Question

aqueous sodium chloride can be electrolysed using inert electrodes. Describe the effect that the position of ions in the electrochemical series has on the products of this electrolysis

Answer 1

The Na⁺ ions migrate toward the negative electrode and the Cl^- ions migrate toward the positive electrode. But, now there are two substances that can be reduced at the cathode: Na⁺ ions and water molecules.

 Electrolysis:

$2NaCl(l)\rightarrow2Na(l)+Cl_2(g)$

Cathode (-):    

$Na^+ + e^-\rightarrow Na$ $E^{\degree}_{red}= -2.71V$

 $2H_2O + 2e^-\rightarrow H_2+2OH^-$ $E^{\degree}_{red}=-0.83V$

Because it is much easier to reduce water than Na⁺ ions, the only product formed at the cathode is hydrogen gas.

So, final product on cathode is:

Cathode (-): $2H_2O(l)+2e^- \rightarrow H_2(g)+2OH^-(aq)$

There are also two substances that can be oxidized at the anode: Cl^- ions and water molecules.

Anode (+):   

  $2Cl^-\rightarrow Cl_2 +2e^-$ $E^{\degree}_{ox}= -1.36V$  

$2H_2O\rightarrow O_2+4H^++4e^-$ $E^{\degree}_{ox}= -1.23V$

The standard-state potentials for these half-reactions are so close to each other that we might expect to see a mixture of Cl₂ and O₂ gas collect at the anode. In practice, the only product of this reaction is Cl₂.

So. final product on anode is:

Anode (+): $2Cl^-\rightarrow Cl_2+ 2e^-$

Electrolysis of aqueous $NaCl$ solutions gives a mixture of hydrogen and chlorine gas and an aqueous sodium hydroxide solution.

  Final Electrolysis:

$2NaCl(aq)+2H_2O(l)\rightarrow 2Na^+(aq)+2OH^-(aq)+H_2(g)+Cl_2(g)$