In the preparation of chlorine by the electrolysis of brine, the reaction taking place at the anode is

Question

In the preparation of chlorine by the electrolysis of brine, the reaction taking place at the anode is (A) Cl -(a q) longrightarrow (1/2) Cl 2(g)+e- (B) Na +(a q)+e- longrightarrow Na ( s ) (C) O 2( g )+4 H ++4 e - longrightarrow 4 H 2 O (I) (D) H +(a q)+ e - longrightarrow (1/2) H 2(g)

Tardigrade Admin · Accepted Answer

Chlorine can be manufactured by electrolysis of a sodium chloride solution (brine). The production of chlorine results in the co-products of caustic soda (sodium hydroxide,

N a O H

) and hydrogen gas

(H_{2})

.

At cathode

2 H^{+} (a q) + 2 e^{-} ⟶ H_{2} (g)

At anode

C l^{-} (a q) ⟶ \frac{l}{2} C l_{2} (a q) + e^{-}

or

2 C l^{-} (a q) ⟶ C l_{2} (g) + 2 e^{-}

Q. In the preparation of chlorine by the electrolysis of brine, the reaction taking place at the anode is

$C l^{-} (a q) ⟶ \frac{1}{2} C l_{2} (g) + e^{-}$

$N a^{+} (a q) + e^{-} ⟶ N a (s)$

$O_{2} (g) + 4 H^{+} + 4 e^{-} ⟶ 4 H_{2} O (I)$

$H^{+} (a q) + e^{-} ⟶ \frac{1}{2} H_{2} (g)$

Q. In the preparation of chlorine by the electrolysis of brine, the reaction taking place at the anode is

Cl−(aq)⟶21​Cl2​(g)+e−

Na+(aq)+e−⟶Na(s)

O2​(g)+4H++4e−⟶4H2​O(I)

H+(aq)+e−⟶21​H2​(g)

$C l^{-} (a q) ⟶ \frac{1}{2} C l_{2} (g) + e^{-}$

$N a^{+} (a q) + e^{-} ⟶ N a (s)$

$O_{2} (g) + 4 H^{+} + 4 e^{-} ⟶ 4 H_{2} O (I)$

$H^{+} (a q) + e^{-} ⟶ \frac{1}{2} H_{2} (g)$