Concept explainers
Calculate the amounts of Cu and Br2 produced in 1.0 h at inert electrodes in a solution of CuBr2 by a current of 4.50 A.
Interpretation:
Need to calculate the amount of Copper deposited and bromine liberated upon electrolysis of CuBr2
Concept introduction:
Electrolysis of CuBr2 resulted in the formation of copper and bromine. The cell reaction can be written as follows.
In the given electrolytic process bromide get oxidized at anode to liberate as bromine gas and Cu2+ ion will get reduced to Cu. By calculating the total number of charges that flow through the circuit, the amount of copper deposited can be calculated in successive steps. Since coulomb is the amount of electric charge flowing in a circuit in 1s, when current is 1A. So the above statement can be represented by the following equation.
On dividing the number of charges with Faraday constant we can get the number of moles of electron
From knowing the number of mole of electrons and using the stoichiometry of the reaction, the number of moles of the substance reduced or oxidized can be determined. This can be explained by the representative reaction as shown below.
2 mole of electrons were needed for the reduction of one mole of copper. So the number of moles of copper reduced can be calculated by the following equation.
To find: Amount of copper and bromine produced during the electrolysis of CuBr2 upon passing 4.5 A of current for 1 hour.
Answer to Problem 18.51QP
The amount of copper deposited during the electrolysis of CuBr2 can be calculated from the given current and time through the following steps.
The half-cell reaction was given below
The amount of bromine liberated during the electrolysis of CuBr2 can be calculated from the given current and time through the following steps.
The half-cell reaction was given below
Explanation of Solution
The amount of copper deposited during the electrolysis of CuBr2 can be calculated from the given current and time through the following steps.
The half-cell reaction was given below
Current = 4.50A
Time = 1 h or 3600s
From the given current and time, the quantity of copper produced by the reduction of Cu2+ can calculated successfully in four steps
- 1) From the given current and time the amount of charges passing the circuit was calculated.
- 2) On dividing the charges with Faraday constant, number of moles of electron involved the reaction was determined.
- 3) On using the stoichiometry of the reaction, number of moles of substance reduced was calculated.
- 4) Multiplication of molar mass with number of moles of the substance that undergone reduction results in the mass of copper produced..
The amount of bromine liberated during the electrolysis of CuBr2 can be calculated from the given current and time through the following steps.
The half-cell reaction was given below
Since
Current = 4.50A
Time = 1 h or 3600s
From the given current and time, the quantity of bromine produced by the oxidation of bromide can calculated successfully in four steps
- 1) From the given current and time the amount of charges passing the circuit was calculated first
- 2) On dividing the charges with Faraday constant, number of moles of electron involved the reaction was determined
- 3) Using the stoichiometry of the reaction, number of moles of oxidized substance was calculated
- 4) Mass of bromine produce can be attained by multiplying the atomic mass with the number of moles of bromine
The amount of copper and bromine produced by the electrolysis of CuBr2 upon passing 4.5A of current for 1 hour was found to be
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Chapter 18 Solutions
Chemistry Atoms First, Second Edition
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