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 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 X?

Short answer

The name of the salt in the second cell is osmium(IV) nitrate with the formula \(\mathrm{Os(NO}_{3}\mathrm{)_4}\). The electron configuration of element X, which we identified as copper (Cu), is [Ar] 3d10 4s1.

Step-by-step solution

Calculate the moles of chromium deposited from the first cell

Using the mass of chromium deposited and its molar mass, we can find the moles of chromium deposited: Moles of \(\mathrm{Cr} = \frac{Mass\ of\ \mathrm{Cr}}{Molar\ mass\ of\ \mathrm{Cr}} = \frac{1.15\mathrm{g}}{51.996\mathrm{g/mol}} = 0.02211\ mol\)

Calculate the charge passed through the first cell using Faraday's Law

Faraday's law states that charge = n * F, where n is the moles of electrons and F is Faraday's constant (96,485 C/mol). Charge passed = moles of electrons * F. For three moles of electrons (as \(Cr^{3+}\) gets reduced to \(Cr\)): Charge passed = 0.02211 mol * 3 * 96,485 C/mol = 6,389 C

Calculate moles of osmium deposited in the second cell

Using the mass of osmium deposited and its molar mass, find moles of osmium deposited: Moles of \(\mathrm{Os} = \frac{Mass\ of\ \mathrm{Os}}{Molar\ mass\ of\ \mathrm{Os}} = \frac{3.15\mathrm{g}}{190.23\mathrm{g/mol}} = 0.01656\ mol\)

Determine the oxidation number of osmium using Faraday's Law

Using the charge passed through the cell (6,389 C) and Faraday's Law, we can find the moles of electrons: Moles of electrons = \frac{6,389\mathrm{C}}{96,485\mathrm{C/mol}} = 0.06622\ mol Oxidation number (n) = \(\frac{\mathrm{Moles\ of\ electrons}}{\mathrm{Moles\ of\ Os}} = \frac{0.06622\ mol}{0.01656\ mol} = 4\)

Name the salt in the second cell

Since the osmium has an oxidation number of 4 and is combined with nitrate ions, the name of the salt is osmium(IV) nitrate with the formula \(\mathrm{Os(NO}_{3}\mathrm{)_4}\).

Calculate moles of X deposited in the third cell

Since element X has a 2+ charge, two moles of electrons are needed to deposit one mole of X. Using Faraday's law and the charge passed, we find the moles of X: Moles of X = \(\frac{1}{2} \times \frac{6,389\mathrm{C}}{96,485\mathrm{C/mol}} = 0.03311\ mol\)

Calculate the molar mass of element X

Using the mass of X deposited and the moles of X, find the molar mass of X: Molar mass of X = \(\frac{2.11\mathrm{g}}{0.03311\ mol} = 63.7\mathrm{g/mol}\)

Identify element X and determine its electron configuration

The molar mass of X (63.7 g/mol) corresponds to copper (Cu). The electron configuration of Cu is [Ar] 3d10 4s1.

Key concepts

Faraday's Law

Faraday's Law plays a crucial role in understanding how electrochemical reactions occur. Specifically, it helps calculate the amount of substance deposited on the electrode during electrolysis. The law states that the amount of chemical change produced by an electric current is proportional to the quantity of electricity used.The equation \[\text{Charge} = n \times F\]binds the relationship between charge, moles of electrons \(n\), and Faraday's constant \(F\), which is approximately 96,485 coulombs per mole. This means that to deposit a certain amount of substance through electrolysis, you need a specific charge that is dependent on the molar quantity of electrons involved.In our exercise, Faraday's Law helps in calculating the charge needed to deposit varying quantities of metals in the electrochemical cells. For instance, the exercise utilizes it to calculate the charge required to deposit chromium by understanding the relation with the moles of electrons used in the transformation of \(Cr^{3+}\) ions to metallic Cr. This understanding bridges chemistry and electricity, showing the quantitative aspect of chemical changes in electrochemical reactions.

Oxidation Number

The oxidation number is a concept that helps in keeping track of electrons in chemical reactions, especially redox (reduction-oxidation) reactions. It indicates the total number of electrons that an atom gains, loses, or shares when it forms chemical bonds. Each element in a chemical compound has an oxidation number, which can be positive, negative, or zero. The rules for assigning oxidation numbers help in determining which element is oxidized and which is reduced in a chemical reaction. In the exercise above, by using Faraday's Law, the oxidation number of osmium is determined as 4. This means Os loses four electrons in forming a compound, here with nitrate ions to form osmium(IV) nitrate. Understanding oxidation numbers is crucial in predicting how elements will react and the kinds of compounds they form.

Electron Configuration

Electron configuration is the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It describes where the electrons are around the nucleus and helps predict the chemical, magnetic, and electrical properties of substances.For example, the electron configuration of copper (Cu), which was identified as element X in the exercise, is \[[\text{Ar}]\ 3d^{10}\ 4s^1\].This configuration indicates that copper has 29 electrons, with the 3d orbital filled before the 4s in its ground state, which is common for transition metals.This concept allows chemists to understand and predict an element's behavior in chemical reactions. With the correct electron configuration, one can infer the reactivity and bonding nature of the element. Electron configuration also explains periodic properties and trends in the periodic table.

Moles Calculation

Understanding moles and their calculations is foundational in chemistry, connecting macroscopic measurements to the atomic scale. A mole denotes Avogadro's number, \(6.022 \times 10^{23}\),of items, typically atoms or molecules, providing a bridge between atoms and grams.Calculating moles involves using a substance's mass and molar mass in the equation:\[\text{Moles} = \frac{\text{Mass}}{\text{Molar\ mass}}\] In our exercise, this calculation helps determine the quantities of chromium, osmium, and element X deposited in electrochemical cells. For example, the mass of chromium deposited was used to calculate moles:\[\frac{1.15\ \text{g}}{51.996\ \text{g/mol}}\]providing 0.02211 moles of chromium.Accurate moles calculation is essential because it sets the stage for other calculations, including those involving Faraday's Law and determination of oxidation states. It translates the amount of a substance into tangible chemical quantities, necessary for practical experimentation and theoretical computations.