Question

Two faraday of electricity is passed through a solution of \(\mathrm{CuSO}_{4}\). The mass of copper deposited at the cathode is (at. mass of \(\mathrm{Cu}=63.5 \mathrm{amu}\) ) (a) \(0 \mathrm{~g}\) (b) \(63.5 \mathrm{~g}\) (c) \(2 \mathrm{~g}\) (d) \(127 \mathrm{~g}\)

Short answer

The mass of copper deposited is 127 g, option (d).

Step-by-step solution

Understanding Faraday's Laws of Electrolysis

Faraday's First Law of Electrolysis states that the mass of a substance deposited at an electrode during electrolysis is directly proportional to the amount of electricity that passes through the solution. The formula is given by: \(m = \frac{Q}{F} \times \frac{M}{z}\), where \(m\) is the mass of the substance deposited, \(Q\) is the total charge, \(F\) is Faraday's constant \((96500 \, C/mol\)), \(M\) is the molar mass of the substance, and \(z\) is the number of electrons exchanged per ion.

Calculate Total Charge Passed

Given that two Faradays of electricity are passed, we calculate the total charge \(Q\) using the relation between moles of electrons and charge: \(Q = nF = 2 \text{ Faradays} \times 96500 \, C/mol = 193000 \, C\).

Determine the Number of Electrons Exchanged

For copper \((\mathrm{Cu}^{2+})\), 2 moles of electrons are required to deposit 1 mole of copper according to the half-reaction: \(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\). Thus, \(z = 2\).

Calculate the Mass of Copper Deposited

Using Faraday's formula \(m = \frac{Q}{F} \times \frac{M}{z}\):\[m = \frac{193000 \, C}{96500 \, C/mol} \times \frac{63.5 \, g/mol}{2} = 127 \, \mathrm{g}\]Thus, 127 g of copper is deposited at the cathode.

Key concepts

Faraday's First Law of Electrolysis

Faraday's First Law of Electrolysis is a fundamental principle in electrochemistry. It tells us that the amount of a substance deposited at an electrode during electrolysis is directly linked to the quantity of electric charge passed through the electrolyte. This is a crucial understanding for predicting how much of a particular substance will be produced or consumed in an electrochemical cell.

The law is mathematically expressed as:

\( m = \frac{Q}{F} \times \frac{M}{z} \)

Here, \(m\) represents the mass of the substance deposited. \(Q\) is the total electric charge passed through the solution, measured in coulombs. \(F\) is Faraday’s constant, approximately 96500 coulombs per mole. This constant represents the charge of one mole of electrons. Lastly, \(M\) is the molar mass of the substance, and \(z\) denotes the valency or the number of electrons involved in the electrode reaction for each ion.

Understanding this formula helps in predicting the mass of a substance that gets deposited based on the electric charge supplied, enabling better control and efficiency in electrochemical processes.

Molar Mass and its Role in Electrolysis

Molar mass is a vital concept in understanding the outcomes of electrolysis processes. It represents the mass of one mole of a substance, often expressed in grams per mole (g/mol). This information is indispensable when calculating how much of a metal or other substance will be deposited at an electrode during electrolysis.

In our exercise, a copper sulfate solution undergoes electrolysis, where copper (Cu) acts as the metal in focus. Copper has a molar mass of 63.5 g/mol. When using Faraday’s First Law of Electrolysis, the molar mass plays a crucial role in determining the mass of copper deposited. The copper ions \((\text{Cu}^{2+})\), which gain electrons and form copper metal during the reaction, participate in this process.

By knowing the copper's molar mass, we can accurately calculate that passing a certain quantity of electricity leads to the deposition of a corresponding mass of copper at the cathode. This connection between molar mass and electrochemical reactions ensures precise predictions and efficient resource use in industrial applications.

Electrode Reactions and Their Importance

Electrode reactions are at the heart of electrolysis, dictating what substances are deposited or dissolved during the process. These reactions occur at the electrodes, which are conductive surfaces where the electric charge interface with the electrolyte.

In the case of the copper sulfate solution, copper ions \((\text{Cu}^{2+})\) are reduced at the cathode, while the sulfate ions remain in the solution. The specific electrode reaction for copper is:

\( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)

Here, copper ions receive two electrons and get deposited as copper metal. This reduction process is balanced by an oxidation process occurring at the anode, which in the case of copper sulfate, can involve oxygen evolution if inert anodes are used.

Recognizing these electrode reactions is crucial because it helps in understanding how materials are transformed and deposited during electrolysis. It also aids in controlling the conditions to optimize desired results, whether that be plating metals, purifying substances, or manufacturing chemicals. The reactions provide insights into the efficiency and feasibility of electrolysis for different applications.