Electrolysis of concentrated aqueous solution of NaCl results in

Question

Electrolysis of concentrated aqueous solution of NaCl results in (A) increase in pH of the solution (B) decrease in pH of the solution (C) O2 liberation at the cathode (D) H2 liberation at the anode

Tardigrade Admin · Accepted Answer

The cathode half-reaction in the process is

2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}

which increases

[O H^{-}]

and, therefore

p H

and the anode half-reaction

2 C l^{-} \to C l_{2} + 2 e^{-}

does not produce or consume

H^{+}

or

O H^{-}

Overall, the reaction can be written as

2 N a c l + 2 H_{2} O \to 2 N a O H + H_{2} + C l_{2}

Q. Electrolysis of concentrated aqueous solution of $N a Cl$ results in

increase in $p H$ of the solution

decrease in $p H$ of the solution

$O_{2}$ liberation at the cathode

$H_{2}$ liberation at the anode

Q. Electrolysis of concentrated aqueous solution of NaCl results in

increase in pH of the solution

decrease in pH of the solution

O2​ liberation at the cathode

H2​ liberation at the anode

The cathode half-reaction in the process is 2H2​O+2e−→H2​+2OH− which increases [OH−] and, therefore pH and the anode half-reaction 2Cl−→Cl2​+2e− does not produce or consume H+ or OH− Overall, the reaction can be written as 2Nacl+2H2​O→2NaOH+H2​+Cl2​

Q. Electrolysis of concentrated aqueous solution of $N a Cl$ results in

increase in $p H$ of the solution

decrease in $p H$ of the solution

$O_{2}$ liberation at the cathode

$H_{2}$ liberation at the anode