How to know the ionization power of an unknown metal? - difficult

Imagine having the 4 following beakers:


[![enter image description here](https://i.stack.imgur.com/rJSt8.png)](https://i.stack.imgur.com/rJSt8.png)


where P, Q, R and S are metallic plates. In beaker 1 and beaker 2 nothing happens. However in beaker 3 and 4 the metalic pieces become thinner.


How would you orden the ions $P^{p+}, Q^{q+}, s^{s+}, R^{r+}$ the following ions by oxidizing power?


I think it is one of the following possibilities:


* $P^{p+} < Q^{q+} < s^{s+} < R^{r+}$


I think that P and Q clearly have the highest oxidizing power as in beaker 3 and 4 the metallic plates become thinner. But I have no idea whether R has a higher oxidizing power than S. Or is it the other way around?


Could someone help me and explain how I can order these ions by oxidizing power?




I am guessing you know when a metal bar is becoming thinner, the metal is being oxidised, i.e $$ \ce{M\_{(s)}} \rightarrow \ce{M^{n+}\_{(aq)}}$$


$P^{p+}, Q^{q+}, s^{s+}, R^{r+}$
Let's start with beaker 3 and beaker 4


In beaker 3, you have $\ce{S\_{(s)}}$ in $\ce{R^{r+}\_{(aq)}}$, and you observer the metal bar becomes thinner. Which means, $\ce{S\_{(s)}}$ is being oxidised by $\ce{R^{r+}\_{(aq)}}$, or in other words, the reduction potential of $\ce{R^{r+}\_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{S^{s+}\_{(aq)}}$ 


$$\ce{R^{r+}\_{(aq)}} > \ce{S^{s+}\_{(aq)}}$$ 


In beaker 4, you have $\ce{Q\_{(s)}}$ in $\ce{S^{s+}\_{(aq)}}$, and again the same reasoning applies. 


$\ce{Q\_{(s)}}$ is being oxidised by $\ce{S^{s+}\_{(aq)}}$, or in other words, the reduction potential of $\ce{S^{s+}\_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{Q^{q+}\_{(aq)}}$ 


$$\ce{S^{s+}\_{(aq)}} > \ce{Q^{q+}\_{(aq)}}$$ 


Thus, so far we have: 


$$\ce{R^{r+}\_{(aq)}} > \ce{S^{s+}\_{(aq)}} > \ce{Q^{q+}\_{(aq)}}$$ 


Let's look at beaker 1 and 2


In beaker 1 you have $\ce{P\_{(s)}}$ in $\ce{Q^{q+}\_{(aq)}}$, and oxidation doesn't take place. Which means the reduction potential of of $\ce{Q^{q+}\_{(aq)}}$ is less than that of $\ce{P^{p+}\_{(aq)}}$


$$\ce{P^{p+}\_{(aq)}} > \ce{Q^{q+}\_{(aq)}}$$ 


And similarly, in beaker 2 you have $\ce{P\_{(s)}}$ in $\ce{R^{r+}\_{(aq)}}$, and nothing happens. Which means that the Which means the reduction potential of of $\ce{R^{r+}\_{(aq)}}$ is less than that of $\ce{P^{p+}\_{(aq)}}$


$$\ce{P^{p+}\_{(aq)}} > \ce{R^{r+}\_{(aq)}}$$ 


So the final order becomes (in terms of reduction potential, or oxidising ability):


$$ \ce{P^{p+}\_{(aq)}} >\ce{R^{r+}\_{(aq)}} > \ce{S^{s+}\_{(aq)}} > \ce{Q^{q+}\_{(aq)}}$$ 


 3