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) \(\mathrm{CuSO}_{4}\). 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

Option (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 directly proportional to the number of electrons added or removed, which is determined by the charge passed. One Faraday ( abla 96485 C/mol) of charge will deposit one mole of electrons.

Step 2: Calculating Moles of Ag from AgNO3

In the electrolysis of AgNO3, silver is deposited according to the reaction: \[ \text{Ag}^+ + \text{e}^- \rightarrow \text{Ag} \]Since the charge of the silver ion is +1, one mole of electrons (1 Faraday) will deposit 1 mole of Ag.

Step 3: Calculating Moles of Sn from SnCl4

For SnCl4, the cathodic reaction is:\[ \text{Sn}^{4+} + 4\text{e}^- \rightarrow \text{Sn} \]It takes 4 Faradays of charge to deposit 1 mole of Sn. Therefore, with 1 Faraday, you will deposit:\[ \frac{1}{4} \text{ mole of Sn} = 0.25 \text{ mole} \]

Step 4: Calculating Moles of Cu from CuSO4

In CuSO4, copper is deposited following the equation:\[ \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \]Thus, it requires 2 Faradays to deposit 1 mole of Cu, so with 1 Faraday, 0.5 moles of Cu is deposited:\[ \frac{1}{2} \text{ mole} = 0.5 \text{ mole} \]

Step 5: Confirming the Correct Option

Summarizing the moles deposited: Ag = 1 mole, Sn = 0.25 mole, Cu = 0.5 mole. The correct choice is (a) 1.0, 0.25, 0.5.

Key Concepts

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis is a fundamental principle in electrochemistry. It tells us that the amount of a substance that gets deposited or dissolved during electrolysis is directly proportional to the number of electrons that flow through the system. This "number of electrons" is actually amount of charge, typically large.**Understanding Faraday's Constant**
This law is effectively captured by Faraday's constant, noted as approximately 96485 Coulombs per mole of electrons. This constant represents the charge carried by one mole of electrons and forms the basis for calculations in electrolysis.
- **Direct Proportionality**: The more charge (or electricity) you pass through an electrolytic cell, the more substance you will deposit or dissolve.- **Equation Formulation**: Using Faraday's Law, you can use the formula: \[ n = \frac{Q}{Fz} \] where \( n \) is the number of moles of substance, \( Q \) is the total charge in Coulombs, \( F \) is Faraday's constant, and \( z \) is the number of electrons involved in the reaction for each ion.
By understanding this concept, it is easy to see how 1 Faraday correlates with the deposition of a different number of moles, depending on the charge of the ions involved in the reaction.

Electrode Reactions

In the electrolysis process, reactions occur at electrodes: cathode and anode. The cathode attracts cations and is the site of reduction (gain of electrons), while the anode attracts anions and is where oxidation (loss of electrons) occurs.**Cathodic Reactions Explained**
For the given problem, each metallic ion in the solutions undergoes a reduction reaction at the cathode:

• AgNO₃: The reaction at the cathode is \( \text{Ag}^+ + \text{e}^- \rightarrow \text{Ag} \), showing that one electron is needed to reduce one silver ion, depositing 1 mole of Ag per mole of electrons.
• SnCl₄: Here, \( \text{Sn}^{4+} + 4\text{e}^- \rightarrow \text{Sn} \) highlights that four electrons are required to reduce one tin ion. Thus, 0.25 moles of tin are deposited by 1 Faraday.
• CuSO₄: The reduction is \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \), needing two electrons per copper ion, which means 0.5 moles of Cu are deposited per Faraday.

**Key Takeaways**
- Each ion type in a solution has different electron demands which dictates how many moles of the metal are deposited at the cathode.- Understanding these reactions helps predict the outcome of electrolysis for metals.

Stoichiometry in Electrolysis

Stoichiometry plays a significant role in electrolysis by helping to calculate how many moles of substance are altered by a given charge. This is based on the concept of mole ratios from balanced chemical equations related to electrode reactions.**Driving the Stoichiometric Calculations**
Stoichiometry relates the moles of reactants to products using the charges involved at the electrodes. Here’s how it relates to each reaction:

- **For AgNO₃, 1:1 Mole Ratio**:
The reaction \( \text{Ag}^+ + \text{e}^- \rightarrow \text{Ag} \) means one mole of electrons reacts with one mole of silver ions, making it simple to calculate the moles of silver deposited.- **For SnCl₄, 4:1 Mole Ratio**:
The reaction \( \text{Sn}^{4+} + 4\text{e}^- \rightarrow \text{Sn} \) involves a 4:1 ratio, so four moles of electrons are needed for each mole of tin deposited. Passing 1 Faraday (or one mole of electrons) deposits a quarter of a mole of tin.- **For CuSO₄, 2:1 Mole Ratio**:
For copper, the \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \) reaction indicates two moles of electrons interact with each copper ion. Thus, 1 Faraday results in half a mole of copper.**Easy Predictability**
By using these mole ratios derived from balanced equations, you can easily forecast product amounts for any deeply engaged electrolysis task, making stoichiometry an essential tool in electrochemistry.