Question

Predict whether the following reactions occur under standard state conditions.
i) Oxidation of ${\text{A}}{{\text{g}}_{\left( {\text{s}} \right)}}$ by ${\text{C}}{{\text{l}}_{\left( {\text{g}} \right)}}$. ${\text{E}}_{{\text{Ag}}}^{\text{0}} = 0.8{\text{ V, E}}_{{\text{C}}{{\text{l}}_2}}^{\text{0}} = 1.36{\text{ V}}$
ii) Reduction of ${\text{F}}{{\text{e}}^{3 + }}$ to ${\text{F}}{{\text{e}}^{2 + }}$ by ${\text{A}}{{\text{u}}_{\left( {\text{s}} \right)}}$. ${\text{E}}_{{\text{F}}{{\text{e}}^{3 + }},{\text{F}}{{\text{e}}^{2 + }}}^{\text{0}} = 0.77{\text{ V, E}}_{{\text{Au}}}^{\text{0}} = 1.4{\text{ V}}$

Answer 1

Hint: We have to predict if the given reactions occur under standard conditions i.e. we have to predict if the given reaction is spontaneous. If the standard potential of the reaction is positive then the reaction is spontaneous and occurs under standard state conditions.

Formula Used: ${\text{E}}_{{\text{reaction}}}^{\text{0}} = {\text{E}}_{{\text{reduction}}}^{\text{0}} - {\text{E}}_{{\text{oxidation}}}^{\text{0}}$

Complete step-by-step solution :
i) We are given that oxidation of ${\text{A}}{{\text{g}}_{\left( {\text{s}} \right)}}$ occurs by ${\text{C}}{{\text{l}}_{\left( {\text{g}} \right)}}$. The reaction is as follows:
\[{\text{2A}}{{\text{g}}_{\left( {\text{s}} \right)}} + {\text{C}}{{\text{l}}_{{\text{2}}\left( {\text{g}} \right)}} \to {\text{2Ag}}_{\left( {{\text{aq}}} \right)}^ + + {\text{2Cl}}_{\left( {{\text{aq}}} \right)}^ - \]
In the given reaction, ${\text{A}}{{\text{g}}_{\left( {\text{s}} \right)}}$ is getting oxidised and ${\text{C}}{{\text{l}}_{\left( {\text{g}} \right)}}$ is getting reduced. The reactions are as follows:
Oxidation: \[{\text{2A}}{{\text{g}}_{\left( {\text{s}} \right)}} \to {\text{2Ag}}_{\left( {{\text{aq}}} \right)}^ + + 2{{\text{e}}^ - }\]
Reduction: \[{\text{C}}{{\text{l}}_{{\text{2}}\left( {\text{g}} \right)}} + 2{{\text{e}}^ - } \to 2{\text{Cl}}_{\left( {{\text{aq}}} \right)}^ - \]
We are given that ${\text{E}}_{{\text{Ag}}}^{\text{0}} = 0.8{\text{ V}}$ and ${\text{E}}_{{\text{C}}{{\text{l}}_2}}^{\text{0}} = 1.36{\text{ V}}$.
The expression for standard potential of the reaction is as follows:
${\text{E}}_{{\text{reaction}}}^{\text{0}} = {\text{E}}_{{\text{reduction}}}^{\text{0}} - {\text{E}}_{{\text{oxidation}}}^{\text{0}}$
Substitute $1.36{\text{ V}}$ for the standard reduction potential, $0.8{\text{ V}}$ for the standard oxidation potential and solve for the standard potential of the reaction. Thus,
${\text{E}}_{{\text{reaction}}}^{\text{0}} = 1.36{\text{ V}} - 0.8{\text{ V}}$
${\text{E}}_{{\text{reaction}}}^{\text{0}} = + 0.56{\text{ V}}$
Thus, the standard potential of the reaction is $ + 0.56{\text{ V}}$. The value of standard potential of the reaction is positive. Thus, the reaction is spontaneous and occurs under standard conditions.
Thus, oxidation of ${\text{A}}{{\text{g}}_{\left( {\text{s}} \right)}}$ by ${\text{C}}{{\text{l}}_{\left( {\text{g}} \right)}}$ occurs under standard conditions.

ii) We are given that reduction of ${\text{F}}{{\text{e}}^{3 + }}$ to ${\text{F}}{{\text{e}}^{2 + }}$ occurs by ${\text{A}}{{\text{u}}_{\left( {\text{s}} \right)}}$. The reaction is as follows:
\[{\text{A}}{{\text{u}}_{\left( {\text{s}} \right)}} + {\text{F}}{{\text{e}}^{3 + }} \to {\text{A}}{{\text{u}}^ + } + {\text{F}}{{\text{e}}^{2 + }}\]
In the given reaction, ${\text{A}}{{\text{u}}_{\left( {\text{s}} \right)}}$ is getting oxidised and ${\text{F}}{{\text{e}}^{3 + }}$ is getting reduced. The reactions are as follows:
Oxidation: \[{\text{A}}{{\text{u}}_{\left( {\text{s}} \right)}} \to {\text{A}}{{\text{u}}^ + } + {{\text{e}}^ - }\]
Reduction: \[{\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ - } \to {\text{F}}{{\text{e}}^{2 + }}\]
We are given that ${\text{E}}_{{\text{F}}{{\text{e}}^{3 + }},{\text{F}}{{\text{e}}^{2 + }}}^{\text{0}} = 0.77{\text{ V}}$ and ${\text{E}}_{{\text{Au}}}^{\text{0}} = 1.4{\text{ V}}$.
The expression for standard potential of the reaction is as follows:
${\text{E}}_{{\text{reaction}}}^{\text{0}} = {\text{E}}_{{\text{reduction}}}^{\text{0}} - {\text{E}}_{{\text{oxidation}}}^{\text{0}}$
Substitute $0.77{\text{ V}}$ for the standard reduction potential, $1.4{\text{ V}}$ for the standard oxidation potential and solve for the standard potential of the reaction. Thus,
${\text{E}}_{{\text{reaction}}}^{\text{0}} = 0.77{\text{ V}} - 1.4{\text{ V}}$
${\text{E}}_{{\text{reaction}}}^{\text{0}} = - 0.63{\text{ V}}$
Thus, the standard potential of the reaction is $ - 0.63{\text{ V}}$. The value of standard potential of the reaction is negative. Thus, the reaction is nonspontaneous and does not occur under standard conditions.
Thus, reduction of ${\text{F}}{{\text{e}}^{3 + }}$ to ${\text{F}}{{\text{e}}^{2 + }}$ by ${\text{A}}{{\text{u}}_{\left( {\text{s}} \right)}}$ does not occur under standard conditions.

Note:The spontaneous reaction does not require an external source to occur. The positive value of ${\text{E}}_{{\text{reaction}}}^{\text{0}}$ indicates that the cell is feasible. The non-spontaneous reaction requires an external source to occur. The negative value of ${\text{E}}_{{\text{reaction}}}^{\text{0}}$ indicates that the cell is not feasible.