Problem 255 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(2 \mathrm{~L}\) of \(0.5 \mathrm{M} \mathrm{NaCl}\) is electrolysed with \(112.10 \mathrm{~F}\) of charge, the \(\mathrm{pOH}\) of resulting solution is close to (assume the concentration of \(\mathrm{Cl}^{-}\)remaining to be very small)

Step-by-Step Solution

Answer

The pOH of the resulting solution is approximately 1.26.

Step 1: Determine Initial Moles of NaCl

Calculate the initial moles of NaCl in the solution using the formula \( ext{Moles} = ext{Concentration} \times ext{Volume} \). Here, \( ext{Concentration} = 0.5 \, ext{M} \) and \( ext{Volume} = 2 \, ext{L} \). So, the initial moles of NaCl is \( 0.5 \times 2 = 1 \, ext{mole} \).

Step 2: Examine Electrolysis Reaction

During the electrolysis of NaCl solution, NaCl dissociates into Na\(^+\) and Cl\(^-\). The hydroxide ions (OH\(^-\)) are formed at the cathode from water, increasing pOH.

Step 3: Calculate Faraday's Requirement for Cl⁻ Conversion

To oxidize Cl\(^-\) to Cl\(_2\), one Faraday (1 F) of electricity is needed per mole of electrons transferred. Since we have 112.1 F of charge, we can convert \(112.1\, / \,1 = 112.1\) moles of Cl\(^-\) to Cl\(_2\). Given that moles of initial Cl\(^-\) is 1, all Cl\(^-\) is converted and excess electricity acts to form OH\(^-\).

Step 4: Compute Excess Charge for OH⁻ Formation

After oxidizing all 1 mole of Cl\(^-\), we used 1 Faraday. There remains \(112.1 - 1 = 111.1\) Faradays for water electrolysis. This excess charge causes the formation of OH\(^-\) ions.

Step 5: Determine Moles of OH⁻ Formed

Each Faraday forms 1 mole of OH\(^-\). Therefore, \( 111.1 \) moles of OH\(^-\) are formed corresponding to the remaining charge.

Step 6: Calculate Concentration of OH⁻

Concentration of OH\(^-\) is given by \( \frac{111.1}{2} \, ext{M} = 55.55 \, ext{M} \) considering the full volume of 2 L.

Step 7: Find pOH from OH⁻ Concentration

The pOH is calculated using \( \text{pOH} = -\log[\text{OH}^-] = -\log[55.55] \approx 1.26 \).

Key Concepts

pOH Calculation

The concept of pOH is closely related to pH, which is a measure of the acidity or alkalinity of a solution. pOH, on the other hand, measures the concentration of hydroxide ions (OH⁻) in a solution. It is calculated using the formula: \[ \text{pOH} = -\log[\text{OH}^-] \]Understanding how to calculate pOH is essential when dealing with solutions that involve changes due to reactions like electrolysis. During electrolysis, substances dissociate and form new ions, which can change the concentration of OH⁻, impacting the pOH. In the solution process, once the concentration of OH⁻ is known, calculating the pOH involves simply taking the negative logarithm of that concentration. For instance, if the concentration of OH⁻ in the solution is found to be 55.55 M, then pOH is approximately 1.26. This low value indicates a highly alkaline solution. Understanding pOH is crucial for assessing the effects of electrochemical reactions and determining the basicity of a solution after electrolysis.

Faraday's Law

Faraday's Law of Electrolysis is a fundamental principle that quantifies the relationship between electric charge and the chemical changes brought about by electrolysis. It is based on the concept that the amount of substance produced or consumed at an electrode is directly proportional to the amount of electric charge passed through the electrolyte.The law can be mathematically expressed as:\[ m = \frac{Q}{F} \cdot \frac{M}{n} \]Where:

• \( m \) represents the mass of the substance transformed (in grams),
• \( Q \) is the total electric charge passed (in Coulombs),
• \( F \) is Faraday's constant (approximately 96485 C/mol),
• \( M \) is the molar mass of the substance (in grams/mol), and
• \( n \) is the number of moles of electrons exchanged per mole of substance.

Electrolysis involves transferring electrons, and Faraday's Law helps predict how much of a substance will undergo transformation. For example, in our problem, with 112.1 Faradays of charge supplied, 112.1 moles of Cl⁻ ions are converted to chlorine gas. Faraday’s Law provides the theoretical understanding and calculations needed to determine how reactions proceed under the influence of electrical currents.

Electrochemical Reactions

Electrochemical reactions involve the processes where chemical reactions are driven by an electric current or where electrical energy is produced by chemical reactions. These reactions are central to many processes, including electrolysis, which separates elements through the application of electricity.Such reactions occur in electrochemical cells and are typically composed of two half-cell reactions: one that involves the oxidation process, and another for the reduction. During electrolysis of NaCl:

• At the anode, oxidation occurs where Cl⁻ ions give up electrons to form chlorine gas, \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \).
• At the cathode, reduction occurs where water molecules reduce, forming OH⁻ ions and hydrogen gas, \( 2\text{H}_2\text{O} + 2e^- \rightarrow 2\text{OH}^- + \text{H}_2 \).

Understanding these reactions is vital for predicting the behavior of components in an electrolytic cell and the changes in solution properties such as pOH. Electrochemical reactions demonstrate the interplay between electricity and chemistry, exemplifying principles like conservation of mass and charge balance across reactions.