Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

Electrolysis of dilute aqueous \(\mathrm{NaCl}\) solution was carried out by passing 10 mili ampere current. The time required to librate \(0.01\) mole of \(\mathrm{H}_{2}\) gas at the cathode? (a) \(9.65 \times 10^{4} \mathrm{Sec}\) (b) \(19.3 \times 10^{4} \mathrm{Sec}\) (c) \(28.95 \times 10^{4} \mathrm{Sec}\) (d) \(38.6 \times 10^{4} \mathrm{Sec}\)

Step-by-Step Solution

Answer

The time required is \( 19.3 \times 10^4 \text{ seconds} \) (Option b).

Step 1: Write the Half-Cell Reaction

In the electrolysis of dilute aqueous \( \mathrm{NaCl} \), \( \mathrm{H}_2 \) gas is produced at the cathode through the reduction of water. The half-cell reaction is: \[ 2\mathrm{H}_2O + 2e^- \rightarrow \mathrm{H}_2 + 2\mathrm{OH}^- \]

Step 2: Calculate Moles of Electrons Needed

The half-reaction shows that it takes 2 moles of electrons to release 1 mole of \( \mathrm{H}_2 \). Therefore, for \( 0.01 \) moles of \( \mathrm{H}_2 \), you need \( 0.02 \) moles of electrons. So, moles of electrons needed: \( 0.02 \) moles.

Step 3: Convert Moles of Electrons to Total Charge

Using Faraday's constant \( F = 96485 \) coulombs/mol, the total charge \( Q \) required is given by: \[ Q = \text{moles of electrons} \times F = 0.02 \times 96485 = 1929.7 \text{ C} \]

Step 4: Calculate Time Using Current

With a current \( I = 10 \) mA (which is \( 0.01 \) A in usual units), use the formula \( t = \frac{Q}{I} \) to find the time: \[ t = \frac{1929.7}{0.01} = 192970 \text{ sec} \]

Step 5: Determine the Closest Option

The calculated time is \( 192970 \text{ sec} \), which matches the option \( 19.3 \times 10^{4} \text{ sec} \), or answer \( (b) \).

Key Concepts

Electrolysis of NaCl

Electrolysis is a process where electric current drives a non-spontaneous chemical reaction. In this case, the solution of aqueous sodium chloride (NaCl) undergoes electrolysis. When NaCl is dissolved in water, it dissociates into sodium ions (Na⁺) and chloride ions (Cl⁻). A low voltage electric current is passed through the solution, causing two main reactions at the electrodes.
At the cathode, water molecules are reduced to produce hydrogen gas (H₂) and hydroxide ions (OH⁻). Concurrently, at the anode, chloride ions (Cl⁻) are oxidized to form chlorine gas (Cl₂). The overall process highlights the separation of ions and formation of gases due to the electric current.
This demonstration of electrolysis serves as an excellent example of applying electrical energy to induce chemical transformations and is widely used in industries for extracting elements and compounds.

Half-cell reactions

Half-cell reactions are the core of understanding electrolysis. They involve smaller, more manageable equations representing reactions at each electrode. During the electrolysis of NaCl, two distinct half-cell reactions occur:
- **At the Cathode**: Water molecules gain electrons to form hydrogen gas and hydroxide ions: \[ 2\mathrm{H}_2\mathrm{O} + 2e^- \rightarrow \mathrm{H}_2 + 2\mathrm{OH}^- \]- **At the Anode**: Chloride ions lose electrons to produce chlorine gas: \[ 2\mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 + 2e^- \]Understanding these half-cell reactions is crucial as they provide insight into how electrons are exchanged, which substances are oxidized or reduced, and which products are formed during the process.

Faraday's constant

Faraday's constant (F) serves as a key factor in converting moles of electrons to an equivalent electrical charge. It is defined as the amount of charge in one mole of electrons, approximately equal to 96,485 coulombs/mol.
In electrochemical calculations, this constant allows us to calculate the total charge required for a given amount of substance produced or consumed at the electrodes. For example, to determine the charge needed to produce hydrogen gas in electrolysis, we multiply the moles of electrons by Faraday's constant:
\[ Q = \text{moles of electrons} \times F \]
This step is essential for linking the chemical changes in electrolysis with the electrical energy applied, and it underscores the interrelationship between chemistry and physics.

Electrochemical calculations

Electrochemical calculations combine chemistry and electricity principles to quantify changes during electrolysis. They involve the following steps:

• A first step involves deducing the moles of electrons needed, based on the half-cell reactions involved. For instance, in hydrogen gas production at the cathode, the reaction requires 2 moles of electrons per mole of hydrogen gas.
• Next, use Faraday's constant to find total charge (Q) required, multiplying the moles of electrons by 96,485 C/mol.
• Finally, calculate the time (t) required using the current applied (I) and charge (Q) through the formula:
  \[ t = \frac{Q}{I} \]

These calculations are integral to designing and optimizing electrolysis processes for practical applications, allowing precise determination of the time and energy needs for chemical transformations.