Step 1: Write down the standing wave equation.
A stationary wave formed between two fixed nodes can be written as $y(x,t) = 2A\sin(kx)\cos(\omega t)$, where $k = \frac{2\pi}{\lambda}$. Every particle in this wave oscillates with the same time part $\cos(\omega t)$; what changes from particle to particle is only the sign and size of the amplitude term $2A\sin(kx)$.
Step 2: Locate two consecutive nodes.
Nodes are points where the amplitude is zero, that is $\sin(kx) = 0$. Two consecutive nodes are separated by $\Delta x = \frac{\lambda}{2}$, since $\sin(kx)$ passes through half a cycle between them.
Step 3: Check the sign of $\sin(kx)$ within one loop.
Between one node and the next, $\sin(kx)$ keeps the same sign throughout, so every particle in that stretch has the same time behaviour $\cos(\omega t)$ and reaches its extremes together, in phase.
Step 4: Compare particles that straddle a node.
Just past a node, $\sin(kx)$ flips sign. This is the same as multiplying the amplitude by $-1$, which turns $\cos(\omega t)$ into $-\cos(\omega t) = \cos(\omega t + \pi)$. So crossing a node adds exactly $\pi$ to the phase, and this is the largest phase gap that can appear for particles bounded by that pair of nodes.
Final Answer:
Across the full stretch between two consecutive nodes, the phase difference reaches its extreme value of $\pi$. \[ \boxed{\pi} \]