a 4 molar aqueous solution of NaCl is prepared and 500 ml of this solution is electrolysed . this leads to evolution of chlorine gas at one of the electrodes. [Hg=200] . if the cathode is Hg electrode the maximum weight of amalgam formed from this solution is? please explain in detail.

To determine the maximum weight of amalgam formed when a 4 molar aqueous solution of NaCl is electrolyzed, we first need to understand the electrolysis process and the reactions occurring at the electrodes. In this scenario, we are particularly interested in the cathodic reaction occurring at the mercury (Hg) electrode.

Understanding the Electrolysis of NaCl Solution

When an aqueous solution of sodium chloride (NaCl) is electrolyzed, it dissociates into sodium ions (Na⁺) and chloride ions (Cl^-). The electrolysis process involves the following key reactions:

• Anode Reaction: At the anode, chloride ions are oxidized to form chlorine gas (Cl₂).
• Cathode Reaction: At the cathode, water is reduced to form hydrogen gas (H₂) and hydroxide ions (OH^-). However, in the presence of mercury, sodium ions can react with mercury to form sodium amalgam (Na-Hg).

Calculating the Amount of Chlorine Gas Produced

First, we need to calculate the total moles of NaCl in the 500 ml of a 4 M solution:

Volume of solution = 500 ml = 0.5 L

Molarity (M) = moles of solute / volume of solution in liters

Thus, moles of NaCl = Molarity × Volume = 4 mol/L × 0.5 L = 2 moles

Since each mole of NaCl produces one mole of Cl₂ at the anode, the moles of chlorine gas produced will also be 2 moles.

Electrolysis and Amalgam Formation

At the cathode, the reduction of sodium ions occurs. The reaction can be represented as follows:

Na⁺ + e^- → Na (sodium metal)

In the presence of mercury, sodium metal can dissolve in mercury to form sodium amalgam:

Na (s) + Hg → Na-Hg (amalgam)

Calculating the Maximum Weight of Amalgam

To find the maximum weight of the amalgam formed, we need to determine how many moles of sodium can be produced from the electrolysis. Since we have 2 moles of NaCl, we can produce 2 moles of sodium ions, which means we can theoretically produce 2 moles of sodium metal.

The molar mass of sodium (Na) is approximately 23 g/mol. Therefore, the mass of sodium produced can be calculated as follows:

Mass of sodium = moles × molar mass = 2 moles × 23 g/mol = 46 g

Now, the amalgam formed will contain this sodium along with mercury. However, the weight of the amalgam will primarily depend on the amount of sodium produced, as the weight of mercury is not typically included in this calculation unless specified.

Final Consideration

Thus, the maximum weight of sodium amalgam formed from the electrolysis of 500 ml of a 4 M NaCl solution, assuming complete reaction and no losses, would be approximately 46 grams. This amalgam can be used in various applications, including in batteries and as a reducing agent in chemical reactions.