NEET 2016 Phase 2 · ChemistryGalvanic & EMFPrevious Year Question

If the $E_{ce l l}^{\circ}$ for a given reaction has a negative value, which of the following gives the correct relationships for the values of $Δ G^{\circ}$ and $K_{e q}$ ?

Answer: (A) $Δ G^{\circ} > 0$ ; $K_{e q} < 1$ . \textbf{Answer:} (A) $Δ G^{\circ} > 0$ ; $K_{e q} < 1$ \textbf{Solution:} The standard Gibbs energy is related to cell potential by $Δ G^{\circ} = - n F E_{ce l l}^{\circ} .$ Since $E_{ce l l}^{\circ}$ is negative, $Δ G^{\circ}$ becomes positive ( $Δ G^{\circ} > 0$ ).

A. $Δ G^{\circ} > 0$ ; $K_{e q} < 1$ ✓
B. $Δ G^{\circ} > 0$ ; $K_{e q} > 1$
C. $Δ G^{\circ} < 0$ ; $K_{e q} > 1$
D. $Δ G^{\circ} < 0$ ; $K_{e q} < 1$

Correct Answer

(A) $Δ G^{\circ} > 0$ ; $K_{e q} < 1$

Solution & Explanation

\textbf{Answer:} (A) $Δ G^{\circ} > 0$ ; $K_{e q} < 1$ \textbf{Solution:} The standard Gibbs energy is related to cell potential by $Δ G^{\circ} = - n F E_{ce l l}^{\circ} .$ Since $E_{ce l l}^{\circ}$ is negative, $Δ G^{\circ}$ becomes positive ( $Δ G^{\circ} > 0$ ). Also, $Δ G^{\circ} = - R T ln K_{e q}$ . A positive $Δ G^{\circ}$ means $ln K_{e q} < 0$ , hence $K_{e q} < 1$ . Therefore $Δ G^{\circ} > 0$ and $K_{e q} < 1$ .

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If the $E_{ce l l}^{\circ}$ for a given reaction has a negative value, which of the following gives the correct relationships for the values of $Δ G^{\circ}$ and $K_{e q}$ ?