Question

For the electrolysis of molten $NaCl$ using inert platinum electrodes, choose the correct option.

• A. Reaction at anode :
  $N{a}^{+}\left(l\right)+e^{-}\rightleftharpoons Na\left(s\right)$
• B. Reaction at cathode :
  $N{a}^{+}\left(l\right)+e^{-}\rightleftharpoons Na\left(s\right)$
• C. Chlorine gas is produced at cathode
• D. All of the above.

Answer 1

The correct option is B Reaction at cathode :

$N{a}^{+}\left(l\right)+e^{-}\rightleftharpoons Na\left(s\right)$
Electrolysis of molten $NaCl$ with $Pt$ electrode:
In molten $NaCl$ , the ions $N{a}^{+}$ and $C{l}^{-}$ are rpesent in molten form
Reaction at anode :
$2C{l}^{-}\left(l\right)\rightleftharpoons C{l}_2\left(g\right)+2e^{-}$
So, chlorine gas is given at anode.

Reaction at cathode :
$N{a}^{+}\left(l\right)+e^{-}\rightleftharpoons Na\left(s\right)$
At cathode, sodium metal is there.

Net reaction:
$2N{a}^{+}\left(l\right)+2C{l}^{-}\left(l\right)\rightleftharpoons 2Na\left(s\right)+C{l}_2\left(g\right)2NaCl\left(l\right)\rightleftharpoons 2Na\left(s\right)+C{l}_2\left(g\right)$
Hence option (b) is correct.