H_2}$ at cathode and $C{l_2}$ at anode. The solution is
  • A
    Copper chloride in water
  • $NaCl$ in water
  • C
    ${H_2}S{O_4}$
  • D
    Water

Answer

Correct option: B.
$NaCl$ in water
b
(b) Since discharge potential of water is greater than that of sodium so water is reduced at cathode instead of $N{a^ + }$

Cathode: ${H_2}O + {e^ - } \to \frac{1}{2}{H_2} + O{H^ - }$

Anode: $C{l^ - } \to \frac{1}{2}C{l_2} + {e^ - }$.

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Similar questions

Arrange reactivity of given alcohol in increasing order of elimination reaction

$(I)$ $\begin{array}{*{20}{c}}
  {C{H_2} - C{H_2} - C{H_2} - C{H_2}OH} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {N{O_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$

$(II)$ $\begin{array}{*{20}{c}}
  {C{H_3} - C{H_2} - CH - C{H_2}OH} \\ 
  {|\,} \\ 
  {N{O_2}} 
\end{array}$

$(III)$ $\begin{array}{*{20}{c}}
  {C{H_3} - CH - C{H_2} - C{H_2}OH} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {N{O_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$

$(IV)$ $CH_3-CH_2-CH_2-CH_2OH$

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