Question

Calculate the weight of copper that will be deposited at the cathode in the electrolysis of a \(0.2 \mathrm{M}\) solution of copper sulphate, when quantity of electricity, equal to the required to liberate \(2.24 \mathrm{~L}\) of hydrogen at STP from a \(0.1 \mathrm{M}\) aqueous sulphuric acid, is passed (Atomic mass of \(\mathrm{Cu}=63.5\) ) (a) \(6.35 \mathrm{~g}\) (b) \(3.17 \mathrm{~g}\) (c) \(12.71 \mathrm{~g}\) (d) \(63.5 \mathrm{~g}\)

Short answer

(a) 6.35 g

Step-by-step solution

Determine electricity for hydrogen liberation

First, calculate the number of moles of hydrogen gas liberated at STP. Given that 1 mole of any gas at STP occupies 22.4 L, if 2.24 L of hydrogen is liberated, the number of moles of hydrogen is given by \( \text{n(H}_2\text{)} = \frac{2.24 \text{ L}}{22.4 \text{ L/mol}} = 0.1 \text{ mol}. \) Each mole of \text\{H}_2\ requires 2 moles of electrons for liberation, thus the total moles of electrons needed are \( 2 \times 0.1 = 0.2 \text{ mol}. \) In terms of Faraday units, \( 0.2 \text{ mol} \) of electrons is equivalent to \( 0.2 \times 96485 = 19297 \text{ C}.\)

Calculate electricity needed for copper deposition

Copper is deposited using the equation \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), meaning 1 mole of copper requires 2 moles of electrons. As the amount of electricity used to deposit copper is the same as that used for hydrogen liberation, i.e., 0.2 mol electrons, we use this quantity to determine copper deposition.

Determine moles of copper deposited

Given that 1 mole of copper requires 2 mol of electrons, the amount of electricity (0.2 mol of electrons) corresponds to \( \frac{0.2}{2} = 0.1 \text{ mol} \) of copper deposited.

Calculate the weight of the deposited copper

The weight of copper deposited is calculated using its atomic mass: \( \text{Weight of Cu} = 0.1 \text{ mol} \times 63.5 \text{ g/mol} = 6.35 \text{ g}.\)

Key concepts

Copper Deposition

Copper deposition is a fascinating aspect of electrochemistry where copper ions (\( \text{Cu}^{2+} \)) in a solution are reduced to solid copper (\( \text{Cu} \)) at the cathode during electrolysis. This process involves the transfer of electrons from the cathode to the copper ions, allowing the pure metal to form. It's widely used in industries for electroplating and purifying copper.

• The cathode is typically made of a conductive material that attracts positive ions.
• As electrons flow towards the cathode, they reduce the copper ions to metallic copper.
• This deposited copper forms a layer on the cathode's surface.

Understanding copper deposition involves recognizing that each copper ion requires two electrons to be reduced to copper metal. This reduction process is essential in various applications, such as improving the electrical conductivity of electronic components. Overall, copper deposition is a crucial technique used in both industrial and laboratory settings.

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis provide foundational understanding of how much substance is deposited during electrolytic processes. These laws link the amount of electrical charge passed through the electrolyte to the quantity of substance deposited or dissolved.
Faraday's First Law states: The mass of the substance deposited or dissolved at an electrode in an electrochemical cell is directly proportional to the total electric charge (\( Q \)) passing through the substance. This means:\[ m = Z \cdot Q \]where \( m \) is the mass of the substance, and \( Z \) is the electrochemical equivalent of the substance.
Faraday's Second Law introduces the concept that different substances have different electrochemical equivalents due to their unique atomic weights and charges required for deposition. This means:\[ Q = n \cdot F \]where \( n \) is the number of moles of electrons, and \( F \) represents Faraday's constant (\( 96485 \, \text{C/mol} \)).In essence, these laws enable precise calculations for determining how much of a substance is deposited during electrolysis based on the known current and time.

Molar Volume at STP

At Standard Temperature and Pressure (STP), the molar volume of an ideal gas is a well-understood concept crucial for calculations in electrochemistry. STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm (101.325 kPa). The molar volume concept tells us that 1 mole of any ideal gas occupies 22.4 liters under these conditions. This knowledge is particularly useful when dealing with gases in electrochemical reactions, such as in the measurement of hydrogen or oxygen volumes produced during electrolysis.
Key points about molar volume at STP:

• It simplifies conversion between volume and moles for gas-phase reactions.
• Enables calculations such as determining the amount of hydrogen liberated during electrolysis.
• Acts as a baseline for understanding gaseous behavior and properties in various conditions.

This concept is vital in accurately determining the quantities needed for reactions and ensuring precise outcomes in chemical processes.

Electrochemistry Calculations

Electrochemistry calculations are essential for predicting the outcomes of electrochemical reactions, such as electrolysis. These calculations involve using basic electrochemical principles to determine quantities like the amount of substance deposited, consumed, or the charge passing through the system.Key elements of electrochemistry calculations include:

• Understanding the stoichiometry of the reaction, like how many moles of a substance react or are formed.
• Using Faraday's laws to relate the charge passed to the amount of a substance deposited or dissolved.
• Employing constants such as Faraday's constant (\( 96485 \, \text{C/mol} \)) for calculations.

These calculations are central to various applications, from industrial processes to laboratory experiments, where precision is crucial. Electrochemistry calculations provide insight into reaction dynamics and efficiency, ensuring optimal operation conditions in processes like copper deposition or hydrogen generation.