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. $O_2$ liberation at the cathode
• D. $H_2$ liberation at the anode

Answer 1

Answer: (A)

The cathode half-reaction in the process is
$2H_{2}O+2e^{-}\to H_{2}+2OH^{-}$
which increases $[OH^{-}]$ and, therefore $pH$ and the anode half-reaction
$2Cl^{-}\to Cl_{2}+2e^{-}$ does not produce or consume $H^{+}$ or $OH^{-}$
Overall, the reaction can be written as
$2Nacl+2H_{2}O\to 2NaOH+H_{2}+Cl_{2}$