Chapter 20.1, Problem 2PP

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 2PP

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

$\text{Cr}\left(\text{s}\right){\text{+Co}}^{\text{+2}}\left(\text{aq}\right)\to {\text{Cr}}^{\text{+3}}\left(\text{aq}\right)\text{+Co}\left(\text{s}\right)$

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ \text{2Cr}\left(\text{s}\right){\text{+3Co}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{2Cr}}^{\text{+3}}\left(\text{aq}\right)\text{+ 3Co}\left(\text{s}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Cr}|{\text{Cr}}^{\text{+3}}\Vert {\text{Co}}^{\text{+2}}|\text{Co}\\ \text{The oxidation is occur at chromium electrode and reduction is occurs at }\\ \text{cobalt electrode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{-0}\text{.28}\right)\text{-}\left(\text{-0}\text{.744}\right)\\ \text{=0}\text{.464}\text{V}\\ \text{The cell potential is 0}\text{.464}\text{V}\end{array}$

Explanation of Solution

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

$\begin{array}{l}\text{Cr}\left(\text{s}\right){\text{+Co}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{Cr}}^{\text{+3}}\left(\text{aq}\right)\text{+ Co}\left(\text{s}\right)\\ \text{Here,}\text{on}\text{the}\text{left}\text{side}\text{oxidation}\text{state}\text{of}\text{chromium}\text{is}\text{zero}\text{and}\text{that}\text{of}\text{cobalt}\text{is}\text{two}\text{but}\\ \text{on}\text{the}\text{right}\text{side}\text{oxidation}\text{number}\text{of}\text{chromium}\text{is}\text{three}\text{and}\text{that}\text{of}\text{cobalt}\text{is}\text{zero}\text{.}\\ \text{so,}\text{we}\text{balancing}\text{it by multiply the Cr by 2 and Co by 3 on left side and multiply the Cr }\\ \text{by 2 and Co by 3 on right side}\text{. So, we get}\\ \text{2Cr}\left(\text{s}\right){\text{+3Co}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{2Cr}}^{\text{+3}}\left(\text{aq}\right)\text{+ 3Co}\left(\text{s}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Cr}|{\text{Cr}}^{\text{+3}}\Vert {\text{Co}}^{\text{+2}}|\text{Co}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{-0}\text{.28}\right)\text{-}\left(\text{-0}\text{.744}\right)\\ \text{=0}\text{.464}\text{V}\\ \text{The cell potential is 0}\text{.464}\text{V}\end{array}$

The oxidation is occurs at chromium electrode and reduction occurs at cobalt electrode because the standard reduction potential of cobalt is greater than that of chromium.

Conclusion

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

$\text{Cr}\left(\text{s}\right){\text{+Co}}^{\text{+2}}\left(\text{aq}\right)\to {\text{Cr}}^{\text{+3}}\left(\text{aq}\right)\text{+Co}\left(\text{s}\right)$

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ \text{2Cr}\left(\text{s}\right){\text{+3Co}}^{\text{+2}}\left(\text{aq}\right)\rightleftharpoons {\text{2Cr}}^{\text{+3}}\left(\text{aq}\right)\text{+ 3Co}\left(\text{s}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ \text{Cr}|{\text{Cr}}^{\text{+3}}\Vert {\text{Co}}^{\text{+2}}|\text{Co}\\ \text{The oxidation is occur at chromium electrode and reduction is occurs at }\\ \text{cobalt electrode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{-0}\text{.28}\right)\text{-}\left(\text{-0}\text{.744}\right)\\ \text{=0}\text{.464}\text{V}\\ \text{The cell potential is 0}\text{.464}\text{V}\end{array}$