Does current lead voltage during discharge in capacitors?
Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is differentiate voltage, according to:

$$ i = C\frac{dv}{dt} $$

From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:

$$ \require{cancel} \begin{align}
1A &amp;= 1F \frac{dv}{dt} \\
1A &amp;= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\
\frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &amp;= \frac{dv}{dt} \\
\frac{1V}{s} &amp;= \frac{dv}{dt}
\end{align} $$

If you consider the case where the applied voltage is sinusoidal, then so too is the current:

$$ \begin{align}
i &amp;= C\frac{dv}{dt} \\
i &amp;= C\frac{d\sin(t)}{dt} \\
i &amp;= C\cos(t)
\end{align} $$

because \$\cos\$ is the derivative of \$\sin\$.

You will also see if you graph these functions, that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it: