Question

Predict the products of electrolysis in a dilute solution of ${\mathrm{H}}_2{\mathrm{SO}}_4$ with platinum electrodes.

Answer 1

Electrolysis:

• Electrolysis is a process by which electrical energy is used to produce a chemical change.
• The electrolysis cell consists of two metallic rods known as electrodes that are connected to a battery and dipped into molten electrolyte or electrolyte solution.
• The electrode connected to the positive side of the battery is called the anode.
• The electrode connected to the negative side of the battery is called the cathode.
• The electrically charged cations and anions presents in the electrolyte move towards the respective electrodes.
• That is, positively charged cations move toward the cathode and negatively charged anions move towards the anode.
• When the anions reach the anode, they are discharged through electron removal(oxidation), liberating neutral atoms or molecules at the electrode.
• Cations are neutralized through electron addition(reduction) when they reach the cathode

Electrolysis in a dilute solution of ${\mathrm{H}}_2{\mathrm{SO}}_4$ with platinum electrodes:

• In an aqueous solution, dilute Sulphuric acid dissociates to form protons and Sulphate ions and the autoionisation of water gives protons and Hydroxide ions.

$\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Protons}}{2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Sulphate}\mathrm{ions}}{{\mathrm{SO}}_2^{4-}\left(\mathrm{aq}\right)}$

$\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Protons}}{{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Hydroxide}\mathrm{ions}}{{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)}$

• Thus, the dilute aqueous Sulphuric acid solution contains protons, Hydroxide ions and Sulphate ions.
• Anode(positive electrode) attracts hydroxide ions and sulphate ions.
• Since the value of the electrode potential for the discharge of Hydroxide ions is less than that of Sulphate ions, the Hydroxide ions will get oxidised preferentially at the anode.
• Thus, the Oxygen gas will liberate in preference to Sulphur dioxide gas at the anode.
  $\underset{\mathrm{Hydroxide}\mathrm{ions}}{4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{electrons}}{4{\mathrm{e}}^{-}}$
• However, if Sulphate ions are discharged, Sulphur dioxide gas is liberated.

$\underset{\mathrm{Sulphate}\mathrm{ions}}{{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)}+\underset{\mathrm{Protons}}{4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Sulphur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{electrons}}{4{\mathrm{e}}^{-}}\mathrm{C}$

• The cathode is a negative electrode and it attracts protons and reduces them to form Hydrogen gas.

$\underset{\mathrm{Hydrogen}\mathrm{ion}\left(\mathrm{Protons}\right)}{2\mathrm{H}{}^{+}(\mathrm{aq})}+\underset{\mathrm{electrons}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Hydrogen}\mathrm{gas}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$

Therefore, in the electrolysis of dilute aqueous Sulphuric acid, using Platinum electrodes, Oxygen gas is liberated at the anode and hydrogen gas is liberated at the cathode.