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

Imagine having the 4 following beakers:

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?

• Hint: pay attention to the solution in the beaker – getafix Aug 25 '16 at 15:19
• @getafix it's not that I don't want to think about it or so. But I have already posted this twice. Have never received an answer untill now. I have to understand this for tomorrow. Would you mind explaining please? – privetDruzia Aug 25 '16 at 15:56
• I guess you made an attempt to answer your question, so it complies with the homework policy on Chemistry.SE. I hope I'm not doing your homework for you – getafix Aug 25 '16 at 16:32

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):

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

• Thank you so much! I have been thinking about it quite some time! Just one clarification. If the reduction potential of the solids in beaker 1 is higher than the liquid. Why doesn't the metal piece become thicker/bigger? – privetDruzia Aug 25 '16 at 17:59
• en.wikipedia.org/wiki/Electroplating You need to supply electrons for that to happen. – getafix Aug 25 '16 at 20:06