How to identify which of two given electrodes with corresponding solutions is reducing more?
The solution is to immerse each rod in both of the solutions (separately of course). Let's imagine that $\ce{A}$ is the more reducing metal. We can consider what happens in each case:


If we immerse $\ce{A}$ in a solution of $\ce{A^m+}$ nothing will happen.
Similarly, if we immerse $\ce{B}$ in a solution of $\ce{B^n+}$ nothing will happen.
If we immerse $\ce{B}$ in a solution of $\ce{A^m+}$ nothing will happen because the electrode potential for $\ce{A|A^m+}$ is more negative (more negative means its a stronger reducing agent) than for $\ce{B|B^n+}$ so the reaction potential is negative and the reaction is not feasible.
However, if we immerse $\ce{A}$ in a solution of $\ce{B^n+}$, $\ce{B}$ will precipitate out on the surface of $\ce{A}$. This is because $\ce{A|A^m+}$ has a more negative electrode potential than $\ce{B|B^n+}$ and so the reaction potential will be positive. $\ce{A}$ will reduce $\ce{B^n+}$ ions to form $\ce{B}$ metal and $\ce{A}$ will be oxidised to form $\ce{A^m+}$:
$$\ce{A (s) + $\frac{m}{n}$B^n+ (aq) -&gt; A^m+ (aq) + $\frac{m}{n}$B (s)}$$


In short, the most reducing metal is the one which causes precipitation of the other metal out of its solution.