Answer to Question #165898 in Organic Chemistry for Jade

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:

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