How to identify which of two given electrodes with corresponding solutions is reducing more? For the sake of the question, let’s assume that you are provided with two labelled rods of the pure metals $\ce{A}$ and $\ce{B}$ and two unlabelled bottles containing $\pu{1.0 mol dm-3}$ aqueous solution of $\ce{A^m+}$ and  the other bottle containing $\pu{1.0 mol dm-3}$ aqueous solution of $\ce{B^n+}$.

I want to find out, which metal out of the given two is more reducing, and identify each solution that is given in the unlabelled bottles. What are the procedures that I should follow?

If I immerse $\ce{A}$ in a solution of  $\ce{A^m+}$, I guess  there will be no change. Because if I do so there would be an equilibrium prevail with the ions in the solution and the electrode, and I think this is true for $\ce{B}$ also.

But I don't know how to identify the one which gets reduced more, yet I know that the anode gets reduced. So I am confused about that.

I feel that the only clue I have is immersing $\ce{A}$ in  $\ce{B^n+}$ and vice versa for $\ce{B}$ also. But  really I don't know what  would happen if I do so.
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.