Question

Brine solution on electrolysis will give:

• A. $NaOH$
• B. $O_2$
• C. $C{l}_2$
• D. $H_2$

Answer 1

Answer: (A), (C), (D)

$NaOH$
$C{l}_2$
$H_2$

Brine is a solution of sodium chloride (NaCl) and water ($H_{2}O$). The electrolysis of brine is a large-scale process used to manufacture chlorine from salt. Two other useful chemicals are obtained during the process, sodium hydroxide (NaOH) and hydrogen (H2). The chlorine and sodium hydroxide produced in the process must be separated they react when they come into contact with each other.

The negative cathode electrode reaction for the electrolysis of brine (sodium chloride solution):

The negative cathode attracts the Na+ (from sodium chloride) and H+ ions
(from water). Only the hydrogen ions are discharged at the cathode. The more reactive a metal, the less readily its ion is reduced on the electrode surface.
The hydrogen ions are reduced by electron (e) gain to form hydrogen molecules at the negative electrode which attracts positive ions.

$2H^{+}\left(aq\right)+2e^{-}\to H_2\left(g\right)$.

other equations:

$2H_{2}O+2e^{-}\to H_2\left(g\right)+2O{H}^{-}\left(aq\right)$

The positive anode electrode reaction for the electrolysis of brine (sodium chloride solution):

The positive anode attracts the negative hydroxide OH ions (from water) and chloride Cl ions (from sodium chloride). Only the chloride ion is discharged in appreciable quantities i.e. it is preferentially oxidized to chlorine.

The chloride ions are oxidized by electron loss to give chlorine molecules at the positive electrode which attracts negative ions.

$C{l}_2\to 2C{l}^{-}+2e^{-}$

The hydroxide ion, with the unchanged sodium ion, forms NaOH

$N{a}^{+}+OH-\to NaOH$.

Hence, the correct options are $A,C$ and $D$ are correct.