Expression for Gibb's free energy change for a cell reaction

Question

According to Chemistry Libretexts, the Gibbs free energy change of a cell reaction is related to the cell voltage as
$$ Delta G = -nFE,$$
where $n$ is the number of moles of electrons passed and $F$ is the charge on a mole of electrons.
But $E$ decreases in value through the course of the cell reaction, so  shouldn't
$$Delta G = -int_{q_mathrm{i}}^{q_mathrm{f}} E,mathrm dq,$$
where $q_mathrm{f} - q_mathrm{i} = nF?$

Rahul Verma · Answer

According to Nernst's equation,
$$E = E^o - frac{RT}{nF}ln Q$$
As you can see, $E$ is independent of $q$, therefore your proposed integral becomes,
$$Delta G = - int_{q_i}^{q_f}{Ecdot dq} = -Eint_{q_i}^{q_f}{dq} = -EDelta q = -nFE$$

Alternatively, you can prove the the relation between Gibbs energy and cell potential as,

Multiply Nernst's equation by $nF$, we get,

$$nFE = nFE^o - RTln Q$$

Swapping the sides, we get,

$$-nFE = -nFE^o + RTln Q$$

As we know,

$$Delta G = Delta G^o + RTln Q$$
$$Rightarrow Delta G = -nFE$$

Expression for Gibb's free energy change for a cell reaction

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