Chapter 20.1, Problem 3PP

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 3PP

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ \text{Cr}\left(\text{s}\right){\text{+ Hg}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Cr}}^{\text{+2}}\left(\text{aq}\right)\text{+ Hg}\left(\text{l}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Cr}|{\text{Cr}}^{\text{+2}}\Vert {\text{Hg}}^{\text{+2}}|\text{Hg}\\ \text{The oxidation is occur at chromium electrode or anode and reduction is occurs at }\\ \text{mercury electrode or cathode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{0}\text{.851}\right)\text{-}\left(\text{-0}\text{.913}\right)\\ \text{= 1}\text{.}\text{764V}\\ \text{The cell potential is 1}\text{.764}\text{V}\end{array}$

Explanation of Solution

The overall reaction on the basis of given half reactions is written as

$\text{Cr}\left(\text{s}\right){\text{+ Hg}}^{\text{+2}}\left(\text{aq}\right)\to {\text{Cr}}^{\text{+2}}\left(\text{aq}\right)\text{+ Hg}\left(\text{l}\right)$

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ \text{Cr}\left(\text{s}\right){\text{+ Hg}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Cr}}^{\text{+2}}\left(\text{aq}\right)\text{+ Hg}\left(\text{l}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Cr}|{\text{Cr}}^{\text{+2}}\Vert {\text{Hg}}^{\text{+2}}|\text{Hg}\\ \text{The oxidation is occur at chromium electrode or anode and reduction is occurs at }\\ \text{mercury electrode or cathode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{0}\text{.851}\right)\text{-}\left(\text{-0}\text{.913}\right)\\ \text{= 1}\text{.}\text{764V}\\ \text{The cell potential is 1}\text{.764}\text{V}\end{array}$

Conclusion

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ \text{Cr}\left(\text{s}\right){\text{+ Hg}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Cr}}^{\text{+2}}\left(\text{aq}\right)\text{+ Hg}\left(\text{l}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Cr}|{\text{Cr}}^{\text{+2}}\Vert {\text{Hg}}^{\text{+2}}|\text{Hg}\\ \text{The oxidation is occur at chromium electrode or anode and reduction is occurs at }\\ \text{mercury electrode or cathode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{0}\text{.851}\right)\text{-}\left(\text{-0}\text{.913}\right)\\ \text{= 1}\text{.}\text{764V}\\ \text{The cell potential is 1}\text{.764}\text{V}\end{array}$