1. The problem states that a pre-image is located entirely in the second quadrant. 2. The second quadrant is where $x < 0$ and $y > 0$. 3. Reflecting a point over the y-axis changes the sign of the $x$-coordinate but keeps the $y$-coordinate the same. 4. If the original point is $(x, y)$ with $x < 0$ and $y > 0$, after reflection over the y-axis, the new point is $(-x, y)$. 5. Since $x < 0$, $-x > 0$, so the new $x$-coordinate is positive and the $y$-coordinate remains positive. 6. Points with $x > 0$ and $y > 0$ lie in the first quadrant. 7. Therefore, the reflected image will be in the first quadrant.