Question

Three electrochemical cells were connected in series so that the same quantity of electrical current passes through all three cells. In the first cell, 1.15 g chromium metal was deposited from a chromium (III) nitrate solution. In the second cell, 3.15 \(\mathrm{g}\) osmium was deposited from a solution made of \(\mathrm{Os}^{n+}\) and nitrate ions. What is the name of the salt? In the third cell, the electrical charge passed through a solution containing \(\mathrm{X}^{2+}\) ions caused deposition of 2.11 \(\mathrm{g}\) metallic \(\mathrm{X}\) . What is the electron configuration of \(\mathrm{X} ?\)

Short answer

The name of the salt in the second cell is osmium (IV) nitrate (Os(NO3)4), and the electron configuration of the metal X in the third cell is [Rn]5f^46d^17s^2, which corresponds to neptunium (Np).

Step-by-step solution

Calculate moles of electrons transferred in the first cell

We know that the same quantity of electric current passes through all three cells. The number of moles of electrons transferred (n) can be calculated using Faraday's law: \(n = \frac{m}{M \times F}\) Where, - m is the mass of chromium deposited (1.15g) - M is the molar mass of chromium (51.996 g/mol) - F is the Faraday constant (96,485 C/mol) Therefore, the moles of electrons transferred (n) in the first cell are: \(n = \frac{1.15}{51.996 \times 96,485} \) Calculating n, we get: \( n = 2.31 \times 10^{-5} \) moles of electrons

Find the oxidation state of osmium and the name of the salt in the second cell

Since the same quantity of electric current passes through all three cells, the number of electron moles in the second cell will be equal to that in the first cell. Therefore, we have "n" moles of electrons for osmium (OsN) deposited: \(n_{Os} = 2.31 \times 10^{-5}\) Now, we can find the oxidation state of osmium (N): \(Mass = n \times M \times F \) Where, - Mass is the mass of osmium deposited (3.15g) - M is the molar mass of osmium (190.23 g/mol) \(3.15 = 2.31 \times 10^{-5} \times 190.23 \times F \) Calculating F, we get: F = 365.17 C/mol Since F is the product of osmium's charge (n+) and the Faraday constant, dividing F by the Faraday constant will give us the oxidation state of osmium: \(Oxidation\: state\: of\: osmium = \frac{365.17}{96,485} \approx 4 \) Therefore, the oxidation state of osmium is 4, and the name of the salt is osmium (IV) nitrate (Os(NO3)4).

Find the electron configuration of the metal X in the third cell

Again, since the same quantity of electric current passes through all three cells, the number of electron moles in the third cell will be equal to that in the first cell. Therefore, we have "n" moles of electrons for metal X deposited: \( n_{X} = 2.31 \times 10^{-5} \) We can find the molar mass of metal X: \( Mass = n \times M \times F \) Where, - Mass is the mass of metal X deposited (2.11g) \( 2.11 = 2.31 \times 10^{-5} \times M \times 96,485 \) Calculating the molar mass of metal X, we get: M = 93.31 g/mol The molar mass (M) of metal X is close to the atomic mass of Neptunium (Np) which is around 93 g/mol. The electron configuration of neptunium is [Rn]5f^46d^17s^2, and that's the electron configuration of metal X.

Key concepts

Faraday's Law

Faraday's Law is a fundamental principle in electrochemistry. It relates the amount of substance transformed at an electrode to the electricity passed through the electrolyte. Specifically, the law states that the amount of substance deposited or dissolved at each electrode is directly proportional to the amount of charge that goes through the cell.
In the given problem, Faraday’s Law is used to calculate the number of moles of electrons transferred. This is key to understanding how the same current affects all three cells. Using the formula, \( n = \frac{m}{M \times F} \), where \( m \) is the mass deposited, \( M \) is the molar mass, and \( F \) is the Faraday constant (approximately 96,485 C/mol).
The moles of electrons calculated establish the relationship between deposited metal mass and transferred charge, helping us explore further properties like oxidation states and configurations.

Electron Configuration

Electron configuration describes the distribution of electrons among the orbitals of an atom. It helps in identifying the chemical properties of the element by revealing electron arrangement.
For metal X in the electrochemical cells, the electron configuration is crucial as it connects the element's identity to its mass and the number of electrons transferred. In this exercise, we determined the element to be Neptunium (Np) with the configuration \([Rn]5f^46d^17s^2\).
Understanding electron configuration allows us to assess how the electrons fill sublevels and orbitals, signifying the element's placement within the periodic table and predicting its chemical behavior.

Oxidation State

The oxidation state indicates the degree of oxidation of an atom within a compound. It represents the electron ownership among atoms within a molecule, often affecting the type of chemical reactions that can occur.
In the exercise, identifying the oxidation state of osmium is integral because it determines the nature of the salt. We calculated it to be +4, giving rise to osmium (IV) nitrate. The oxidation state provides insight into electron transfer processes that occur during electrochemical reactions, highlighting how oxidation contributes to chemical transformations.
Such understanding informs predictions about reactivity and helps in balancing redox equations, vital for any chemical analysis.

Molar Mass

Molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). It is a critical property that links the amount of substance to the mass measured during reactions.
In this scenario, molar mass was used to help identify metal X as Neptunium. By using the measured mass from the deposition and the moles of electrons, the molar mass was found to be approximately 93.31 g/mol. This value matched closely with the known atomic mass of Neptunium.
Knowledge of molar mass allows chemists to convert between grams and moles easily, facilitating the stoichiometric calculations necessary for predicting reaction yields and compositions.