Q21.122 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question: To examine the effect of ion removal on cell voltage, a chemist constructs two voltaic cells, each with a standard hydrogen electrode in one compartment. One cell also contains a  ${\text{Pb/Pb}}^{\text{2 + }}$half-cell; the other contains a ${\text{Cu/Cu}}^{\text{2 + }}$half-cell. 

(a) What is ${\text{E}}^{\text{o}}$ of each cell 298 K at ? 

(b) Which electrode in each cell is negative? 

(c) When ${\text{Na}}_{\text{2}}\text{S}$ solution is added to the ${\text{Pb}}^{\text{2 + }}$electrolyte, solid $\text{PbS}$ forms. What happens to the cell voltage? 

(d) When sufficient ${\text{Na}}_{\text{2}}\text{S}$ is added to the ${\text{Cu}}^{\text{2 + }}$electrolyte,  $\text{CuS}$ forms and  ${\text{[Cu}}^{\text{2 + }}\text{]}$drops to $1\times {10}^{-6}$ . Find the cell voltage.

Step-by-Step Solution

Answer

(a) The ${\text{E}}^{\text{o}}$ of each cell at 298k is ${\text{E}}_{\text{cell}}^{\text{o}}\text{ = 0.13V}$and ${\text{E}}_{\text{cell}}^{\text{o}}\text{ = 0.34V}$ .

(b) In $\text{Pb}$ cell and $\text{Cu}$ cell the electrodes which are negative is $\text{Pb}$ and Hydrogen respectively.

(c) The cell voltage increases when $\text{PbS}$ solid forms.

(d) The cell voltage is found to be ${\text{E}}_{\text{cell}}\text{ = - 0.133V}$ .

Step 1: Concept Introduction

An oxidation reaction in which electrons are lost or a reduction reaction in which electrons are gained is known as a half-cell reaction. The reactions take place in an electrochemical cell, where electrons are lost at the anode via oxidation and consumed at the cathode via reduction.

Step 2: (a) The E degree of each cell

In the voltaic cell with ${\text{Pb/Pb}}^{\text{2 + }}$, the reaction is –

${\text{Pb}}_{\text{(s)}}{\text{ + 2H}}_{\text{(aq)}}^{\text{ + }}\to {\text{Pb}}_{\text{(aq)}}^{\text{2 + }}{\text{ + H}}_{\text{2(g)}}$

The value for ${\text{E}}^{\text{o}}$can be calculated as –

$$\begin{gathered} {\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{Pb}}^{\text{2 + }}}^{\text{o}}\text{ = - 0.13V} \\ {\text{E}}_{\text{reduction }}^{\text{o}}{\text{ = E}}_{{\text{H}}^{\text{ - }}}^{\text{o}}\text{ = 0V} \\ {\text{E}}_{\text{cell }}^{\text{o}}{\text{ = E}}_{\text{reduction }}^{\text{o}}{\text{ - E}}_{\text{oxidation }}^{\text{o}} \\ {\text{ = E}}_{{\text{H}}^{\text{ - }}}^{\text{o}}{\text{ - E}}_{{\text{Pb}}^{\text{2 + }}}^{\text{o}} \\ \text{ = 0 - ( - 0.13)} \\ \text{ = 0.13V} \end{gathered}$$

In the voltaic cell with  ${\text{Cu/Cu}}^{\text{2 + }}$ , the reaction is –

${\text{H}}_{\text{2(g)}}{\text{ + Cu}}^{\text{2 + }}\to {\text{2H}}_{\text{(ag)}}^{\text{ + }}{\text{ + Cu}}_{\text{(s)}}$

The value for ${\text{E}}^{\text{o}}$can be calculated as –

$$\begin{gathered} {\text{E}}_{\text{oxidation }}^{\text{o}}{\text{ = E}}_{{\text{H}}^{\text{ + }}}^{\text{o}}\text{ = 0V} \\ {\text{E}}_{\text{reduction }}^{\text{o}}{\text{ = E}}_{{\text{Cu}}^{\text{2 + }}}^{\text{o}}\text{ = 0.34V} \\ {\text{E}}_{\text{cell }}^{\text{o}}{\text{ = E}}_{\text{reduction }}^{\text{o}}{\text{ - E}}_{\text{oxidation }}^{\text{o}} \\ {\text{ = E}}_{{\text{Cu}}^{\text{2 + }}}^{\text{o}}{\text{ - E}}_{{\text{H}}^{\text{ + }}}^{\text{o}} \\ \text{ = 0.34 - 0} \\ \text{ = 0.34V} \end{gathered}$$

 

Therefore, the values for ${\text{E}}_{\text{cell}}^{\text{o}}$ are obtained as $\text{0.13V}$and $\text{0.34V}$ .

Step 3: (b) The Negative Electrode

In the voltaic cell with $\text{Pb}$, the anode is the $\text{Pb}$ . While in the voltaic cell with copper, the negative electrode is the standard hydrogen electrode. Both are negative electrodes because they lose electrons in the voltaic cell.

 

Therefore, in first voltaic cell, the negative electrode is $\text{Pb}$ electrode and in the second voltaic cell the negative electrode ${\text{H}}_{\text{2}}$ is electrode.

Step 4: The Cell Voltage

When solid $\text{PbS}$ forms, the concentration of ${\text{Pb}}^{\text{2 + }}$ in solution decreases. This will drive the reaction forward and cause an increase in the cell voltage.

 

Therefore, there is an increase in the cell voltage.

Step 5: (d) The Cell Voltage

When ${\text{[Cu}}^{\text{2 + }}\text{]}$ is $1\times {10}^{-6}$  the Nernst equation can be used to determine the new cell voltage.

The given information is: ${\text{F = 96500F,n = 2 electrons and L}}_{\text{ccll}}^{\text{o}}\text{ = 0.31V,T = 298K}$

$$\begin{gathered} {\text{E}}_{\text{cell}}{\text{=E}}_{\text{cell }}^{\text{o}}\text{ - }\frac{\text{RT}}{\text{nF}}\text{×In}\frac{\left[{\text{H}}^{\text{ + }}\right]}{\left[{\text{Cu}}^{\text{2 + }}\right]} \\ {\text{E}}_{\text{cell}}\text{ = 0.34 - }\frac{\text{(8.314)(298)}}{\text{(2)(96500)}}\text{×In}\frac{\left[1\right]}{1\times {10}^{-6}} \\ {\text{E}}_{\text{cell}} =-0.133V \end{gathered}$$

 

Therefore, the value for cell voltage is obtained as ${\text{E}}_{\text{cell}}\text{ = - 0.133V}$ .