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Chapter 18: Problem 133

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0\) min. If \(2.618 \mathrm{~g}\) Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

Short Answer

Expert verified
The charge on the ruthenium ions in solution can be calculated by following these steps: 1. Calculate the total charge transferred during the electrolysis: Total charge (Q) = Current (I) 脳 Time (t) = \(2.50 A \times 3000 s = 7500 C\). 2. Calculate the number of moles of electrons transferred: Number of moles of electrons (n) = Total charge (Q) / Faraday constant (F) = \(7500 C / 96485 C/mol = 0.0777 mol\). 3. Calculate the number of moles of ruthenium produced at the cathode: Number of moles of Ru = Mass of Ru produced (m) / Molar mass of Ru (M) = \(2.618 g / 101.07 g/mol = 0.0259 mol\). 4. Determine the charge on the ruthenium ions (z) in solution: Charge on ruthenium ions (z) = Number of moles of electrons (n) / Number of moles of Ru = \(0.0777 mol / 0.0259 mol = 3\). Thus, the charge on the ruthenium ions in solution is \(3+\).

Step by step solution

01

Calculate the number of electrons transferred in total

We need to find out how many electrons are transferred in total throughout the electrolysis process. We can do this by calculating the total charge transferred during the electrolysis using the formula: Total charge (Q) = Current (I) 脳 Time (t) where I = 2.50 A and t = 50 min = 50 脳 60 s = 3000 s Total charge (Q) = 2.50 A 脳 3000 s Calculate the value of Q.
02

Calculate the number of moles of electrons transferred

Now, let's find the number of moles of electrons (n) transferred by dividing the total charge by the Faraday constant (F), which is approximately \(96485\) coulombs per mole of electrons. Number of moles of electrons (n) = Total charge (Q) / F Calculate the value of n using Q from Step 1.
03

Calculate the number of moles of ruthenium

To calculate the number of moles of ruthenium (Ru) produced at the cathode, we use the formula: Number of moles of Ru = Mass of Ru produced (m) / Molar mass of Ru (M) where m = 2.618 g and M = 101.07 g/mol (molar mass of ruthenium) Number of moles of Ru = 2.618 g / 101.07 g/mol Calculate the value of number of moles of Ru.
04

Determine the charge on the ruthenium ions

Now that we have the number of moles of electrons and ruthenium, we can find the charge on the ruthenium ions (z) in solution using the following formula: Charge on ruthenium ions (z) = Number of moles of electrons (n) / Number of moles of Ru Calculate the value of z using n from Step 2 and the number of moles of Ru from Step 3. The result will give you the charge on the ruthenium ions in solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Current
Current is the flow of electric charge, typically measured in amperes (A). In the context of electrolysis, current reflects the number of charged particles, like electrons, that pass through the solution over time.
It plays a crucial role because it determines the total amount of charge that participates in the reaction.
During electrolysis, the current applied helps drive the movement of ions in the solution, which leads to the formation of new substances.
  • Formula: Total charge (\( Q \)) = Current (\( I \)) 脳 Time (\( t \)).
  • Example in Practice: If you apply a current of 2.50 A for 50 minutes, it's key to calculate in seconds (\( 50 \times 60 \)) to align with standard units.
This concept helps understand the role of electricity in converting ions into solid elements in a controlled manner.
Charge
Charge refers to the quantity of electricity transported through the solution. It is measured in coulombs (C). The charge passed through during electrolysis is a vital parameter because it affects the extent of the reaction.
Consider the formula given by:
  • Total Charge: Calculated using the current and time, \( Q = I \times t \).
  • Impact: It determines how many moles of substance are formed or consumed at the electrodes.
Understanding charge helps predict how much of a material will get deposited or dissolved, which is especially useful in industrial applications.
Faraday constant
The Faraday constant (\( F \)) is a crucial value used in electrolysis to connect the amount of electric charge to moles of electrons. It is approximately equal to 96485 coulombs per mole of electrons.
This constant allows us to transition from the realm of charge to the molar aspect, which combines with other chemical data for precise calculations.
  • The Fundamental Link: Describes how many electrons鈥攐r rather, how much charge鈥攊s required to deposit or dissolve a mole of ions.
  • Practical Use: In calculations, using the Faraday constant helps determine the moles of electrons exchanged during the reaction based on measured charge (\( n = Q / F \)
The Faraday constant bridges physics and chemistry, facilitating calculations that underpin many practical electrochemical processes.
Moles of electrons
The concept of moles of electrons refers to the amount of electrons involved in the electrolysis reaction. It is typically found using the total charge and the Faraday constant.
By calculating moles of electrons, you can understand how many electrons were exchanged during the process.
This aspect directly correlates with how many atoms of a material like ruthenium will be deposited or dissolved.
  • Calculation: \( n = \text{Total Charge } (Q) / \text{Faraday Constant } (F) \).
  • Role in Electrolysis: Helps identify the stoichiometry of the reaction and predict product formation.
Accurate knowledge of moles of electrons is essential for precise quantitative electrochemical analysis. It ultimately defines the efficiency and magnitude of chemical changes achieved through electrolysis.

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Most popular questions from this chapter

Consider only the species (at standard conditions) $$ \mathrm{Br}^{-}, \mathrm{Br}_{2}, \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd} $$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s)\) ?

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Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous \(\mathrm{NO}\). Write a balanced equation for the dissolution of gold in aqua regia.

The measurement of \(\mathrm{pH}\) using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$ \mathscr{E}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH} $$ where \(\mathscr{E}_{\text {ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{~V}\) and that \(\mathscr{B}_{\text {meas }}=0.480 \mathrm{~V}\). a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}\) \((\pm 0.001 \mathrm{~V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?
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