Chapter 20.1, Problem 4PP

Interpretation Introduction

Interpretation:

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

Concept introduction:

The half cell reactions are the reactions which occur at anode and cathode electrode. The difference between the electrode potential of the two electrodes consisting an electrochemical cell is known as cell potential

Answer to Problem 4PP

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ {\text{2No}}_{\text{3}}{}^{\text{-1}}\left(\text{aq}\right){\text{+ 10H}}^{\text{+}}{\text{+ 2O}}_{\text{2}}\left(\text{l}\right)\to \text{2NO}\left(\text{g}\right){\text{+ 4OH}}^{\text{-}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ {\text{O}}_{\text{2}}|{\text{OH}}^{\text{-1}}\Vert {\text{NO}}_{\text{3}}{}^{{}^{\text{-1}}}|\text{NO}\\ \text{The oxidation is occur at oxygen electrode or anode and reduction is occurs at }\\ \text{nitrate electrode or cathode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{0}\text{.957}\right)\text{-}\left(\text{-0}\text{.401}\right)\\ \text{= 0}\text{.556 V}\\ \text{The cell potential is 0}\text{.556 V}\end{array}$

Explanation of Solution

Given information:

$\begin{array}{l}\text{The}\text{standard}\text{potential}\text{of}{\text{NO}}_{\text{3}}{}^{\text{-1}}{\text{+4H}}^{\text{+}}{\text{+3e}}^{\text{-}}\to \text{NO+}{\text{2H}}_{\text{2}}\text{O}\text{=0}\text{.957}\text{V}\\ \text{The standard}\text{potential}\text{of}{\text{O}}_{\text{2}}\left(\text{g}\right){\text{+2H}}_{\text{2}}{\text{O+ 4e}}^{\text{-}}\to {\text{4OH}}^{\text{-}}\left(\text{aq}\right)\text{=}\text{0}\text{.401}\text{V}\end{array}$

The half cell reaction at anode

${\text{O}}_{\text{2}}\left(\text{g}\right){\text{+2H}}_{\text{2}}{\text{O+ 4e}}^{\text{-}}\to {\text{4OH}}^{\text{-}}\left(\text{aq}\right)$

The half cell reaction at cathode is

${\text{NO}}_{\text{3}}{}^{\text{-1}}{\text{+4H}}^{\text{+}}{\text{+3e}}^{\text{-}}\to \text{NO+}{\text{2H}}_{\text{2}}\text{O}$

In the reaction, balancing the nitrate ion by multiplying the number 2 on left side as well as on right side. Now balance $\text{O}$ atoms by adding ${\text{OH}}^{\text{-}}$ ion on the side deficient in oxygen atom and to balance $\text{H}$ atom, add one molecule of water for each H atom needed on the side deficient in H atom. At the same time add an equal number of ${\text{OH}}^{\text{-}}$ ions to the opposite side.

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ {\text{2No}}_{\text{3}}{}^{\text{-1}}\left(\text{aq}\right){\text{+ 10H}}^{\text{+}}{\text{+ 2O}}_{\text{2}}\left(\text{l}\right)\to \text{2NO}\left(\text{g}\right){\text{+ 4OH}}^{\text{-}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ {\text{O}}_{\text{2}}|{\text{OH}}^{\text{-1}}\Vert {\text{NO}}_{\text{3}}{}^{{}^{\text{-1}}}|\text{NO}\\ \text{The oxidation is occur at oxygen electrode or anode and reduction is occurs at }\\ \text{nitrate electrode or cathode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{0}\text{.957}\right)\text{-}\left(\text{-0}\text{.401}\right)\\ \text{= 0}\text{.556 V}\\ \text{The cell potential is 0}\text{.556 V}\end{array}$

Conclusion

$\begin{array}{l}\text{The}\text{balanced}\text{over}\text{all}\text{reaction}\text{is}\\ {\text{2No}}_{\text{3}}{}^{\text{-1}}\left(\text{aq}\right){\text{+ 10H}}^{\text{+}}{\text{+ 2O}}_{\text{2}}\left(\text{l}\right)\to \text{2NO}\left(\text{g}\right){\text{+ 4OH}}^{\text{-}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ \text{The}\text{cell}\text{reprsentation}\text{of}\text{this}\text{over}\text{}\text{all}\text{reaction}\text{is}\\ {\text{O}}_{\text{2}}|{\text{OH}}^{\text{-1}}\Vert {\text{NO}}_{\text{3}}{}^{{}^{\text{-1}}}|\text{NO}\\ \text{The oxidation is occur at oxygen electrode or anode and reduction is occurs at }\\ \text{nitrate electrode or cathode}\text{.}\\ \text{The cell potental of the reaction is calculated as}\\ {\text{E°}}_{\text{cell}}{\text{=E°}}_{\text{cathode}}{\text{-E°}}_{\text{anode}}\\ \text{=}\left(\text{0}\text{.957}\right)\text{-}\left(\text{-0}\text{.401}\right)\\ \text{= 0}\text{.556 V}\\ \text{The cell potential is 0}\text{.556 V}\end{array}$