Say I have the following (where <space> indicates an actual whitespace character)
Line 1
Line 2<space>
Line 3
Line 4<space>
Line 5
and I wanted it to be
Line 1/
Line 2/
Line 3/
Line 4/
Line 5/
I initially thought that maybe s/\s*$/\//g would work but then I ended up with the following as the solution
Line 1/
Line 2//
Line 3/
Line 4//
Line 5/
I can't think of a way to include the case where there is a whitespace character at the end that I want replaced.
\s*$ can match a string ending in a space twice; once to capture the actual space character, and again to match the end of string (see for example this Q&A). This is why you are getting two replacements on the strings ending in space. To avoid this problem, use a positive lookbehind for a non-space character; that prevents the end of string that is preceded by a space being matched again. In normal regex you would match with:
(?<=\S)\s*$
and replace with a /. Demo on regex101.
In vim you can use the \zs marker to define the start of the match e.g.
:%s/\S\zs\s*$/\//
