Bihar BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry2 Marks
Question
Write a note on reaction between Gibbs free energy and cell potential for cell reaction.
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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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