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Consider the electrolysis of molten barium chloride, \(\mathrm{BaCl}_{2}\). (a) Write the half-reactions. (b) How many grams of barium metal can be produced by supplying 0.50 A for 30 min?

Short Answer

Expert verified
The half-reactions for the electrolysis of molten barium chloride are \nReduction: \(Ba^{2+} + 2e^− \rightarrow Ba\) \nOxidation: \(2Cl^− \rightarrow Cl_{2} + 2e^−\). Given 0.50 A for 30 min, 0.64 g of barium metal can be produced.

Step by step solution

01

Write the half-reactions

First, separate the BaCl2 into its ions: Ba2+ and 2Cl-. Then, write the half-reactions. For the reduction half-reaction at the cathode, Ba2+ gains 2e- to become Ba. For the oxidation half-reaction at the anode, 2Cl- lose 2e- to become Cl2. Therefore, the half-reactions are as follows: \nReduction: \( Ba^{2+} + 2e^- \rightarrow Ba \). \nOxidation: \(2Cl^- \rightarrow Cl_{2} + 2e^-\).
02

Conversion of units

The current 0.50 A is equivalent to 0.50 Coulombs/second. Also, 30 minutes is equivalent to 1800 seconds. Thus, total charge passed Q is equal to current supplied (I) x time (t). Therefore, Q = 0.50 A x 1800 s = 900 Coulombs.
03

Calculate the amount of Barium produced

Using Faraday’s law of electrolysis, the amount of substance produced is proportional to the quantity of electricity supplied. Now, to deposit one mole of Ba, we need 2 moles of electrons according to the half-reactions, which is 2 x 96500 C = 193000 C given that 1 mole of electrons equals 96500 C (Faraday's constant). Therefore, the moles of Ba produced would be 900 C / 193000 C/mol = 0.00466 mol. Knowing that the molar mass of Ba is 137.3 g/mol, the mass of Barium produced would be 0.00466 mol x 137.3 g/mol = 0.64 g.

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Key Concepts

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

Faraday's Law of Electrolysis
Faraday's Law of Electrolysis is a fundamental principle that quantifies how electric charge is used to enact chemical changes during electrolysis. In layman's terms, this law helps us predict exactly how much of a substance will be produced or consumed at an electrode during electrolysis, using the amount of electric current passed through the substance.

The law states that the amount of chemical change is directly proportional to the quantity of electricity that flows through the electrolyte. This relationship depends on two critical factors: the quantity of electricity (measured in coulombs) and a proportionality constant known as the electrochemical equivalent of the substance.
  • 1st Faraday's Law: The mass (m) of the substance altered at an electrode during electrolysis is directly proportional to the quantity of electricity (Q) that passes through the electrolyte. This is stated mathematically as: \( m = EQ \), where E is a constant.
  • 2nd Faraday's Law: The masses of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.

The practical application comes through when we want to calculate the mass of a substance formed during electrolysis, like in the example of molten barium chloride, where we calculate how many grams of barium are deposited by a known quantity of electric charge.
Half-reaction Method
The half-reaction method is an approach to balancing chemical equations used specifically for redox (reduction-oxidation) reactions. This method separates the overall reaction into two half-reactions: one for reduction and one for oxidation.

Each half-reaction involves either the gain or loss of electrons. For the reduction half-reaction, species gain electrons and is said to be reduced. Conversely, the oxidation half-reaction involves the loss of electrons, and the species that loses electrons is said to be oxidized. The key to using the half-reaction method is balancing the number of electrons transferred between the oxidation and reduction parts of the reaction.

Steps to Balance Half-reactions:

  • Separate the overall reaction into two half-reactions.
  • Balance all the atoms except for hydrogen and oxygen.
  • Balance the oxygen atoms by adding water molecules.
  • Balance the hydrogen atoms by adding hydrogen ions.
  • Balance the charge by adding electrons.
  • Ensure that the electrons lost in the oxidation half-reaction are equal to the electrons gained in the reduction half-reaction.

Following these steps, we ensure that our half-reactions are balanced and ready to be combined to present the overall balanced redox reaction.
Calculating Substance Amount via Electrolysis
Calculating the amount of a substance produced or consumed during electrolysis involves understanding the relationship between the electric current, the duration the current is applied, and the substance's properties, such as its valence and molar mass.

To calculate this quantitatively, we use Faraday's Law of electrolysis alongside the knowledge that 1 Faraday of charge (approximately 96500 Coulombs) is required to deposit or dissolve 1 gram-equivalent of a substance. The number of Faradays can be calculated by dividing the total charge passed through the electrolyte by Faraday's constant.

Steps for Calculation:

  • Identify the total electric charge passed, which is the product of the current (in amperes) and the time (in seconds).
  • Determine how many moles of electrons corresponds to a mole of the substance, referred to as the n-factor.
  • Calculate the amount in moles of substance produced using the charge and the n-factor.
  • Multiply the moles of the substance by its molar mass to find the mass produced.

By carefully applying these steps, as shown in the Barium Chloride example, one can adeptly determine the practical outcomes of electrolysis reactions in a laboratory or industry setting.

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Most popular questions from this chapter

What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M}\), \(\left[\mathrm{H}^{+}\right]=0.050 \mathrm{M},\) and \(P_{\mathrm{H}}=2.0 \mathrm{~atm} ?\)

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The passage of a current of 0.750 A for \(25.0 \mathrm{~min}\) deposited \(0.369 \mathrm{~g}\) of copper from a \(\mathrm{CuSO}_{4}\) solution. From this information, calculate the molar mass of copper.

The magnitudes (but not the signs) of the standard reduction potentials of two metals \(X\) and \(Y\) are $$ \begin{array}{ll} \mathrm{Y}^{2+}+2 e^{-} \longrightarrow \mathrm{Y} & \mid E^{\mathrm{O}} \mathrm{I}=0.34 \mathrm{~V} \\ \mathrm{X}^{2+}+2 e^{-} \longrightarrow \mathrm{X} & \mid E^{\circ} \mathrm{I}=0.25 \mathrm{~V} \end{array} $$ where the II notation denotes that only the magnitude (but not the sign) of the \(E^{\circ}\) value is shown. When the half-cells of \(\mathrm{X}\) and \(\mathrm{Y}\) are connected, electrons flow from \(X\) to \(Y\). When \(X\) is connected to a SHE, electrons flow from \(X\) to \(\mathrm{SHE}\). (a) Are the \(E^{\circ}\) values of the half-reactions positive or negative? (b) What is the standard emf of a cell made up of \(\mathrm{X}\) and \(\mathrm{Y} ?\)

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