# Gibbs Free Energy And Emf Of A cell

Gibbs Free Energy is defined as the thermodynamic potential that signifies the maximum or reversible work performed by a thermodynamic system at constant temperature and pressure. Work done by electrical power in one second is given as the product of emf of the cell and the total charge passed.

Therefore,

$W$ = $nFE_{(cell)}$

Where,

$w$ = work done,

$nF$ = total charge passed and

$E_{(cell)}$ = emf of the cell

When the charge is passed reversibly through the galvanic cell, it is observed that maximum work is done by the galvanic cell. This reversible work done by the galvanic cell is related to the decrease in Gibbs energy of the reaction.

$Δ_r~G$ = $-W$

$Δ_r~G$ = $-nFE_{(cell)}$

The above equation can also be used to calculate the standard cell potential. When concentration of all the reacting species is unity then E(cell) = Eo(cell) (emf of a cell is equal to standard cell potential).

$\Rightarrow~~Δ_r~G^o$ = $-nFE^o_{(cell)}$

Gibbs energy of a reaction is an Extensive Thermodynamic Property which means its value depends on n. Thus, for two cell reactions involving different values of n we observe different values of Gibbs free energy of the reaction. For example:

$Zn(s)~ +~ Cu^{2+}~(aq) ~\rightarrow ~Zn^{2+}~(aq)~ + ~Cu(s)$

$Δ_r~G$ = $- 2FE_{(cell)}$

Whereas after multiplying RHS and LHS by 2 we notice,

$2 ~Zn(s)~ + 2Cu_{2+}~(aq) ~\rightarrow~2 Zn^{2+}~(aq)~ +2 ~Cu(s)$

$Δ_r~G$ = $- 4FE_{(cell)}$

According to thermodynamics, Gibbs energy of a reaction can be related to the reaction quotient and when the reaction is at equilibrium it can be related to the equilibrium constant. Since, Gibbs energy of a reaction also depends on the emf of a cell we can relate the equilibrium constant of a cell reaction with the standard cell potential. At equilibrium,

$Δ_r~G^o$ = $-nFE^o_{(cell)}$

$Δ_rG^o$$-~ RT~ ln~ K$

$-nFE^o_{(cell)}$ = $-~ RT~ ln~ K$

$E^o_{(cell)}$ = $\frac{RT}{nF}~ln ~K$

Where,

$K$ = equilibrium constant

$R$ = universal gas constant

From the above equation, we can infer:

For K > 1,$E^o_{(cell)}$> 0, the reaction is spontaneous and favors the formation of product.

For K < 1,$E^o_{(cell)}$&<0, the reaction is non-spontaneous and favors the formation of reactant.

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#### Practise This Question

The half-cell reactions for rusting of iron are,
2H++12O2+2e2H2O,  E=+1.23V,
Fe2++2eFe(s);   E=0.44V.
ΔG (in kJ) for the reaction is :