Why does the Gibbs free energy only correspond to non-expansion work?
The association of the Gibbs free energy with “additional”, or “non-expansion” work, is simply a mathematical result of its definition: $G = H - TS = U + pV - TS$.

From the First Law, for a closed system, we have $\mathrm{d}U = \delta q + \delta w$.

For a reversible change, the Second Law tells us that $\mathrm{d}S = \delta q_\text{rev}/T$; so $\delta q_\text{rev} = T\,\mathrm{d}S$.

We can split $\mathrm{d}w$ into two parts, expansion work and additional work: expansion work is given by the formula $\mathrm{d}w_{\text{exp}} = -p\,\mathrm{d}V$, and for now we can just label additional work as $\mathrm{d}w_{\text{add}}$. In this context, additional simply means anything that isn’t expansion work. Putting all this together, we have

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}}$$

and from the definition of the Gibbs free energy in the first paragraph,

$$\begin{align}
\mathrm{d}G &amp;= \mathrm{d}(U + pV - TS)\\
&amp;= \mathrm{d}U + (p\,\mathrm{d}V + V\,\mathrm{d}p) - (T\,\mathrm{d}S + S\,\mathrm{d}T)\\
&amp;= T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}} + p\,\mathrm{d}V + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T\\
&amp;= V\,\mathrm{d}p - S\,\mathrm{d}T + \delta w_{\text{add}}
\end{align}$$

Now, if a closed system is kept at constant pressure and temperature, $\mathrm{d}p = 0$ and $\mathrm{d}T = 0$. Therefore $\mathrm{d}G = \delta w_{\text{add}}$.

The simplest application of this interpretation of $G$ is in electrochemistry, where one could move $|\mathrm{d}n|$ moles of electrons through a certain potential difference $E$; the “additional work” done in this case is given by $-FE\,|\mathrm{d}n|$ (you could consult a physics text for details). This gives us $\mathrm{d}G = -FE\,|\mathrm{d}n|$, and (after several more steps) the familiar equation $\Delta_\mathrm{r} G = -zFE$.