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) \[{\text{$\Delta$ }}{{\text{G}}^{\text{0}}} > 0;{{\text{K}}_{{\text{eq}}}} > 1\]

Hint: 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.