Question

Four half-reactions, I to IV are shown below:

1. $$2\mathrm{Cl}{\phantom{A}}^{-}⟶\mathrm{Cl}{\phantom{A}}_2+2\mathrm{e}{\phantom{A}}^{-}$$
2. $$4\mathrm{OH}{\phantom{A}}^{-}⟶\mathrm{O}{\phantom{A}}_2+2\mathrm{H}{\phantom{A}}_2\mathrm{O}+2\mathrm{e}{\phantom{A}}^{-}$$
3. $$\mathrm{Na}{\phantom{A}}^{+}+\mathrm{e}{\phantom{A}}^{-}⟶\mathrm{Na}$$
4. $$2\mathrm{H}{\phantom{A}}^{+}+2\mathrm{e}{\phantom{A}}^{-}⟶\mathrm{H}{\phantom{A}}_2$$

Which two of these reactions are most likely to occur when concentrated brine is electrolysed?

Options

• I and III

• II and III

• I and IV

• II and IV

Answer 1

I and III

Explanation:

When concentrated brine (sodium chloride solution) is electrolysed, the following two half-reactions are most likely to happen:

$$2\mathrm{Cl}{\phantom{A}}^{-}⟶\mathrm{Cl}{\phantom{A}}_2+2\mathrm{e}{\phantom{A}}^{-}$$

Chloride ions (Cl⁻) lose electrons to generate chlorine gas (Cl₂) in this process, which takes place at the anode.

$$\mathrm{Na}{\phantom{A}}^{+}+\mathrm{e}{\phantom{A}}^{-}⟶\mathrm{Na}$$

Sodium ion (Na⁺) gains electrons during this process at the cathode, forming sodium (Na).