Question

If the ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ for a given reaction has a negative value, then which of the following gives the correct relationships for the values of ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}$and ${{\text{K}}_{{\text{eq}}}}$?
A) ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}} < 0;{{\text{K}}_{{\text{eq}}}} > 1$
B) ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}} < 0;{{\text{K}}_{{\text{eq}}}} < 1$
C) ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}} > 0;{{\text{K}}_{{\text{eq}}}} < 1$
D) ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}} > 0;{{\text{K}}_{{\text{eq}}}} > 1$

Answer 1

Using the equation related to ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ and ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}$ determine the sign of ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}$ value for a negative value of ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ . The negative sign of ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}$ indicates it is less than 0 while the positive sign indicates it is greater than 0.
Similarly using the relation between ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ and ${{\text{K}}_{{\text{eq}}}}$ determine the sign of ${\text{ln}}{{\text{K}}_{{\text{eq}}}}$value. Using the sign ${\text{ln}}{{\text{K}}_{{\text{eq}}}}$ value predicts the value of ${{\text{K}}_{{\text{eq}}}}$. Negative value of ${\text{ln}}{{\text{K}}_{{\text{eq}}}}$indicates ${{\text{K}}_{{\text{eq}}}} < 1$while the positive value of ${\text{ln}}{{\text{K}}_{{\text{eq}}}}$ indicates ${{\text{K}}_{{\text{eq}}}} > 1$

Formula Used: ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}{\text{ = - nF}}{{\text{E}}^{\text{0}}}_{{\text{cell}}}$
${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}{\text{ = - RTln}}{{\text{K}}_{{\text{eq}}}}$

Complete step by step answer:
The standard electrode potential of a metal may be defined as the potential difference in volts developed in a cell consisting of two electrodes, the pure metal in contact with a molar solution of one of its own ions and the normal hydrogen electrode.
The equation related to standard electrode potential and Gibbs free energy is as follows:
${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}{\text{ = - nF}}{{\text{E}}^{\text{0}}}_{{\text{cell}}}$
Here,
${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}$ = Gibbs free energy
n= number of electrons transfer
F = Faraday constant
${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ = standard electrode potential
So, if the ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ for a given reaction has a negative value then the value of ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}$ would be positive.
The positive value of ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}$ indicates that it is greater than 0.
So, for the given reaction ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}} > 0$.

Now, using the relation between ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ and ${{\text{K}}_{{\text{eq}}}}$ we can determine the sign of ${{\text{K}}_{{\text{eq}}}}$ as follows:
We know,
${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}{\text{ = - nF}}{{\text{E}}^{\text{0}}}_{{\text{cell}}}$
And
${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}{\text{ = - RTln}}{{\text{K}}_{{\text{eq}}}}$
So,
${\text{ - nF}}{{\text{E}}^{\text{0}}}_{{\text{cell}}}{\text{ = - RTln}}{{\text{K}}_{{\text{eq}}}}$
So, if the ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ for a given reaction has a negative value then the value of ${\text{ln}}{{\text{K}}_{{\text{eq}}}}$ would be negative.
A negative value of ${\text{ln}}{{\text{K}}_{{\text{eq}}}}$ indicates ${{\text{K}}_{{\text{eq}}}} < 1$.
Thus, if the ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ for a given reaction has a negative value then ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}} > 0$ and ${{\text{K}}_{{\text{eq}}}} < 1$.

Hence, the correct option is (C) ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}} > 0;{{\text{K}}_{{\text{eq}}}} < 1$

Note: The sign of Gibbs free energy indicates the spontaneity of the reaction. The signs of ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ and ${\text{$\Delta$ }}{{\text{G}}^{\text{0}}}$ are always opposite to each other. The value of ${{\text{K}}_{{\text{eq}}}} < 1$ when ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ is negative and ${{\text{K}}_{{\text{eq}}}} > 1$ when ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ is positive.