Chapter 18, Problem 18.54QP

A constant electric current flows for 3.75 h through two electrolytic cells connected in series. One contains a solution of AgNO₃ and the second a solution of CuCl₂. During this time, 2.00 g of silver is deposited in the first cell. (a) How many grams of copper are deposited in the second cell? (b) What is the current flowing (in amperes)?

Interpretation Introduction

Interpretation:

Need to calculate the amount of copper deposited and current needed to electrolyze the cell containing CuCl₂ connected in series with the cell containing AgNO₃.

Concept introduction:

In the given case two electrolytic cell containing AgNO₃ and CuCl₂ were connected in series, so the quantity of electricity flowing through both of the cell will be same

Half-cell reaction for cell-1 was given below.

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

The amount of the silver deposited was given, from which the amount of electricity flows though both of the cell can be calculated in steps. Number of moles of silver deposited was calculated first.  Since one mole of electron is needed to reduce one mole of Ag⁺_. Therefore number of moles of electron is equal to number of moles of sliver. The coulombs of electron passing can be calculated by the equation given below.

$\text{Number}\text{of}\text{moles}\text{of}\text{silver}\text{=}\frac{\text{weight}\text{(g)}}{\text{Atomic}\text{mass}\text{(g}{\text{mole}}^{\text{-1}}\text{)}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons = Number of moles of Silver}\\ \text{then}\\ \text{Charges}\text{=}\text{Number}\text{of}\text{moles of electrons×Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}\end{array}$

Half-cell reaction for cell-2 was given below.

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

Since they are connected in series, same amount of charges will be passing through both the cell. From the calculated charges the number of moles of electrons utilized can be calculated.

$\text{Number}\text{of}\text{moles of electrons (}n\text{)}\text{=}\text{Charges}\text{×Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}$

From the cell reaction it was known that two mole of electron will be needed to produce one mole of copper,

So

$\begin{array}{l}numberofmolesof\text{copper}=nmoleof{e}^{-}\times \frac{\text{1mole}\text{Cu}}{2mole{e}^{-}}\\ \text{Weight}\text{of}\text{Copper}=numberofmolesof\text{copper }\times \text{ atomic mass}\end{array}$

Since time was given the amperes of current flowing through the circuit can be calculated by the equation given below

$\text{Current (A) =}\frac{\text{charges (C)}}{\text{time(s)}}$

To find: The amount of copper deposited and current consumed in a CuCl₂ cell connected in series with the cell containing AgNO_3.

Answer to Problem 18.54QP

Since cells containing AgNO₃ and CuCl₂ were connected in series, the same amount of electricity will pass through both of them. So first let us calculate the coulombs of charges passed through the AgNO₃ cell, from the quantity of silver deposited.

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{silver}\text{=}\frac{2.00\text{g}}{\text{107}\text{.9 g}{\text{mole}}^{\text{-1}}}\\ =1.8536\times {10}^{-2}\text{mole}\end{array}$

Since one mole of electron is needed for the reduction of one mole of Ag⁺

Then

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons = 1}\text{.8536 }\times {10}^{-2}\text{mole}\\ \text{Charges}\text{=}\text{1}\text{.8536 }\times {10}^{-2}\text{mole×96500 C/mole}\\ \text{= 1}\text{.7887 }\times {10}^3\text{C}\end{array}$

As state above the coulombs of current passing through the CuCl₂ cell also will be $\text{1}\text{.7887 }\times {10}^3\text{C}$. The number of moles of electron passing through the second cell as calculated as shown below.

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons}\text{=}\frac{1.7887\times {10}^3\text{C}}{96500\text{C/mole}}\\ \text{=}1.8536{\text{×10}}^{\text{-2}}\text{mole}\end{array}$

Then

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{copper}\text{=}\frac{\text{1}{\text{.8536×10}}^{\text{-2}}\text{mole}}{\text{2}}\\ \text{=}\text{9}\text{.268}{\text{×10}}^{\text{-3}}\text{mole}\\ \text{Mass}\text{of}\text{copper}\text{=}\text{9}\text{.268}{\text{×10}}^{\text{-3}}\text{mole ×}\text{63}\text{.55 g/mole}\\ \text{=}\text{0}\text{.589}\text{g}\end{array}$

Explanation of Solution

Since cells containing AgNO₃ and CuCl₂ were connected in series, the same amount of electricity will pass through both of them. So first let us calculate the coulombs of charges passed through the AgNO₃ cell, which resulted in the deposition of 2.0 g of silver.

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{silver}\text{=}\frac{\text{weight(g)}}{\text{Atomic}\text{mass}\text{(g/mole)}}\\ \text{=}\frac{\text{2}\text{.00}\text{g}}{\text{107}\text{.9 g}{\text{mole}}^{\text{-1}}}\\ \text{=}\text{1}\text{.8536}{\text{×10}}^{\text{-2}}\text{mole}\end{array}$

Since one mole of electron is needed for the reduction of one mole of Ag⁺

Then

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons = Number}\text{of}\text{moles of silver deposited }\\ \text{= 1}\text{.8536 }\times {10}^{-2}\text{mole}\end{array}$

$\begin{array}{l}\text{Charges}\text{= number of moles of electron}\text{×}\text{Faraday}\text{constant}\\ \text{= 1}{\text{.8536 ×10}}^{\text{-2}}\text{mole×96500 C/mole}\\ \text{= 1}{\text{.7887 ×10}}^{\text{3}}\text{C}\end{array}$

As state above the coulombs of charges passing through the CuCl₂ cell also will be $\text{1}\text{.7887 }\times {10}^3\text{C}$.

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons}\text{=}\frac{1.7887\times {10}^3\text{C}}{96500\text{C/mole}}\\ \text{=}1.8536{\text{×10}}^{\text{-2}}\text{mole}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{copper}\text{=}\frac{\text{1}{\text{.8536×10}}^{\text{-2}}\text{mole}}{\text{2}}\\ \text{=}\text{9}\text{.268}{\text{×10}}^{\text{-3}}\text{mole}\\ \text{Mass}\text{of}\text{copper}\text{=}\text{9}\text{.268}{\text{×10}}^{\text{-3}}\text{mole ×}\text{63}\text{.55 g/mole}\\ \text{=}\text{0}\text{.589}\text{g}\end{array}$

In the present case the quantity of silver deposited upon electrolysis of AgNO₃ was given, from that the quantity of charges consumed was calculated and utilized for the determination of mass of Cu deposited in the second cell.

Interpretation Introduction

Interpretation:

Need to calculate the amount of copper deposited and current needed to electrolyze the cell containing CuCl₂ connected in series with the cell containing AgNO₃.

Concept introduction:

In the given case two electrolytic cell containing AgNO₃ and CuCl₂ were connected in series, so the quantity of electricity flowing through both of the cell will be same

Half-cell reaction for cell-1 was given below.

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

The amount of the silver deposited was given, from which the amount of electricity flows though both of the cell can be calculated in steps. Number of moles of silver deposited was calculated first.  Since one mole of electron is needed to reduce one mole of Ag⁺_. Therefore number of moles of electron is equal to number of moles of sliver. The coulombs of electron passing can be calculated by the equation given below.

$\text{Number}\text{of}\text{moles}\text{of}\text{silver}\text{=}\frac{\text{weight}\text{(g)}}{\text{Atomic}\text{mass}\text{(g}{\text{mole}}^{\text{-1}}\text{)}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons = Number of moles of Silver}\\ \text{then}\\ \text{Charges}\text{=}\text{Number}\text{of}\text{moles of electrons×Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}\end{array}$

Half-cell reaction for cell-2 was given below.

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

Since they are connected in series, same amount of charges will be passing through both the cell. From the calculated charges the number of moles of electrons utilized can be calculated.

$\text{Number}\text{of}\text{moles of electrons (}n\text{)}\text{=}\text{Charges}\text{×Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}$

From the cell reaction it was known that two mole of electron will be needed to produce one mole of copper,

So

$\begin{array}{l}numberofmolesof\text{copper}=nmoleof{e}^{-}\times \frac{\text{1mole}\text{Cu}}{2mole{e}^{-}}\\ \text{Weight}\text{of}\text{Copper}=numberofmolesof\text{copper }\times \text{ atomic mass}\end{array}$

Since time was given the amperes of current flowing through the circuit can be calculated by the equation given below

$\text{Current (A) =}\frac{\text{charges (C)}}{\text{time(s)}}$

To find: The amount of copper deposited and current consumed in a CuCl₂ cell connected in series with the cell containing AgNO_3.

Answer to Problem 18.54QP

The amount of current passing through the CuCl₂ cell for 3.75 h can calculate as follows

Time = 3.75 h or 13500 s

$\begin{array}{l}\text{Current (A) =}\frac{1.7887\times {10}^3\text{C}}{13500\mathrm{s}}\\ =0.1325\text{A}\end{array}$

Explanation of Solution

The amount of current passing through the CuCl₂ cell for 3.75 h can calculate as follows

Time = 3.75h or 13500s.

$\begin{array}{l}\text{Current (A) =}\frac{1.7887\times {10}^3\text{C}}{13500\mathrm{s}}\\ =0.1325\text{A}\end{array}$

On dividing the coulombs of charges produced by time in seconds the amount of current passing through the second cell was calculated as 0.1325A.