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Question

Write a note on reaction between Gibbs free energy and cell potential for cell reaction.

Vidyadip · Accepted Answer

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

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