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Q2.18(iii):
Predict the products of electrolysis in each of the following: A dilute solution of \(H_{2}SO_{4}\) with platinum electrodes.

Solution :

Step 1: Identifying Ionic Species in Solution

In an aqueous dilute solution of sulfuric acid ($H_{2}SO_{4}$), the following dissociation and ionization processes occur:

  • Sulfuric acid (strong electrolyte): $H_{2}SO_{4}(aq) \rightarrow 2H^{+}(aq) + SO_{4}^{2-}(aq)$
  • Water (weak electrolyte/solvent): $H_{2}O(l) \rightleftharpoons H^{+}(aq) + OH^{-}(aq)$

The ions present in the electrolytic cell are: $H^{+}$, $SO_{4}^{2-}$, and $OH^{-}$. [Per the Arrhenius theory of dissociation].

Step 2: Analysis of Cathodic Reactions

At the cathode, reduction occurs. The competing species are $H^{+}$ ions and $H_{2}O$ molecules. Since only $H^{+}$ ions are present (from the acid), the following reduction half-reaction occurs:

$H^{+}(aq) + e^{-} \rightarrow \frac{1}{2}H_{2}(g)$

Product at Cathode: Hydrogen gas ($H_{2}$). [By standard electrode potential convention, $E^{\circ}_{H^{+}/H_{2}} = 0.00 \, V$].

Step 3: Analysis of Anodic Reactions

At the anode, oxidation occurs. The competing species are $SO_{4}^{2-}$ and $OH^{-}$ (from water). The competing half-reactions are:

Reaction $E^{\circ}$ (Volts)
$2SO_{4}^{2-}(aq) \rightarrow S_{2}O_{8}^{2-}(aq) + 2e^{-}$ $+2.01 \, V$
$2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-}$ $+1.23 \, V$

[Per the principle of selective discharge, the ion/molecule with the lower oxidation potential (or higher reduction potential) is preferred]. Since $E^{\circ}_{ox}$ for water ($+1.23 \, V$) is significantly lower than that of the sulfate ion ($+2.01 \, V$), water is preferentially oxidized.

Step 4: Visualization of the Electrolytic Process

Cathode Anode $H_{2}(g)$ ↑ $O_{2}(g)$ ↑ Electrolyte: $dil. H_{2}SO_{4}$

Step 5: Conclusion of Electrolysis Products

The net cell reaction is the electrolysis of water:

$2H_{2}O(l) \xrightarrow{electricity} 2H_{2}(g) + O_{2}(g)$

Final Solution: The product at the cathode is Hydrogen gas ($H_{2}$) and the product at the anode is Oxygen gas ($O_{2}$).


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