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Step-by-Step Explanation
Step 1: Recall the relationship between standard cell potential and standard Gibbs energy
The standard Gibbs free energy change $ \Delta G^o $ is related to the standard cell potential $ E^o_{\text{cell}} $ by the equation:
$ \Delta G^o = -\,n F\,E^o_{\text{cell}} $
Here,
$n$ is the number of moles of electrons transferred,
$F$ is the Faraday constant ($ \approx 96500\, \text{C mol}^{-1} $),
$E^o_{\text{cell}}$ is the standard cell potential.
A negative value of $ E^o_{\text{cell}} $ makes $ -n F E^o_{\text{cell}} $ positive, hence $ \Delta G^o $ becomes positive.
Step 2: Understand the sign of $ \Delta G^o $ and spontaneity
When $ \Delta G^o > 0 $, the reaction is non-spontaneous under standard conditions.
Step 3: Relate $ \Delta G^o $ to the equilibrium constant
The standard Gibbs free energy change can also be written as:
$ \Delta G^o = -\,R T \ln K_{\text{eq}} $
where
$R$ is the universal gas constant,
$T$ is the absolute temperature,
$K_{\text{eq}}$ is the equilibrium constant of the reaction.
If $ \Delta G^o $ is positive ($ > 0 $), the term $ -RT \ln K_{\text{eq}} $ is positive. This implies that $ \ln K_{\text{eq}} $ is negative, hence $ K_{\text{eq}} < 1 $.
Step 4: Conclude the correct relationship
Because $ E^o_{\text{cell}} $ is negative, $ \Delta G^o $ becomes positive and $ K_{\text{eq}} < 1 $. Therefore, the correct relationship is:
$ \Delta G^o > 0 \quad \text{and} \quad K_{\text{eq}} < 1.
