Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

One faraday of electricity is passed separately through one litre of one molar aqueous solutions of (i) \(\mathrm{AgNO}_{3}\) (ii) \(\mathrm{SnCl}_{4}\) and (iii) \(\mathrm{CuSO}_{4^{\prime}}\). The number of moles of \(\mathrm{Ag}, \mathrm{Sn}\), and \(\mathrm{Cu}\) deposited at cathode are respectively (a) \(1.0,0.25,0.5\) (b) \(1.0,0.5,0.25\) (c) \(0.5,1.0,0.5\) (d) \(0.25,0.25,0.5\)

Step-by-Step Solution

Answer

The correct answer is (a) \(1.0, 0.25, 0.5\).

Step 1: Understanding Faraday's Law of Electrolysis

Faraday's law states that the amount of substance deposited at an electrode during electrolysis is proportional to the quantity of electricity passed. The number of moles of electrons (q) required to deposit one mole of an element is given by the formula \[n = \frac{F}{z}\], where \(F\) is the Faraday constant (1 Faraday = 1 mole of electrons = 96500 C) and \(z\) is the number of electrons required to deposit one mole of the substance from its ions.

Step 2: Calculate Moles for \(\mathrm{Ag}\)

Silver (\(\mathrm{Ag}\)) from \(\mathrm{AgNO}_3\) requires one electron to deposit, as the reaction \(\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}\) shows. Therefore, \(z = 1\). If 1 Faraday is passed, the moles of \(\mathrm{Ag}\) deposited is \(n = \frac{1}{1} = 1.0\ moles\).

Step 3: Calculate Moles for \(\mathrm{Sn}\)

Tin (\(\mathrm{Sn}\)) from \(\mathrm{SnCl}_4\) requires four electrons to deposit, as the reaction \(\mathrm{Sn}^{4+} + 4e^- \rightarrow \mathrm{Sn}\) shows. Hence, \(z = 4\). By passing 1 Faraday, the moles of \(\mathrm{Sn}\) deposited is \(n = \frac{1}{4} = 0.25\ moles\).

Step 4: Calculate Moles for \(\mathrm{Cu}\)

Copper (\(\mathrm{Cu}\)) from \(\mathrm{CuSO}_4\) requires two electrons to deposit, as indicated by the reaction \(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\). Therefore, \(z = 2\). With 1 Faraday of electricity, the moles of \(\mathrm{Cu}\) deposited is \(n = \frac{1}{2} = 0.5\ moles\).

Step 5: Conclusion

The moles of \(\mathrm{Ag}\), \(\mathrm{Sn}\), and \(\mathrm{Cu}\) deposited are respectively 1.0, 0.25, and 0.5, matching option (a).

Key Concepts

Electrolysis

Electrolysis is an important chemical process where electrical energy is used to drive a non-spontaneous chemical reaction. Imagine a setup with two electrodes placed into a solution or molten ionic compound. When an electric current is passed through, ions in the electrolyte move towards the electrodes of opposite charge, causing chemical transformations.

This process is fascinating because it can decompose substances and form new substances. For example, water can be decomposed into hydrogen and oxygen gas. This phenomenon is extensively used in various industries, from electroplating to the production of chemicals like chlorine.

Key points to remember about electrolysis are:

• An external power source is needed to conduct the current.
• The solution or molten compound must contain ions that can react at the electrodes.
• Electrolysis can be used to purify metals, among other applications.

Moles of Electrons

Faraday's law of electrolysis introduces the concept of moles of electrons, which links electricity to the amount of substance altered or deposited in an electrochemical reaction. In practical terms, it means that the amount of material deposited or consumed in electrolysis is proportional to the current passed and the time it flows for.

To calculate the moles of a substance deposited at an electrode, you need to know the number of Faradays passed. One Faraday is equivalent to one mole of electrons, which equals 96500 coulombs. The formula to determine the moles of electrons can be expressed as:

• \[{n} = \frac{F}{z}\]
• Where \( n \) = moles of substance deposited, \( F \) = Faraday constant, and \( z \) = number of electrons required per mole of the substance.

The path a setup takes in depositing material across electrolysis experiments relies on interpreting this straightforward relationship.

Electrode Reactions

In electrolysis, electrode reactions play a crucial role as they determine which substances are deposited or dissolved at each electrode. Each setup involves two electrodes: the anode, where oxidation occurs, and the cathode, where reduction takes place.

To understand how they work, consider the reactions at the cathode for some common ions:

• Silver ions, \( \mathrm{Ag}^+ \), receive one electron to form metallic silver: \[\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}\]
• Tin ions, \( \mathrm{Sn}^{4+} \), need four electrons to become tin metal: \[\mathrm{Sn}^{4+} + 4e^- \rightarrow \mathrm{Sn}\]
• Copper ions, \( \mathrm{Cu}^{2+} \), require two electrons to be reduced into copper metal: \[\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\]

The efficiency of these reactions is determined by the amount of current passed and the time allowed for the current to flow, which directly affects the mass of material deposited.