Question

If $E_{cell}^o$ for a given reaction is negative, which gives the correct relationships for the values of $\Delta G^o$ and $K_{eq.}$?

• A. $\Delta G^o>0,K_{eq.}<1$
• B. $\Delta G^o>0,K_{eq.}>1$
• C. $\Delta G^o<0,K_{eq.}>1$
• D. $\Delta G^o<0,K_{eq.}<1$

Answer 1

Answer: (A)

$\Delta G^o>0,K_{eq.}<1$

$Explanation:$

From the relation between change in free energy$\left(\Delta G\right)$ and equilibrium constant $\left(K_{eq}\right)$ we have :

$\Delta G=-RTln{K}_{eq}$ $\to \left(1\right)$

where:

$\Delta G=$ The change in free energy

$R=$ Gas constant

$T=$ The absolute temperature

$K_{eq}=$ Equilibrium constant

From the relation between change in energy $\Delta G$ and $E_{cell}\left(E^o\right)$ we have :

$\Delta G=-nF{E}^o$ $\to \left(2\right)$

$\Delta G=$ The change in free energy

$E^o=E_{cell}$

$n=$ moles of e- from balanced redox reaction

$F=$ Faraday's constant

From equation $\left(2\right)$ if $E^o<0$ then $\Delta G>0$$\to \left(3\right)$

If then $lnK<1$ and then $K_{eq}<1$ that is positive $\to \left(4\right)$

From $\left(3\right)$ and $\left(4\right)$ we get that

$\Delta G>0$ and $K_{eq}<1$

Hence the correct answer is option $A$.