Question

The electrolysis of \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is conducted in two separate half-cells joined by a salt bridge, as suggested by the cell diagram \(\mathrm{Pt}\left|\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right|\left|\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right| \mathrm{Pt}\) (a) In one experiment, the solution in the anode compartment becomes more acidic and that in the cathode compartment, more basic during the electrolysis. When the electrolysis is discontinued and the two solutions are mixed, the resulting solution has \(\mathrm{pH}=7\). Write half-equations and the overall electrolysis equation. (b) In a second experiment, a 10.00 -mL sample of an unknown concentration of \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) and a few drops of phenolphthalein indicator are added to the \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) in the cathode compartment. Electrolysis is carried out with a current of \(21.5 \mathrm{mA}\) (milliamperes) for 683 s, at which point, the solution in the cathode compartment acquires a lasting pink color. What is the molarity of the unknown \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) ?\)

Short answer

The overall electrolysis equation is \(2 \mathrm{H}_{2} \mathrm{O}\left(\mathrm{l}\right) \rightarrow 2 \mathrm{H}_{2}\left(\mathrm{g}\right)+\mathrm{O}_{2}\left(\mathrm{g}\right)\). The molarity of the sulfuric acid solution is \(3.81 \times 10^{-3} M\)

Step-by-step solution

Write Half-Reactions and the Overall Equation

For the observed changes in acidity, the reactions involved in the two compartments are:\[2 \mathrm{H}_{2} \mathrm{O}\left(\mathrm{l}\right)+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\left(\mathrm{g}\right)+2 \mathrm{OH}^{-}\left(\mathrm{a q}\right)\] at the cathode where the solution becomes more basic and \[2 \mathrm{H}_{2} \mathrm{O}\left(\mathrm{l}\right) \rightarrow \mathrm{O}_{2}\left(\mathrm{g}\right)+4 \mathrm{H}^{+}\left(\mathrm{a q}\right)+4 \mathrm{e}^{-}\] at the anode where solution becomes more acidic. Combining these half-reactions and simplifying, we obtain the overall equation: \[2 \mathrm{H}_{2} \mathrm{O}\left(\mathrm{l}\right) \rightarrow 2 \mathrm{H}_{2}\left(\mathrm{g}\right)+\mathrm{O}_{2}\left(\mathrm{g}\right)\]

Apply Faraday's Law

The number of moles of electrons transferred is calculated from the charge passed thanks to Faraday's laws. The total charge which has passed through the cell during the electrolysis is given by multiplying the current (in amps) and the time (in seconds). The number of moles of e− can then be calculated by dividing the total charge by the charge of a mole of electrons (i.e. Faraday's constant). Quoting Faraday's constant as 96,500 coulombs per mole of electrons, we get that the total charge passed was equal to \(Q = I \times t = 0.0215 A \times 683 s = 14.7 C\). As 4 moles of e− are needed per mole of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), the number of moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) that reacted at the cathode is \((14.7 C) / (4 \times 96,500 C mol^{-1}) = 3.81 \times 10^{-5} mol\)

Calculate the Molarity

The molarity of the \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution can be calculated by dividing the number of moles by the volume of the solution in litres. As the volume is given as 10.00 ml, corresponding to 0.01 L, the molarity of the \( \mathrm{H}_{2}\mathrm{SO}_{4} \)solution is \((3.81 \times 10^{-5} mol) / (0.01 L) = 3.81 \times 10^{-3} M\)

Key concepts

Electrochemical Cells

Electrochemical cells are the foundations of battery technology and industrial processes like electrolysis. They work by converting chemical energy into electrical energy through redox reactions. In simple terms, these cells have two electrodes—an anode where oxidation occurs and a cathode where reduction takes place. When we talk about the electrolysis of sodium sulfate aq), it's imperative to visualize it occurring in an electrochemical cell setup. The sodium sulfate solution acts as an electrolyte, facilitating the movement of ions, which are crucial for the redox reactions at the electrodes. The cells are connected by a salt bridge that helps maintain charge balance by permitting the flow of ions, thus completing the electrical circuit without mixing the two solutions.

Faraday's Laws of Electrolysis

Understanding Faraday's laws of electrolysis is essential when investigating electrochemical reactions like the electrolysis of sodium sulfate aq). Faraday's first law tells us that the amount of chemical change produced at each electrode during electrolysis is proportional to the quantity of electricity that passes through the electrolyte. This directly correlates to the second law, which states that the amount of chemical change is also proportional to the equivalents of the substance undergoing the change.

This means that with a known current and time we can calculate the total amount of charge passed and thereby determine the moles of electrons transferred using the constant 96,500 C/mol (one Faraday), which represents the charge of one mole of electrons. This enables us to tie in the stoichiometry of the reaction with the electrical components, a core principle in exercises dealing with electrochemistry.

Stoichiometry Calculations

Stoichiometry calculations are pivotal in chemistry, providing a quantitative bridge between the reactants and products in a chemical reaction. For the electrolysis scenario, stoichiometry helps us understand the relationships between the water molecules being electrolyzed and the hydrogen and hydroxide ions being formed at the cathode, and oxygen and hydrogen ions at the anode.

The stoichiometry of the overall reaction is balanced based on the conservation of mass and charge, ensuring that the number of atoms and the electric charge are the same on both sides of the reaction equation. This balance is what allows precise calculations, such as determining the number of moles of sulfuric acid neutralized during the process, which is the first step towards finding its molarity.

Acid-base Indicators

Acid-base indicators such as phenolphthalein are substances that exhibit different colors depending on the pH of the environment they are in. Phenolphthalein is particularly helpful in titrations and electrolysis experiments like the one described because it turns from colorless in acidic solutions to pink in basic solutions.

When added to the cathode compartment during electrolysis, it stays colorless until a sufficient amount of hydroxide ions (OH-) are produced to raise the pH above about 8.3, at which point the solution acquires a lasting pink color. This color change signifies the endpoint of the reaction, allowing us to calculate the time taken for a certain amount of electric charge to pass through the solution, which is subsequently used for molarity calculations based on Faraday's laws.

Molarity Calculations

Molarity, the concentration of a solution expressed in moles of solute per liter of solution, is a critical concept in stoichiometry and chemistry at large. Calculations involving molarity are often used to determine the strength of an unknown solution or the amount of reagent required for a reaction.

In the context of the electrolysis exercise, the molarity of an unknown concentration of sulfuric acid is calculated using the moles of acid reacted (which we have determined from the stoichiometry and Faraday's laws) and the volume of sulfuric acid sample. Precise molarity calculations allow chemists to prepare solutions with accurate concentrations, necessary for quantitative analyses and chemical reactions.