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

Solution :

Given Variables & Initial Setup

The system consists of a dilute aqueous solution of sulfuric acid ($H_{2}SO_{4}$) undergoing electrolysis using inert platinum (Pt) electrodes. The ionic species present in the electrolyte are:

  • From the dissociation of water: $H_{2}O(l) \rightleftharpoons H^{+}(aq) + OH^{-}(aq)$
  • From the dissociation of sulfuric acid: $H_{2}SO_{4}(aq) \rightarrow 2H^{+}(aq) + SO_{4}^{2-}(aq)$

The total concentration of $H^{+}$ ions is significantly high due to the presence of both the acid and water dissociation equilibrium. The potential competitors for oxidation at the anode and reduction at the cathode are identified as follows:

Step 1: Identifying Cathode Reactions (Reduction)

At the cathode, the cations present are $H^{+}$. Since there is only one type of cation, the reduction reaction is straightforward:

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

[Per the principle of competitive discharge, $H^{+}$ is the only species competing for reduction at the cathode in this acidic medium.]

Step 2: Identifying Anode Reactions (Oxidation)

At the anode, the competing anions are $OH^{-}$ and $SO_{4}^{2-}$. However, we must also consider the oxidation of water molecules. The standard electrode potentials ($E^{\circ}$) are as follows:

Reaction $E^{\circ}$ (V)
$2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-}$ $+1.23\text{ V}$
$2SO_{4}^{2-}(aq) \rightarrow S_{2}O_{8}^{2-}(aq) + 2e^{-}$ $+1.96\text{ V}$

[By comparing the standard oxidation potentials, the reaction with the lower positive potential (for oxidation) occurs preferentially.]

Since the oxidation potential of water ($1.23\text{ V}$) is significantly lower than that of the sulfate ion ($1.96\text{ V}$), water molecules are oxidized at the anode instead of the sulfate ions.

Step 3: Visual Representation of the Electrolytic Cell

Anode (+) Cathode (-) Product: $O_{2}(g)$ Product: $H_{2}(g)$

Step 4: Summary of Cell Reactions

Combining the reactions at both electrodes:

  • Cathode: $4H^{+}(aq) + 4e^{-} \rightarrow 2H_{2}(g)$
  • Anode: $2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-}$
  • Net Reaction: $2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)$

[This confirms that the electrolysis of dilute $H_{2}SO_{4}$ is essentially the electrolysis of water, as the $H^{+}$ and $SO_{4}^{2-}$ ions act as charge carriers but are not consumed in the net redox process.]

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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