Chapter 18, Problem 18.51QP

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 Introduction

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.

$\begin{array}{l}\text{Anode(oxidation) }{\text{2Br}}^{\text{-}}{}_{\text{(l)}}\stackrel{}{\to }{\text{Br}}_{\text{2(g)}}{\text{+2e}}^{\text{-}}\\ {\text{Cathode(reduction)Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Cu}}_{\text{(s)}}\\ \text{Overallreaction}{\text{Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2Br}}^{\text{-}}{}_{\text{(l)}}\stackrel{}{\to }{\text{ Cu}}_{\text{(s)}}{\text{+ Br}}_{\text{2(g)}}\end{array}$

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.

$\text{charges(coulombs)=current(ampheres)×time(s)}$

On dividing the number of charges with Faraday constant we can get the number of moles of electron

$\text{Number}\text{of}\text{moles}\text{of}\text{electron(n)=charges/Faraday constant (96500 C/mole of electron)}$

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.

${\text{Cathode(reduction) Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Cu}}_{\text{(s)}}$

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.

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{copper=}\text{n mole}{\text{e}}^{\text{-}}\text{×}\frac{\text{1mole}{\text{Cu}}^{\text{2+}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{Mass}\text{}\text{of}\text{copper=}\text{Number}\text{of}\text{mole}\text{of}\text{copper×Atomic}\text{mass}\text{of}\text{copper}\end{array}$

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

${\text{Cathode(reduction)Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Cu}}_{\text{(s)}}$

$\begin{array}{l}\text{charges(in}\text{coulombs)}\text{=}\text{current(ampheres)×time(s)}\\ =\text{4}\text{.5×3600}\\ =1.62\text{×}{10}^4\text{C}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{electron(n)}\text{=}\text{charges/Faraday constant (96500 C/mole of electron)}\\ \text{=}\frac{\text{1}{\text{.62×10}}^{\text{4}}\text{C}}{\text{96500}{\text{C/mole e}}^{\text{-}}}\\ \text{=}\text{1}{\text{.6788×10}}^{\text{-}}{}^{\text{1}}\text{mole}{\text{e}}^{\text{-}}\\ \text{Number}\text{of}\text{moles}\text{of}\text{copper}\text{=}\text{n mole}{\text{e}}^{\text{-}}\text{×}\frac{\text{1mole}{\text{Cu}}^{\text{2+}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{=}\text{1}{\text{.6788×10}}^{\text{-}}{}^{\text{1}}\text{mole}{\text{e}}^{\text{-}}\text{×}\frac{\text{1mole}{\text{Cu}}^{\text{2+}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{=}\text{8}{\text{.394×10}}^{\text{-}}{}^{\text{2}}\text{mole Cu}\\ \text{Mass}\text{}\text{of}\text{copper=}\text{Number}\text{of}\text{mole}\text{of}\text{copper×Atomic}\text{mass}\text{of}\text{copper}\\ \text{ =}\text{8}{\text{.394×10}}^{\text{-}}{}^{\text{2}}\text{ mole Cu×63}\text{.55g/mole}\\ \text{=}\text{5}\text{.3344g}\end{array}$

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

${\text{Anode(oxidation) 2Br}}^{\text{-}}{}_{\text{(l)}}\stackrel{}{\to }{\text{Br}}_{\text{2(g)}}{\text{+2e}}^{\text{-}}$

$\begin{array}{l}\text{charges(in}\text{coulombs)}\text{=}\text{4}\text{.5×3600}\\ =1.62\text{×}{10}^4\text{C}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{electron(n)=charges/Faraday constant (96500 C/mole of electron)}\\ \text{=}\frac{\text{1}{\text{.62×10}}^{\text{4}}\text{C}}{\text{96500}{\text{C/mole e}}^{\text{-}}}\\ \text{=}\text{1}{\text{.6788×10}}^{\text{-}}{}^{\text{1}}\text{mole}{\text{e}}^{\text{-}}\\ \text{Number}\text{of}\text{moles}\text{of}\text{Bromine}\text{=}\text{n mole}{\text{e}}^{\text{-}}\text{×}\frac{\text{1mole}{\text{Br}}^{\text{-}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{=1}{\text{.6788×10}}^{\text{-}}{}^{\text{1}}\text{mole}{\text{e}}^{\text{-}}\text{×}\frac{{\text{1moleBr}}^{\text{-}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{=}\text{8}{\text{.394×10}}^{\text{-}}{}^{\text{2}}{\text{mole Br}}_{\text{2}}\\ \text{Mass}\text{}\text{of}\text{bromine=}\text{Number}\text{of}\text{mole}\text{of}\text{bromine×Atomic}\text{mass}\text{of}\text{bromine}\\ \text{ =}\text{8}{\text{.394×10}}^{\text{-}}{}^{\text{2}}\text{ mole Cu×159}\text{.8g/mole}\\ \text{=}1\text{3}\text{.4136}\text{g}\end{array}$

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

${\text{Cathode(reduction)Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Cu}}_{\text{(s)}}$

$\text{charges(in}\text{coulombs)=current(in}\text{ampheres)×time(s)}$

Current = 4.50A

Time = 1 h or 3600s

$\begin{array}{l}\text{charges(in}\text{coulombs)}\text{=}\text{4}\text{.5×3600}\\ =1.62\text{×}{10}^4\text{C}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{electron(n)=charges/Faraday constant (96500 C/mole of electron)}\\ \text{=}\frac{\text{1}{\text{.62×10}}^{\text{4}}\text{C}}{\text{96500}{\text{C/mole e}}^{\text{-}}}\\ \text{=}\text{1}{\text{.6788×10}}^{\text{-}}{}^{\text{1}}\text{mole}{\text{e}}^{\text{-}}\\ \text{Number}\text{of}\text{moles}\text{of}\text{copper}\text{=}\text{n mole}{\text{e}}^{\text{-}}\text{×}\frac{\text{1mole}{\text{Cu}}^{\text{2+}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{=1}{\text{.6788×10}}^{\text{-}}{}^{\text{1}}\text{mole}{\text{e}}^{\text{-}}\text{×}\frac{\text{1mole}{\text{Cu}}^{\text{2+}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{=}\text{8}{\text{.394×10}}^{\text{-}}{}^{\text{2}}\text{mole Cu}\\ \text{Mass}\text{}\text{of}\text{copper=}\text{Number}\text{of}\text{mole}\text{of}\text{copper×Atomic}\text{mass}\text{of}\text{copper}\\ \text{ =}\text{8}{\text{.394×10}}^{\text{-}}{}^{\text{2}}\text{ mole Cu×63}\text{.55g/mole}\\ \text{=}\text{5}\text{.3344g}\end{array}$

From the given current and time, the quantity of copper produced by the reduction of Cu²⁺ can calculated successfully in four steps

1. 1) From the given current and time the amount of charges passing the circuit was calculated.

1. 2) On dividing the charges with Faraday constant, number of moles of electron involved the reaction was determined.

1. 3) On using the stoichiometry of the reaction, number of moles of substance reduced was calculated.

1. 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

${\text{Anode(oxidation) 2Br}}^{\text{-}}{}_{\text{(l)}}\stackrel{}{\to }{\text{Br}}_{\text{2(g)}}{\text{+2e}}^{\text{-}}$

Since

Current = 4.50A

Time = 1 h or 3600s

$\begin{array}{l}\text{charges(in}\text{coulombs)}\text{=}\text{4}\text{.5×3600}\\ =1.62\text{×}{10}^4\text{C}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{electron(n)=charges/Faraday constant (96500 C/mole of electron)}\\ \text{=}\frac{\text{1}{\text{.62×10}}^{\text{4}}\text{C}}{\text{96500}{\text{C/mole e}}^{\text{-}}}\\ \text{=}\text{1}{\text{.6788×10}}^{\text{-}}{}^{\text{1}}\text{mole}{\text{e}}^{\text{-}}\\ \text{Number}\text{of}\text{moles}\text{of}\text{Bromine}\text{=}\text{n mole}{\text{e}}^{\text{-}}\text{×}\frac{\text{1mole}{\text{Br}}^{\text{-}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{=1}{\text{.6788×10}}^{\text{-}}{}^{\text{1}}\text{mole}{\text{e}}^{\text{-}}\text{×}\frac{{\text{1moleBr}}^{\text{-}}}{\text{2mole}{\text{e}}^{\text{-}}}\\ \text{=}\text{8}{\text{.394×10}}^{\text{-}}{}^{\text{2}}{\text{mole Br}}_{\text{2}}\\ \text{Mass}\text{}\text{of}\text{bromine=}\text{Number}\text{of}\text{mole}\text{of}\text{bromine×Atomic}\text{mass}\text{of}\text{bromine}\\ \text{ =}\text{8}{\text{.394×10}}^{\text{-}}{}^{\text{2}}\text{ mole Cu×159}\text{.8g/mole}\\ \text{=}1\text{3}\text{.4136}\text{g}\end{array}$

From the given current and time, the quantity of bromine produced by the oxidation of bromide can calculated successfully in four steps

1. 1) From the given current and time the amount of charges passing the circuit was calculated first
2. 2) On dividing the charges with Faraday constant, number of moles of electron involved the reaction was determined
3. 3) Using the stoichiometry of the reaction, number of moles of oxidized substance was calculated
4. 4) Mass of bromine produce can be attained by multiplying the atomic mass with the number of moles of bromine

Conclusion

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 $\text{5}\text{.3344g}$ and $1\text{3}\text{.4136}\text{g}$ respectively.