MCQ
If the current is passed into the solution of an electrolyte
  • Anions move towards anode, cations towards cathode
  • B
    Anions and cations both move towards anode
  • C
    Anions move towards cathode, cations towards anode
  • D
    No movement of ions takes place

Answer

Correct option: A.
Anions move towards anode, cations towards cathode
a
When an electric current is passed through a cell having an electrolyte, the positive ions move towards the cathode and the negative ions towards the anode. If the cathode is pulled out of the solution, the circuit will not be complete and the ions will have no direction and thus they will move randomly.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The more negative the standard potential, the .............. is its ability to displace hydrogen from acids.
Which two of the following salts are used for preparing iodized salt ?

$(i)\,\,KIO_3$                      $(ii)\,\,KI$

$(iii)\,\,I_2$                          $(iv)\,\, HI$

The octahedral diamagnetic low spin complex among the following is
At $298\,K$, the standard reduction potential for $Cu ^{2+} / Cu$ electrode is $0.34\,V$.

Given : $K _{ sp } Cu ( OH )_2=1 \times 10^{-20}$

$\operatorname{Take} \frac{2.303 RT }{ F }=0.059 \,V$

The reduction potential at $pH =14$ for the above couple is $(-) x \times 10^{-2}\,V$. The value of $x$ is $........$.

Aniline reacts with alkyl halide to give
The boiling point of benzene is $353.23\,\, K$. When $1.80\,gm$ of a nonvolatile solute was dissolved in $90\,gm$ of benzene, the boiling point is raised to $354.11\,K$. the molar mass of the solute is ........ $g\,mo{l^{ - 1}}$ . $[{K_b}$ for benzene $= 2.53 \,K\,mol^{-1}$]
product $(B)$ is
The reagent that can distinguish benzaldehyde from propionaldehyde is
In the following reactions,

$(i) \,\,\begin{array}{*{20}{c}}
  {C{H_3}\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3} - C - CH - C{H_3}} \\ 
  {\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,OH} 
\end{array}\,\xrightarrow{{{H^ + }/heat}}\,\,\mathop A\limits_{[Major\,product]} \, + \,\mathop B\limits_{[Minor\,product]} $

$(ii)\,\, A\xrightarrow[{in\,\,absence\,\,\,of\,peroxide}]{{HBr,\,dark}}\,\,\mathop C\limits_{[Major\,product]} \, + \,\mathop D\limits_{[Minor\,product]} $

the major products $(A)$ and $(C)$ are respectively

The formation of $PH_4^+$ is difficult compared to $NH_4^+$ because