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One of the half-reactions for the electrolysis of water is: $$2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g)$$ If \(0.845 \mathrm{~L}\) of \(\mathrm{H}_{2}\) is collected at \(25^{\circ} \mathrm{C}\) and \(782 \mathrm{mmHg}\), how many faradays of electricity had to pass through the solution?

Short Answer

Expert verified
Approximately 0.071 faradays of electricity passed through the solution.

Step by step solution

01

Convert Volume of Gas to Moles

First, we need to calculate the number of moles of hydrogen gas, \(\mathrm{H}_2\), using the ideal gas law equation: \(PV = nRT\). To do this, convert all values to SI units. The pressure \(P\) is given as \(782\) mmHg, which can be converted to atm by dividing by \(760\), giving approximately \(1.029\) atm. The volume \(V\) of \(\mathrm{H}_2\) is \(0.845\) L. Recall that \(R\), the gas constant, is \(0.0821\,\text{L atm mol}^{-1} \text{K}^{-1}\), and temperature \(T\) is \(25 + 273 = 298\) K. Substitute these values into the equation: \(n = \frac{PV}{RT}\). Calculate \(n\).
02

Calculate Moles of Electrons Required

From the half-reaction, \(2\) moles of electrons are needed for every mole of \(\mathrm{H}_2\) gas produced. Using the moles of \(\mathrm{H}_2\) calculated in Step 1, determine the moles of electrons: \(\text{Moles of electrons} = 2 \cdot \text{Moles of } \mathrm{H}_2\).
03

Convert Moles of Electrons to Faradays

A Faraday is defined as the charge on one mole of electrons, which is approximately \(96485\) coulombs. Therefore, the number of faradays is equal to the number of moles of electrons. Simply take the moles of electrons calculated in Step 2, which is also equal to the number of faradays since 1 Faraday = 1 mole of electrons.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is an essential concept in chemistry and physics. It is represented by the equation \( PV = nRT \), where \(P\) stands for pressure, \(V\) is volume, \(n\) is the amount in moles, \(R\) is the ideal gas constant (0.0821 L atm mol-1 K-1), and \(T\) is the temperature in Kelvin. This equation is used to relate the physical properties of gases.
To use the Ideal Gas Law, it is crucial to have all units in the correct form. Pressure often needs to be converted to atmospheres, volume in liters, and temperature must always be in Kelvin.
  • To convert pressure from mmHg to atm, divide by 760.
  • Temperature in Celsius should be converted to Kelvin by adding 273.
Understanding this law helps us calculate the number of moles of a gas when given certain conditions, such as the pressure and temperature of hydrogen gas collected during electrolysis.
Faraday's Constant
Faraday's Constant is a fundamental number in electrochemistry, representing the charge of one mole of electrons. It is roughly \(96485\) coulombs per mole. This constant connects the quantities of electric charge and the chemical change in an electrochemical reaction.
When you know the total charge passed through a solution during an electrolysis process, you can use Faraday's Constant to determine the number of moles of electrons transferred, which is vital for calculating the amount of chemical change occurring in the system.
In practical terms, during electrolysis, if you know how many moles of electrons are produced, you can easily calculate how many Faradays correspond to that quantity. Thus, it is crucial in converting electrical information into stoichiometric data for various reactions.
Half-Reaction
A half-reaction is part of a redox (reduction-oxidation) reaction that either involves reduction or oxidation, showing either the loss or gain of electrons. Half-reactions help you understand the electron transfer process separately for reduction and oxidation.
For the electrolysis of water, the half-reaction given is: \(2 \text{H}^{+} (aq) + 2e^{-} \rightarrow \text{H}_{2} (g)\). This particular half-reaction illustrates the reduction process, where hydrogen ions \(\text{H}^{+}\) gain electrons to form hydrogen gas \(\text{H}_{2}\).
Understanding half-reactions is essential for determining how many electrons are involved in the equation and it helps you calculate the moles of electrons required for forming a particular amount of product, like hydrogen gas in this case.
Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate the amounts of reactants and products.
When performing electrolysis, stoichiometry allows us to determine how many moles of electrons are required to produce a certain amount of gas. For instance, if 1 mole of \(\text{H}_{2}\) requires 2 moles of electrons, stoichiometry enables us to calculate the precise quantity of electricity needed.
  • It provides a way to relate the moles of reactants like \(\text{H}^{+}\) ions to the moles of electrons.
  • Helps predict the volume of gas produced at specific conditions.
Thus, stoichiometry is crucial for ensuring that an electrochemical process is efficiently calculated and performed.
Electrochemistry
Electrochemistry is the study of chemical processes that cause the movement of electrons. This branch of chemistry explores the interplay between electrical energy and chemical change. Central to electrochemistry are the concepts of oxidation and reduction as well as the stoichiometry of charged species.
In the electrolysis of water, electrical energy is used to drive a non-spontaneous chemical reaction, splitting water into hydrogen and oxygen gases. This process serves as a practical application of electrochemical principles.
  • It involves the direct conversion of electrical energy into chemical energy.
  • Tools like Faraday's Constant are used to quantify electron transfer in these reactions.
Understanding electrochemistry enables us to design systems like batteries and other electrochemical cells that store and harness chemical energy as practical power sources.

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Most popular questions from this chapter

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

The zinc-air battery shows much promise for electric cars because it is lightweight and rechargeable: The net transformation is \(\mathrm{Zn}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnO}(s)\) (a) Write the half-reactions at the zinc-air electrodes, and calculate the standard emf of the battery at \(25^{\circ} \mathrm{C}\). (b) Calculate the emf under actual operating conditions when the partial pressure of oxygen is 0.21 atm. (c) What is the energy density (measured as the energy in kilojoules that can be obtained from \(1 \mathrm{~kg}\) of the metal) of the zinc electrode? (d) If a current of \(2.1 \times 10^{5} \mathrm{~A}\) is to be drawn from a zinc-air battery system, what volume of air (in liters) would need to be supplied to the battery every second? Assume that the temperature is \(25^{\circ} \mathrm{C}\) and the partial pressure of oxygen is 0.21 atm.

What is a cell diagram? Write the cell diagram for a galvanic cell consisting of an Al electrode placed in a \(1 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution and an Ag electrode placed in a \(1 \mathrm{M} \mathrm{AgNO}_{3}\) solution.

The equilibrium constant for the reaction: $$\operatorname{Sr}(s)+\mathrm{Mg}^{2+}(a q) \rightleftharpoons \mathrm{Sr}^{2+}(a q)+\mathrm{Mg}(s)$$ is \(2.69 \times 10^{12}\) at \(25^{\circ} \mathrm{C}\). Calculate \(E^{\circ}\) for a cell made up of \(\mathrm{Sr} / \mathrm{Sr}^{2+}\) and \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) half-cells.

Given the following standard reduction potentials, calculate the ion-product, \(K_{\mathrm{w}},\) for water at \(25^{\circ} \mathrm{C}:\) $$ \begin{array}{ll} 2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(\mathrm{~g}) & E^{\circ}=0.00 \mathrm{~V} \\ 2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(a q) & E^{\circ}=-0.83 \mathrm{~V} \end{array} $$

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