Question
Write a note on reaction between Gibbs free energy and cell potential for cell reaction.
✓
Answer
$\rightarrow$ Electrical work done in one second is equal to electrical potential multiplied by total charge passed.
$\rightarrow$ If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly.
$\rightarrow$ The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy.
$\rightarrow$ If the $\text{EMF}$ of the cell is $E$ and $nF$ is the amount of charge passed and $\Delta_{ r } G$ is the Gibbs energy of the reaction, $\Delta_{ r } G =- nF E_{\text {cell }}$
$\rightarrow E ($cell$)$ is an intensive parameter but $\Delta_{ r } G$ is an extensive thermodynamic property and the value depends on $n$ .
$Zn ( s )+ Cu ^{2+}( aq ) \rightarrow Zn ^{2+}( aq )+ Cu ( s )$
$\Delta_{ r } G =-2 FE _{\text {(cell) }}$
but when we write the reaction
$2 Zn ( s )+2 Cu ^{2+}( aq ) \rightarrow 2 Zn ^{2+}( aq )+2 Cu ( s )$
$\Delta_{ r } G =-4 FE _{\text {(cell) }}$
$\rightarrow$ If the concentration of all the reacting species is unity
$\Delta_{ r } G ^{\ominus}=- nF E _{\text {(cell) }}^{\ominus}$
$\rightarrow$ We can calculate Equilibrium constant from the value of $\Delta_r G^{\ominus}$
Where, $ \Delta_{ r } G ^{\ominus}=- RT \ln K =-2.303 RT \log K$
$\Delta G ^{\ominus}=$ Change in gibbs free energy
$R =$ Gas constant
$T =$ Temperature
$K =$ Equilibrium constant
$\rightarrow$ If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly.
$\rightarrow$ The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy.
$\rightarrow$ If the $\text{EMF}$ of the cell is $E$ and $nF$ is the amount of charge passed and $\Delta_{ r } G$ is the Gibbs energy of the reaction, $\Delta_{ r } G =- nF E_{\text {cell }}$
$\rightarrow E ($cell$)$ is an intensive parameter but $\Delta_{ r } G$ is an extensive thermodynamic property and the value depends on $n$ .
$Zn ( s )+ Cu ^{2+}( aq ) \rightarrow Zn ^{2+}( aq )+ Cu ( s )$
$\Delta_{ r } G =-2 FE _{\text {(cell) }}$
but when we write the reaction
$2 Zn ( s )+2 Cu ^{2+}( aq ) \rightarrow 2 Zn ^{2+}( aq )+2 Cu ( s )$
$\Delta_{ r } G =-4 FE _{\text {(cell) }}$
$\rightarrow$ If the concentration of all the reacting species is unity
$\Delta_{ r } G ^{\ominus}=- nF E _{\text {(cell) }}^{\ominus}$
$\rightarrow$ We can calculate Equilibrium constant from the value of $\Delta_r G^{\ominus}$
Where, $ \Delta_{ r } G ^{\ominus}=- RT \ln K =-2.303 RT \log K$
$\Delta G ^{\ominus}=$ Change in gibbs free energy
$R =$ Gas constant
$T =$ Temperature
$K =$ Equilibrium constant
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