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A certain current liberated \(0.50 \mathrm{~g}\) of hydrogen in \(2 \mathrm{~h}\). How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution? \((\mathrm{Cu}=63.5)\) (a) \(12.7 \mathrm{~g}\) (b) \(15.88 \mathrm{~g}\) (c) \(31.75 \mathrm{~g}\) (d) \(63.5 \mathrm{~g}\)

Short Answer

Expert verified
15.88 g

Step by step solution

01

Use Faraday’s Laws of Electrolysis

According to Faraday’s first law of electrolysis, the mass (m) of a substance liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (Q) that passes through the solution. That is, m = E * Q, where E is the electrochemical equivalent of the substance.
02

Determine the Electrochemical Equivalent of Hydrogen

The electrochemical equivalent (E) of hydrogen is its atomic weight divided by the charge number, which is approximately 1 g/mol divided by 1 (since 1 mole of hydrogen gas, which contains Avogadro’s number of atoms, has a charge equivalent to 1 mole of electrons). Therefore, the electrochemical equivalent of hydrogen is roughly 1 g/Faraday.
03

Calculate the Quantity of Electricity for Hydrogen Liberation

Using m = E * Q, we find that 0.5 g of hydrogen were liberated by a certain quantity of electricity Q. So, Q = m/E, which means Q = 0.5 g / (1 g/Faraday) = 0.5 Faraday.
04

Calculate the Electrochemical Equivalent of Copper

The electrochemical equivalent of copper is its atomic weight divided by the valency (which is 2 for copper as it usually forms Cu ions). Thus, E(Cu) = atomic weight of Cu / 2 = 63.5 g/mol / 2 = 31.75 g/Faraday.
05

Calculate the Mass of Copper That Can Be Liberated

Using the same quantity of electricity Q = 0.5 Faraday, we can find the mass of copper liberated: mass (Cu) = E(Cu) * Q = 31.75 g/Faraday * 0.5 Faraday = 15.875 g.
06

Choose the Correct Option

The mass of copper that can be liberated is 15.875 g, which, after rounding, gives 15.88 g. The correct option is (b) 15.88 g.

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Key Concepts

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

Electrochemical Equivalent
Understanding the electrochemical equivalent is vital when delving into the realm of electrochemistry and electrolysis. It is a constant that relates the mass of a substance to the charge passed during electrolysis. Essentially, it translates the quantity of electricity into actual mass of substance deposited or liberated at an electrode.

For an element, the electrochemical equivalent (E) can be calculated by dividing the atomic mass by the valency and then further dividing by the Faraday constant (approximately 96485 Coulombs per mole), which represents the charge of one mole of electrons. The units for electrochemical equivalent are typically grams per Coulomb or grams per Faraday. For example, hydrogen, with an atomic mass of approximately 1 gram/mole and a valency of 1, has an electrochemical equivalent of roughly 1 gram/Faraday.

Understanding this concept is crucial for electrolysis calculations, as it establishes a direct relationship between the amount of electric charge passed through a substance and the mass of that substance that is electrolyzed.
Electrolysis Calculations
Calculations in electrolysis allow us to predict the outcomes of electrochemical reactions based on the amount of electricity used. These calculations are grounded upon Faraday's laws, where the first law states that the amount of chemical change is directly proportional to the amount of electricity that flows through the electrolyte.

To perform these calculations, we use the formula:
\[ m = EQ \]
where \( m \) is the mass of the substance liberated or deposited, \( E \) is the electrochemical equivalent of the substance, and \( Q \) is the quantity of electricity in Faradays. To find the mass of a particular substance released during electrolysis, it's essential to know the electrochemical equivalent of the substance and the total charge that passes through the electrolyte.

For instance, in the given example, we determined the amount of electricity required to liberate 0.5 grams of hydrogen and used the same amount to calculate the mass of copper that could be liberated under similar conditions. The ability to perform these electrolysis calculations becomes a powerful tool in predicting and quantifying electrochemical reactions in practical applications.
Stoichiometry in Electrolysis
Stoichiometry in electrolysis involves the quantitative relationship between reactants and products in an electrochemical cell. It is the balancing act between the amount of electrons transferred during the electrochemical reaction and the stoichiometry of the substances involved.

Understanding stoichiometry is crucial for calculating how much product will form given a certain amount of reactants or vice versa. The amount of a substance produced or consumed in electrolysis is proportional to its molar mass, valency, and the amount of charge passed.

One core principle is that for a given quantity of electricity, the amount of substance liberated is proportional to its equivalent weight, which is its molar mass divided by the number of electrons transferred (or its valency). This relationship is pivotal in determining the exact quantities of substances involved in electrochemical processes.

The completion of the electrolysis process depends on the stoichiometry of the reaction taking place, which correlates with the amount of charge and time applied to the electrochemical cell. Being adept with stoichiometry in electrolysis is essential for anyone studying or working with electrochemical reactions and processes.

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Most popular questions from this chapter

The conductivity of a saturated solution of \(\mathrm{AgCl}\) at \(298 \mathrm{~K}\) was found to be \(3.40\) \(\times 10^{-6} \Omega^{-1} \mathrm{~cm}^{-1} ;\) the conductivity of water used to make up the solution was \(1.60\) \(\times 10^{-6} \Omega^{-1} \mathrm{~cm}^{-1} .\) Determine the solubility of \(\mathrm{AgCl}\) in water in mole per litre at \(298 \mathrm{~K}\). The equivalent conductivity of \(\mathrm{AgCl}\) at infinite dilution is \(150.0 \Omega^{-1} \mathrm{~cm}^{-2} \mathrm{eq}^{-1}\). (a) \(1.44 \times 10^{-10}\) (b) \(1.2 \times 10^{-5}\) (c) \(3.33 \times 10^{-5}\) (d) \(1.2 \times 10^{-8}\)

The same quantity of electricity is passed through one molar solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and one molar solution of \(\mathrm{HCl}\). The amount of hydrogen evolved from \(\mathrm{H}_{2} \mathrm{SO}_{4}\) as compared to that from \(\mathrm{HCl}\) is (a) the same (b) twice as such (c) one half as such (d) dependent on size of electrode

The molar conductance of a strong electrolyte at infinite dilution (a) tends to a finite value, which is above that at higher concentration (b) tends to a finite value, which is below that at higher concentration (c) tends to zero (d) tends to a finite value, which is equal

Indicator electrode is (a) SHE (b) Calomel electrode (c) \(\mathrm{Ag} / \mathrm{AgCl}\) electrode (d) Quinhydrone electrode

The standard reduction potential for the process: \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\mathrm{e}^{-} \rightarrow\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is \(1.8 \mathrm{~V}\). The standard reduction potential for the process: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}+\mathrm{e}^{-}\) \(\rightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is \(0.1 \mathrm{~V} .\) Which of the complex ion, \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) or \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) can be oxidized to the corresponding cobalt (III) complex, by oxygen, in basic medium, under standard condition? \(\left[\right.\) Given: \(\left.E_{\mathrm{O}_{2} / \mathrm{OH}^{-}}^{\circ}=0.4 \mathrm{~V}\right]\) (a) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) (c) both (d) none of these

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