Question

A volume of \(100 \mathrm{~m}\) of a buffer of \(1 \mathrm{M}\) \(-\mathrm{NH}_{3}\) and \(1 \mathrm{M}-\mathrm{NH}_{4}^{+}\) is placed in two half-cells connected by a salt bridge. \(A\) current of \(1.5 \mathrm{~A}\) is passed through the cell for \(20 \mathrm{~min}\). If electrolysis of water takes place only and the electrode reactions are: Right: \(2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 \mathrm{e} \rightarrow 4 \mathrm{OH}^{-}\) and Left: \(2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{H}^{+}+\mathrm{O}_{2}+4 \mathrm{e}\), then, the \(\mathrm{pH}\) of the (a) right electrode will increase (b) left electrode will increase (c) both electrode will increase (d) both electrode will decrease

Short answer

The pH of the right electrode will increase, and the pH of the left electrode will decrease.

Step-by-step solution

Understanding the Electrode Reactions

The exercise describes the electrolysis of water which involves two half-reactions occurring at the anode (right) and the cathode (left). No reaction is mentioned for the presence of the -H_3 and H_4^+ in the buffer solution, so we assume they don't participate in the electrode reactions. This will not be considered further in the calculations.

Calculating the Moles of Electrons

Using the current (I) and the time (t), calculate the total charge (Q) passed using Q = I × t. Then, calculate the moles of electrons (n) transferred using Q = n × F, where F is the Faraday constant (F = 96,485 C/mol).

Determining the Moles of Water Electrolyzed

From the half-reactions, we know that 4 electrons are required to electrolyze 2 water molecules. So, the moles of water electrolyzed can be calculated by dividing the moles of electrons by 4.

Assessing Changes in pH

For the right electrode (anode), oxygen gas is produced and OH- ions are released into the solution, which will cause the pH to increase. For the left electrode (cathode), H+ ions are produced alongside oxygen gas, which will cause the pH to decrease.

Selecting the Correct Answer

Since the pH at the right electrode will increase due to production of OH- ions, and the pH at the left electrode will decrease due to production of H+ ions, the correct answer is that the pH of the right electrode will increase.

Key concepts

Electrode Reactions Chemistry

Electrolysis is a chemical process that uses electricity to drive a non-spontaneous reaction. Specifically, in the electrolysis of water, electricity is used to decompose water into oxygen and hydrogen gases. The reactions at the electrodes are fundamental to understanding the overall process.

At the anode, water oxidizes to form oxygen gas, protons, and electrons. The half-reaction is written as: \[2 \text{H}_2\text{O} (l) \rightarrow \text{O}_2 (g) + 4 \text{H}^+ (aq) + 4 \text{e}^-\]. This process is known as the oxidation reaction because it involves the loss of electrons.

At the cathode, the reduction half-reaction occurs, where water is reduced to hydroxide ions while producing hydrogen gas: \[2 \text{H}_2\text{O} (l) + 2 \text{e}^- \rightarrow \text{H}_2 (g) + 2 \text{OH}^- (aq)\]. This process is known as the reduction reaction as it involves the gain of electrons.

The electrode reactions illustrate the changes occurring in the chemical composition of the solution, ultimately affecting the concentration of hydrogen (H+) and hydroxide (OH-) ions, which in turn influences the pH of the solution.

Calculation of Moles of Electrons

The charge passed through an electrolytic cell over time is directly related to the number of moles of electrons that have been transferred during the reaction. To determine this relationship, one must understand the basic principles of electrochemistry.

First, the total charge (Q) passed through the system is calculated using the formula: \[Q = I \times t\], where 'I' is the current in amperes and 't' is the time in seconds.

Once the total charge is known, the number of moles of electrons (n) involved can be calculated using Faraday's constant (F), which is approximately 96,485 coulombs per mole of electrons: \[n = \frac{Q}{F}\].

Using these equations allows us to quantitatively relate the electrical current to the chemical changes occurring during electrolysis. It is crucial for understanding how much of a substance is produced or consumed at the electrodes during the process.

pH Assessment in Electrolysis

The pH of a solution during electrolysis is a measure of the concentration of hydrogen ions (H+) present; a low pH indicates an acidic solution, while a high pH indicates a basic solution. During the electrolysis of water, the reactions at the electrodes affect the pH by either increasing the concentration of H+ ions or OH- ions.

At the anode, where oxidation occurs, oxygen gas is produced and excess H+ ions are released into the solution. As the concentration of H+ ions increases, this leads to a lower pH, thereby making the solution more acidic.

Conversely, at the cathode, reduction produces hydroxide (OH-) ions, increasing the solution's pH and making it more basic. The increase in OH- ions surpasses the concentration of H+ ions, leading to a higher, more alkaline pH.

Therefore, by assessing the pH changes at each electrode, one can determine the direction of the chemical reactions and their effects on the solution's properties. This is critical for processes where the pH must be controlled, such as in the production of certain chemicals or in environmental treatment applications.