Question

One litre of \(0.5 \mathrm{M} \mathrm{KCI}\) solution is electrolysed for one minute in a current of \(16.08 \mathrm{~mA}\). Considering \(100 \%\) efficiency, the \(\mathrm{pH}\) of resulting solution will be: (a) 7 (b) 9 (c) 8 (d) 10

Short answer

(b) 9

Step-by-step solution

Understand Current and Time Relation

The electrolysis process is driven by current, which causes the chemical reactions at the electrodes. Here, the given current is 16.08 mA (or \(0.01608 \mathrm{~A}\)), applied for a duration of 1 minute (or 60 seconds). The total charge \(Q\) transferred is calculated as \(Q = I \times t\), where \(I\) is current and \(t\) is time.

Calculate Total Charge Transferred

Using the formula for total charge \(Q = I \times t\), we substitute \(I = 0.01608 \mathrm{~A}\) and \(t = 60 \mathrm{~s}\). Thus, \(Q = 0.01608 \times 60 = 0.9648 \mathrm{~C}\).

Relate Charge to Amount of Substances Formed

In electrolysis, charge \(Q\) can be used to find the amount of substance formed based on Faraday's laws of electrolysis. Given that 1 Faraday (\(F\)) corresponds to 1 mole of electrons and is approximately 96485 coulombs, we calculate moles of electrons as \(\text{moles} = \frac{Q}{F}\). Here, \(\text{moles} = \frac{0.9648}{96485} \approx 10^{-5}\) moles of electrons.

Write the Electrolysis Reaction

In the solution, water is primarily reduced according to the half-reaction \(2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow \mathrm{H}_2 + 2\mathrm{OH}^-\). Thus, the electrons reduce water, producing hydroxide ions (\(\mathrm{OH}^-\)). Each mole of electrons produces one mole of \(\mathrm{OH}^-\) ions.

Calculate Accidentally Formed Hydroxide Ions

From step 3, there are \(10^{-5}\) moles of \(\mathrm{OH}^-\) ions formed. In a 1 liter solution, this translates directly to a concentration of \(10^{-5} \mathrm{~M}\) for \(\mathrm{OH}^-\).

Derive the pH of the Solution

Using the relation \(\mathrm{pOH} = -\log \text{[OH}^-\text{]}\), we find \(\text{[OH}^-\text{]} = 10^{-5} \mathrm{~M}\). Therefore, \(\mathrm{pOH} = 5\). The pH is calculated using \(\mathrm{pH} + \mathrm{pOH} = 14\), implying \(\mathrm{pH} = 14 - 5 = 9\).

Conclude with the Correct Answer

Based on the calculations, the resulting pH of the solution after electrolysis is 9.

Key concepts

Faraday's laws of electrolysis

Faraday's laws of electrolysis are fundamental principles in electrochemistry that quantify the relationship between the charge passed through an electrolyte and the amount of substance liberated at the electrodes. The first law states that the mass of a substance deposited or dissolved at an electrode is directly proportional to the amount of electric charge passed through the electrolyte. This means that more charge leads to more material being produced or consumed.

• Mass deposited \( ext{m} \propto \text{Q}\)

The second law states that the mass of substances produced by the same amount of charge is proportional to their equivalent weights. This explains how different substances deposit or dissolve at different rates depending on their chemical nature.
Understanding these laws helps us calculate the number of moles of electrons involved in an electrolysis process, as we use in the exercise to determine the moles of hydroxide ions produced.
Faraday's constant, approximately 96485 coulombs per mole, is crucial as it links the charge passed to the moles of electrons transferred. This was utilized in our problem to deduce the appearance of ions affecting the solution's pH.

pH calculation

Calculating pH is a key concept when dealing with aqueous solutions like the one in the exercise. The pH scale ranges from 0 to 14 and indicates the acidity or basicity of a solution. It is calculated as the negative logarithm of the hydrogen ion concentration: \( ext{pH} = -\log[ ext{H}^+]\).
In the context of electrolysis, as demonstrated in our example, reactions can create hydroxide ions, leading to a change in the solution's basicity. The resulting pOH, derived from the hydroxide ion concentration, helps determine the pH. The relation between pH and pOH is given by the equation: \( ext{pH} + ext{pOH} = 14\).
In our example, hydroxide ions were formed, making the solution more basic than neutral water. We calculated the concentration of resulting hydroxide ions, found the pOH value, and then derived the pH of the solution. Understanding this calculation ensures that students can predict whether a solution becomes more acidic or basic due to the reactions occurring during electrolysis.

Electrolysis reactions

Electrolysis reactions are core chemical processes where electrical energy is used to drive non-spontaneous reactions. In an electrolytic cell, electrical current flows through a solution, causing chemical changes at the electrodes. These changes include oxidation at the anode and reduction at the cathode.
In the context of aqueous \( ext{KCl}\) solution, water molecules can undergo reduction at the cathode, as seen in the half-reaction: \[2 ext{H}_2 ext{O} + 2e^- ightarrow ext{H}_2 + 2 ext{OH}^- \] This reaction explains how hydroxide ions are formed in the solution.
Certain ions might deposit at electrodes, but in the electrolysis of \( ext{KCl}\), water reduction is more favorable given the low concentration of any other ions compared to water.
By understanding these reactions, students learn how electron flow facilitates chemical transformations, explaining the change in solution composition during the electrolysis process. The production of gases and ions during electrolysis can significantly affect things like pH, providing a practical application of the exercise's reactions.

Current and charge relationship

The relationship between electric current and charge is fundamental in understanding electrolysis processes. Electric current, measured in amperes (A), is a flow of electric charge. During electrolysis, this charge facilitates chemical reactions.
Specifically, the total charge \(Q\) transferred is the product of current \(I\) and time \(t\): \[Q = I \times t \]
This relationship allows us to quantify how much charge has been used during the process. In the presented exercise, we applied 16.08 mA of current over one minute. Converting this to amperes and seconds enables the calculation of total charge, leading to the determination of the moles of electrons involved.

• Current: \( ext{16.08 mA} = 0.01608 A\)
• Time: \( ext{1 min} = 60 s\)
• Charge: \[Q = 0.01608 \times 60 = 0.9648 ext{ C}\]

By understanding this relationship, students can connect the dots between electrical input and chemical output, as shown by the production of hydroxide ions in our electrolysis example.