Question
Derive the relationship between Gibbs energy of cell reaction and cell potential.

Answer

1. The electrical work done in a galvanic cell is the electricity (charge) passed multiplied by the cell potential.
Electrical work $=$ amount of charge $\times$ cell potential
2. Charge of one mole electrons is $F$ coulombs.
For the cell reaction involving $n$ moles of electrons, Charge passed $= nF$ coulombs
3. Hence, electrical work $= nFE _{\text {cell }}$
Electrical work done in a galvanic cell is equal to the decrease in Gibbs energy, $-\Delta G$, of cell reaction.
It then follows that
Electrical work done $=-\Delta G$
4. Therefore, $-\Delta G=n F_{\text {cell }}$
or
$\Delta G =- nFE _{\text {cell }}$
Under standard state conditions, we write
$\Delta G ^{\circ}=- nFE _{\text {cell }}^0$

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