Spontaneity of reversible chemical reactions - Both sides?
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We know the formula $\Delta G=\Delta H-T\Delta S$ and also know that if for the left reaction $\Delta H_\text{r}=c$, then for the other-way reaction $\Delta H_\text{r}=-c$ (because $\Delta H_\text{r}=\sum\Delta H_\text{f} \text{(product)}-\sum\Delta H_\text{f}\text{(reactant)}$)

Same for the entropy change $\Delta S$: If for the left reaction $\Delta S_\text{r}=c$, then for the other-way reaction $\Delta S_\text{r}=-c$ (because $\Delta S_\text{r}=\sum\Delta S_\text{f}\text{(product)}-\sum\Delta S_\text{f}\text{(reactant)}$).

So, for the reverse of a reaction, since both $\Delta H_\text{r}$ and $\Delta S_\text{r}$ are opposite, the whole $\Delta G_\text{r}$ will be the opposite, hence at equilibrium $\Delta G=0$, so that neither way is thermodynamically favoured.

In fact, how much a reaction (warning: reversible reactions isn’t the same as reversible processes, not to be confused) is reversible and the direction it proceeds is determined not only from thermodynamics (I would say much more from other factors), but from Le Chatelier’s principle (temperature, pressure, concentration), catalysts and of course activation energy.

So, we have seen the relationship of $\Delta G_\text{r}$ of a reaction and its reverse and the fact that at equilibrium $\Delta G=0$.

It must be noted that:

$\Delta G_\text{r}$, $\Delta H_\text{r}$, $\Delta S_\text{r}$ refer to “reaction” (r)

and

$\Delta G_\text{f}$, $\Delta H_\text{f}$, $\Delta S_\text{f}$ refer to “formation” (f) (of chemical compounds and elements)