Chapter 20.1, Problem 6PP

Interpretation Introduction

Interpretation:

The given cell reaction is to be balanced and the cell potential is to be determined.

Concept introduction:

An electrochemical reaction is balanced by balancing the oxidation number on both side of the reaction. The difference between the electrode potential of the two electrodes consisting an electrochemical cell is known as cell potential

Answer to Problem 6PP

The half cell reaction at anode

$\text{Mg}\left(\text{s}\right)\to {\text{Mg}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}$

The half cell reaction at cathode is

${\text{Pb}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to \text{Pb}\left(\text{s}\right)$

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ \text{Mg}\left(\text{s}\right){\text{+ Pb}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Mg}}^{\text{+2}}\left(\text{aq}\right)\text{+ Pb}\left(\text{s}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Mg}|{\text{Mg}}^{\text{+2}}\Vert {\text{Pb}}^{\text{+2}}|\text{Pb}\\ \text{The oxidation is occur at Mg electrode or anode and reduction is occurs at }\\ \text{Pb electrode or cathode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{-0}\text{.1262}\right)\text{-}\left(\text{-2}\text{.372}\right)\\ \text{= 2}\text{.246V}\\ \text{The cell potential is 2}\text{.246}\text{V}\end{array}$

Explanation of Solution

The half cell reaction at anode

$\text{Mg}\left(\text{s}\right)\to {\text{Mg}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}$

The half cell reaction at cathode is

${\text{Pb}}^{\text{+2}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}\to \text{Pb}\left(\text{s}\right)$

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ \text{Mg}\left(\text{s}\right){\text{+ Pb}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Mg}}^{\text{+2}}\left(\text{aq}\right)\text{+ Pb}\left(\text{s}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Mg}|{\text{Mg}}^{\text{+2}}\Vert {\text{Pb}}^{\text{+2}}|\text{Pb}\\ \text{The oxidation is occur at Mg electrode or anode and reduction is occurs at }\\ \text{Pb electrode or cathode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{-0}\text{.1262}\right)\text{-}\left(\text{-2}\text{.372}\right)\\ \text{= 2}\text{.246V}\\ \text{The cell potential is 2}\text{.246}\text{V}\end{array}$

Conclusion

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ \text{Mg}\left(\text{s}\right){\text{+ Pb}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Mg}}^{\text{+2}}\left(\text{aq}\right)\text{+ Pb}\left(\text{s}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Mg}|{\text{Mg}}^{\text{+2}}\Vert {\text{Pb}}^{\text{+2}}|\text{Pb}\\ \text{The oxidation is occur at Mg electrode or anode and reduction is occurs at }\\ \text{Pb electrode or cathode}\text{.}\\ \text{The cell potential is 2}\text{.246}\text{V}\end{array}$