Problem 70 - General Chemistry: Principles and Modern Applications

Question

Calculate the quantity indicated for each of the following electrolyses. (a) \(\left[\mathrm{Cu}^{2+}\right]\) remaining in \(425 \mathrm{mL}\) of a solution that was originally \(0.366 \mathrm{M} \mathrm{CuSO}_{4},\) after passage of \(2.68 \mathrm{A}\) for 282 s and the deposition of Cu at the cathode (b) the time required to reduce \(\left[\mathrm{Ag}^{+}\right]\) in \(255 \mathrm{mL}\) of \(\mathrm{AgNO}_{3}(\mathrm{aq})\) from 0.196 to \(0.175 \mathrm{M}\) by electrolyzing the solution between \(\mathrm{Pt}\) electrodes with a current of \(1.84 \mathrm{A}\)

Step-by-Step Solution

Answer

The remaining amount of copper ions in the solution is 0.14775 mol. The time required to reduce the silver ions in the solution from 0.196M to 0.175M is 28126.99 seconds.

Step 1: Calculation of remaining copper ions

The quantity of copper deposited is given by the relation: \( moles_{Cu} = \frac{I*t}{F} \). Hence, \( moles_{Cu} = \frac{2.68*282}{96485} = 0.00780 \, mol \) . The initial amount of copper ions in the solution is given by \( moles_{Cu_{initial}} = conc_{initial} * V \), hence, \( moles_{Cu_{initial}} = 0.366 * 0.425 = 0.15555 \, mol \). Therefore the remaining amount of copper ions is \( moles_{Cu_{remaining}} = moles_{Cu_{initial}} - moles_{Cu} = 0.15555 - 0.00780 = 0.14775 \, mol \)

Step 2: Calculation of required time to reduce silver ions

The moles of silver ions to be reduced is given by \( moles_{Ag} = (conc_{initial} - conc_{final}) * V \). Hence, \( moles_{Ag} = (0.196 - 0.175) * 0.255 = 0.005355 \, mol \). Thus the time required is given by \( t = \frac{moles_{Ag} * F}{I} = \frac{0.005355 * 96485}{1.84} = 28126.99 \, s \)

Key Concepts

Faraday's Laws of Electrolysis

Understanding the Faraday's laws of electrolysis is fundamental to solve a wide range of problems in electrochemistry. Faraday's laws allow us to relate the amount of substance altered at an electrode during electrolysis to the amount of electricity used.

The first law states that the amount of chemical change or the mass of a substance liberated during electrolysis is directly proportional to the quantity of electricity (charge) that passes through the electrolyte. This can be represented by the simple formula:
\[ m = (Q/F) \times M \]
where \( m \) is the mass of the substance altered at an electrode, \( Q \) is the total charge in coulombs, \( F \) is Faraday's constant (96485 C/mol), and \( M \) is the molar mass of the substance.

The second law states that for a given quantity of electricity, the amounts of different substances liberated by electrolysis are directly proportional to their chemical equivalent weights.

These principles are applied in exercises to determine quantities such as the remaining concentration of ions in a solution after electrolysis, as seen in our example problem.

Molarity and Volume Relationship

To tackle exercises involving electrolysis, one must comprehend the relationship between molarity (M) and volume (V). Molarity is defined as the number of moles (\( n \)) of solute per liter of solution. Thus, the number of moles can be found using the equation:
\[ n = M \times V \]
where \( V \) must be in liters to adhere to the definition of molarity. Once we know the initial and remaining moles from the electrolysis problem, we can determine the new concentration by rearranging the formula as:
\[ M = \frac{n}{V} \]
This relationship is pivotal in calculating the changes in concentration of a solution after passing a certain amount of electric current through it.

Stoichiometry of Electrolysis

Stoichiometry plays a key role in electrolysis calculations. It involves the quantitative relationships between reactants and products in chemical reactions, and in the context of electrolysis, the electrons transferred during the process.

In electrolytic reactions, the stoichiometric coefficients show the number of electrons involved in the reduction or oxidation reactions at the electrodes, and this must be considered alongside Faraday's laws. For instance, copper ions (\( Cu^{2+} \)) require two electrons to be reduced to copper metal. So we must account for this in our calculations of moles of copper deposited during the electrolysis process from the textbook problem.

The adjustment for stoichiometry is implicitly accounted for in the use of Faraday's constant—each mole of electrons has a charge equal to the value of Faraday's constant, and the stoichiometry determines how many moles of electrons are needed per mole of substance involved.

Electrochemical Cell Calculations

Electrochemical cell calculations involve determining various aspects of electrolytic processes, such as the quantity of electricity needed to produce a certain amount of substance at an electrode, or the duration required for a given electrolytic reaction to occur.

Using Faraday's laws, we can calculate the quantity of electricity (\( Q \)) using:\[ Q = n \times F \times z \]where \( n \) is the number of moles of ions, \( F \) is Faraday's constant, and \( z \) is the number of electrons transferred per ion in the reaction. The knowledge of stoichiometry helps us find \( z \). With \( Q \), we can go further to find out other parameters like the current (\( I \)), using \( I = Q/t \), where \( t \) is the time in seconds. This is useful in problems that ask for the time needed to deposit a metal or reduce an ion concentration, as in our second textbook example.