Question

Calculate the quantity indicated for each of the following electrolyses. (a) the mass of \(\mathrm{Zn}\) deposited at the cathode in 42.5 min when 1.87 A of current is passed through an aqueous solution of \(\mathrm{Zn}^{2+}\) (b) the time required to produce \(2.79 \mathrm{g} \mathrm{I}_{2}\) at the anode if a current of \(1.75 \mathrm{A}\) is passed through \(\mathrm{KI}(\mathrm{aq})\)

Short answer

The mass of \(Zn\) deposited at the cathode in 42.5 min when 1.87 A of current is passed through an aqueous solution of \(Zn^{2+}\) ions is approximately 3.24 g. The time required to produce 2.79 g of \(I_2\) at the anode if a current of 1.75 A is passed through \(KI(aq)\) is approximately 10.1 min.

Step-by-step solution

Understanding Faraday's Laws

The conversion of charge to moles of electrons can be done using Faraday’s Laws of electrolysis. According to the first law of Faraday, the mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the solution. The formulas derived from Faraday's law that will be used in this exercise are \(Q = I \times t\) to calculate charge, where Q is the total electric charge, I is the current, and t is time; and \(Q = n \times F\), where n is the amount of substance produced (in moles) and F is Faraday's constant, approximately equal to 96485.332 C/mol.

Calculating Mass of \(Zn\) in Case (a)

We first calculate the total electric charge passed through the solution using the formula \(Q = I \times t\), where I is 1.87 A and t is 42.5 min converted to seconds, since 1 min equals 60 seconds. This gives us \(Q = 1.87 \, A \times 42.5 \, min \times 60 \, sec/min = 4785 \, C\). Then we calculate the number of moles of \(Zn^{2+}\) ions reduced using the formula \(Q = n \times F\), which rearranges to \(n = Q / F\). This gives us the moles of \(Zn\) deposited, \(n = 4785 \, C / 96485.332 \, C/mol = 0.0496 \, mol\). Then we multiply the moles of \(Zn\), \(n = 0.0496 \, mol\), with the molar mass of \(Zn\), 65.38 g/mol, to obtain the mass of \(Zn\), which is \(m = 0.0496 \, mol \times 65.38 \, g/mol = 3.24 \, g\).

Calculating Time in Case (b)

We first need to find the number of moles of \(I_2\) produced. We do this by dividing the mass, 2.79 g, by the molar mass of \(I_2\), 253.8 g/mol. This gives us \(n = 2.79 \, g / 253.8 \, g/mol = 0.011 \, mol\). To get the charge, we use Faraday's constant and multiply it by the number of moles, \(Q = n \times F\). This gives us \(Q = 0.011 \, mol \times 96485.332 \, C/mol = 1061.34 \, C\). Finally, to get the time, we divide the total charge by the current, \(t = Q / I\), where I is 1.75 A. This gives \(t = 1061.34 \, C / 1.75 \, A = 606 \, sec\). As the final answer makes more sense in minutes, we convert it back to minutes by dividing by 60, thus \(t = 606 \, sec / 60 = 10.1 \, min\).

Key concepts

Electrochemical Calculations

Electrochemical calculations are essential for understanding the process of electrolysis, a chemical reaction driven by an external electric current. To perform these calculations, one needs to establish the relationship between the amount of substance reacted and the electric current applied. This relationship is grounded in Faraday's Laws of electrolysis. The core formula used here is \( Q = I \times t \), where \( Q \) is the total electric charge in coulombs (C), \( I \) is the current in amperes (A), and \( t \) is the time in seconds (s) during which the current flows. For example, if you pass a current of 1.87 A for 42.5 minutes through a solution, you would first convert the time to seconds and calculate the total charge using this formula. Understanding this relationship allows for the calculation of the mass of substances deposited at the electrodes during electrolysis.

Another vital part of electrochemical calculations involves using the second part of the formula \( Q = n \times F \), combining the calculated electric charge with Faraday's constant, approximately 96485.332 C/mol, to find the number of moles \( n \) of electrons transferred. The practical application of these formulas was demonstrated in the given exercise, which guided you through the calculation of mass and time based on the charge passed through the solution.

Molar Mass

Molar mass, the mass of one mole of a substance, is a fundamental concept not just in stoichiometry but also in electrochemical calculations where it connects the microscopic world of atoms and molecules to the macroscopic world that we can measure. The molar mass is usually expressed in grams per mole (g/mol) and can be found on the periodic table as the atomic or molecular weight. For instance, zinc (\( Zn \) has a molar mass of 65.38 g/mol.

In the exercise, the molar mass allows us to convert the moles of \( Zn \) deposited (\( n = 0.0496 \) mol) into a tangible mass (\( m = n \times \text{molar mass of } Zn \) equivalent to mass in grams. This conversion is fundamental in being able to conduct electrochemical calculations that result in practical, measurable amounts of substances, showing the real-world application of chemical principles.

Electric Charge

The Role of Electric Charge in Electrolysis

Electric charge is central to the process of electrolysis, acting as the driving force that enables the chemical reactions at the electrodes. Charge, denoted by \( Q \), is measured in coulombs (C) and is a product of current (in amperes, A) and time (in seconds, s). For electrolytic processes, it's essential to determine the total charge that passes through the electrolyte as it's directly related to the amount of substance involved in the reaction at each electrode.

In our textbook example, for case (a), we calculate the charge that passed through the \( Zn^{2+} \) solution to ultimately determine the mass of \( Zn \) deposited. This demonstrates the quantitative aspect of charge in electrolysis and shows its significant role in practical applications.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In the context of electrolysis, stoichiometry is applied to find out how much product is formed or how much reactant is consumed when a certain amount of electric charge is applied. Stoichiometry involves using balanced chemical equations to determine these relationships.

For example, in the electrolysis of aqueous \( KI \), stoichiometry helps in determining the quantity of \( I_2 \) produced at the anode when a specific current is applied. By understanding the molar ratios of reactants to products in the balanced equation for the reaction, one can determine the amount of electricity required to produce a desired amount of product, as seen in case (b) of the exercise. Such stoichiometric calculations are invaluable for predicting the outcomes of electrochemical reactions.