Question

The same quantity of electricity is passed through one molar solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and one molar solution of \(\mathrm{HCl}\). The amount of hydrogen evolved from \(\mathrm{H}_{2} \mathrm{SO}_{4}\) as compared to that from \(\mathrm{HCl}\) is (a) the same (b) twice as such (c) one half as such (d) dependent on size of electrode

Short answer

The amount of hydrogen evolved from \(\mathrm{H}_2\mathrm{SO}_4\) as compared to that from \(\mathrm{HCl}\) is the same (a).

Step-by-step solution

Understand the Electrolysis Process

Electrolysis involves passing an electric current through an electrolyte to effect a chemical change. During this process, the amount of a substance liberated at an electrode is directly proportional to the quantity of electricity passed and is given by Faraday's laws of electrolysis.

Analyze the Electrolysis of \(\mathrm{H}_2\mathrm{SO}_4\) and \(\mathrm{HCl}\)

Both sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)) and hydrochloric acid (\(\mathrm{HCl}\)) dissolve in water to release hydrogen ions (\(\mathrm{H}^+\)) which are reduced at the cathode to form hydrogen gas (\(\mathrm{H}_2\)). This reaction will consume two moles of electrons (2 Faradays of electricity) to produce one mole of hydrogen gas according to the equation \,\(2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2\).

Compare the Amount of Hydrogen Evolved

Since the amount of electricity passed is the same through both solutions and the number of electrons required to release hydrogen gas is the same for both \(\mathrm{H}_2\mathrm{SO}_4\) and \(\mathrm{HCl}\), the amount of hydrogen gas evolved will also be the same.

Key concepts

Faraday's Laws of Electrolysis

Understanding the fundamentals of electrolysis is critical when examining the behavior of electrochemical reactions. With Faraday's laws of electrolyysis, we unlock the quantitative aspects of these reactions. Simply put, the first Faraday's law states that the amount of substance discharged at an electrode during electrolysis is directly proportional to the quantity of electric charge passing through the electrolyte. This can be expressed mathematically as:
\( M = (Q / F) \times (M_w / z) \),
where \(M\) is the mass of the substance liberated, \(Q\) is the quantity of electricity, \(F\) is Faraday's constant (approximately 96485 C/mol), \(M_w\) is the molar mass of the substance, and \(z\) is the valency number of ions of the substance (number of electrons transferred per ion).

Faraday's second law complements the first by stating that when the same amount on electricity passes through different electrolytes, the masses of substances liberated are proportional to their equivalent weights (molar mass/valency).

Electrochemical Reactions

Electrochemical reactions are at the heart of electrolysis. These involve the transfer of electrons from one species to another, driven by the flow of an electrical current. In the context of our exercise, when electricity is passed through solutions of \(\mathrm{H}_2\mathrm{SO}_4\) and \(\mathrm{HCl}\), an electrochemical reaction takes place at the cathode (negative electrode). This reaction reduces \(\mathrm{H}^+\) ions to form hydrogen gas \(\mathrm{H}_2\) through the following half-reaction:
\(2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2\).

• Both sulfuric acid and hydrochloric acid release \(\mathrm{H}^+\) ions into the solution.
• These positively charged ions migrate towards the cathode.
• They gain electrons (are reduced) and form hydrogen gas.

Importantly, this electrochemical reaction is not influenced by the presence of other ions in the solution, such as sulfate \(\mathrm{SO}_4^{2-}\) from sulfuric acid, since they do not participate in the reaction occurring at the cathode.

Stoichiometry of Electrolysis

Stoichiometry is a term that refers to the quantitative relationships within a chemical reaction. When we apply stoichiometry to electrolysis, we calculate the amounts of reactants and products based on the conservation of charge and mass. Here, the stoichiometry of the electrolysis process ensures that two moles of electrons are needed to produce one mole of hydrogen gas from \(\mathrm{H}^+\) ions.

In our exercise, since the number of moles of electrons required to produce a certain volume of hydrogen gas is consistent for both acids, the stoichiometry implies that the same quantity of electricity will evolve an equal amount of hydrogen, regardless of which solution is used. Hence, the presence of the sulfate ion \(\mathrm{SO}_4^{2-}\) in sulfuric acid doesn't affect the stoichiometry of hydrogen evolution, because the reduction of hydrogen ions is a distinct process, solely dependent on the number of electrons, which is the same in both cases.