Question

In the electrolysis of aqueous sodium chloride with inert electrodes the products obtained at anode and cathode respectively are

• A. $Cl_2$ and $Na$
• B. $O_2$ and $Na$
• C. $Cl_2$ and $H_2$
• D. $Na$ and $Cl_2$

Answer 1

Answer: (C)

In the electrolysis of aqueous sodium chloride with inert electrodes, the products obtained at anode and cathode respectively are $Cl _{2}$ and $H _{2}$.

The solution contains four ions $Na ^{+}, Cl^- , H ^{+}, \overline{ O } H .$

Following reaction takes place:

At cathode $Na ^{+}( aq )+ e ^{-} \longrightarrow Na ( s ) ; $

$E^-_{ Na ^{+} / Na } =-2.71 V $

$2 H _{2} O (l)+2 e^{-} \longrightarrow H _{2}(g)+2 \overline{ O } H ( aq ) ;$

$E_{ H _{2} O / H _{2}}^{\circ}=-0.83 V$

At anode

$2 Cl^-(a q) \longrightarrow Cl _{2}(g)+2 e^{-} ; E_{ Cl ^{-} / Cl _{2}}^{\circ}=-1.36 V$

$2 H _{2} O _{2}(l) \longrightarrow O (g)+4 H ^{+}( aq )+4 e^{-}$

$E_{ H _{2} O / O _{2}}^{\circ}=-1.23 V$

$E_{ H _{2} O / H _{2}}^{\circ} > E_{ Na ^{+} / Na }^{\circ} .$ So, $H _{2}$ is formed at cathode.

Whereas $E_{ H _{2} O/O _{2}}^{\circ}>E_{ Cl ^{-} / Cl _{2}}^{\circ} .$ So, $Cl _{2}$ is formed at anode.