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) \(0.25,0.25,0.5\)

Short answer

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

Step-by-step solution

Understand Faraday's First Law of Electrolysis

Faraday's First Law of Electrolysis states that the amount of substance deposited at an electrode is directly proportional to the quantity of electricity (in coulombs) passed through the electrolyte. Mathematically, it's given by \( m = \frac{Q}{nF} \) where \( m \) is the mass of the substance deposited, \( Q \) is the charge in coulombs, \( n \) is the number of electrons exchanged per ion, and \( F \) is Faraday's constant (approximately 96500 C/mol). For one faraday, \( Q = F \), therefore \( m = \frac{1}{n} \).

Determine the Valency (n) for Each Solution

For each compound, determine how many electrons are involved in the reduction process at the cathode:- \( \text{AgNO}_3 \): Ag\(^+\) + e\(^-\) \rightarrow \text{Ag} \Rightarrow n = 1- \( \text{SnCl}_4 \): Sn\(^{4+}\) + 4e\(^-\) \rightarrow \text{Sn} \Rightarrow n = 4- \( \text{CuSO}_4 \): Cu\(^{2+}\) + 2e\(^-\) \rightarrow \text{Cu} \Rightarrow n = 2

Calculate the Moles of Metal Deposited

Using the formula from Step 1, calculate the moles for each metal:- For Ag: \( \text{moles} = \frac{1}{1} = 1.0 \)- For Sn: \( \text{moles} = \frac{1}{4} = 0.25 \)- For Cu: \( \text{moles} = \frac{1}{2} = 0.5 \)

Match Calculation with Options

Compare the calculated moles: Ag = 1.0, Sn = 0.25, Cu = 0.5, which matches the option (a). The correct sequence is option (a): \(1.0, 0.25, 0.5\).

Key concepts

Electrochemistry

Electrochemistry is a branch of chemistry that studies the relationship between electrical energy and chemical reactions. It is essentially about how electricity can cause chemical changes, and vice-versa. In electrochemistry, reactions primarily occur at the interface between an electrode and an electrolyte. This field is crucial because it allows us to capture or release energy from chemical changes.
Electrochemistry is fundamental for a host of applications, such as batteries, fuel cells, and electrolysis. In electrolysis, electricity is used to drive a chemical reaction that would not occur on its own. This electrochemical process can break down compounds or deposit elements onto electrodes. Understanding how electricity interacts with chemical substances helps in the design of more efficient energy-storing devices and industrial processes.

Valency in Electrolysis

Valency is a measure that indicates the number of electrons an atom can gain, lose, or share. In electrolysis, understanding valency is critical because it tells us how many electrons are involved in a particular reaction. Each ion undergoing electrolysis will either release or accept electrons, determining how much substance is produced at an electrode.
For example:

• Silver in (\(\text{AgNO}_3\)) has a valency of 1, meaning each ion needs one electron to become metallic silver.
• Tin in (\(\text{SnCl}_4\)) requires four electrons to turn into metallic tin, hence a valency of 4.
• Copper in (\(\text{CuSO}_4\)) has a valency of 2, needing two electrons to deposit as copper metal.

Understanding the valency helps in calculating how much material will be deposited during electrolysis, allowing predictions about the outcome of the process.

Electrode Reactions

Electrode reactions are the oxidation and reduction reactions that occur at the electrodes during electrolysis. In an electrochemical cell, these reactions are necessary for the flow of electrons, which constitutes the electrical current.
Generally, two types of electrodes are involved:

• The cathode, where reduction happens. It attracts cations and gains electrons.
• The anode, where oxidation occurs. It attracts anions and loses electrons.

In the context of the given solutions:

• For (\(\text{AgNO}_3\)), the reaction at the cathode is (\(\text{Ag}^+ + e^- \rightarrow \text{Ag}\)), depositing silver metal.
• With (\(\text{SnCl}_4\)), the reaction is (\(\text{Sn}^{4+} + 4e^- \rightarrow \text{Sn}\)), producing tin.
• For (\(\text{CuSO}_4\)), (\(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\)) results in copper being deposited.

These electrode reactions are the central events enabling the conversion from electrical to chemical energy, ultimately leading to the deposition of pure metal on the cathode.