A uniform field of 2.0 N C^ -1 exists in space in the x -direction....

The relation between electric potential and electric field is given by $dV = -\vec{E} \cdot d\vec{r}$.

Given $\vec{E} = 2.0 \hat{i} \text{ N C}^{-1}$, the potential difference between a point $(x, y, z)$ and the origin $(0, 0, 0)$ is:

$V(x, y, z) - V(0, 0, 0) = -\int_{0}^{x} E_{x} dx$

$V(x, y, z) - 0 = -\int_{0}^{x} 2.0 dx$

$V(x, y, z) = -2.0 x$

We are given that the potential is $25 \text{ V}$. Substituting this value:

$25 = -2.0 x$

$x = -12.5 \text{ m}$

This equation represents a plane perpendicular to the $x$-axis. Thus, the potential is $25 \text{ V}$ at all points on the plane $x = -12.5 \text{ m}$.