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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 $1.0\ \mathrm{mol\ dm^{-3}}$ aqueous solution of $\ce{A^{$m$+}}$ and the other bottle containing $1.0\ \mathrm{mol\ dm^{-3}}$ aqueous solution of $\ce{B^{$n$+}}$.

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


These are my thoughts:

  • 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 get reduced more, yet I know that anode get 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.

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  • $\begingroup$ @Burak Ulgut : No I am allow to use only those given things. $\endgroup$ – On the way to success Jun 15 '15 at 12:56
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    $\begingroup$ This seems like a homework question. We ‎have a policy which states that you should show your thoughts and/or efforts into solving the ‎problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this ‎question may get closed. $\endgroup$ – bon Jun 15 '15 at 16:37
  • $\begingroup$ @ bon: This is not totally a home-work question.But some sort of a thing that I want to do .Like an assignment in lab. $\endgroup$ – On the way to success Jun 16 '15 at 0:48
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    $\begingroup$ Also, just for the record, this is a homework question but it is a good homework question :). A "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself. This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc. $\endgroup$ – bon Jun 16 '15 at 14:41
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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 $\ce{A^{m+}}$

$$\ce{A(s) + \frac{m}{n}B^{n+}(aq) -> A^{m+}(aq) + \frac{m}{n}B(s)}$$

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

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  • $\begingroup$ Thanks a lot. But what happen if I immerse the two metal rods in same solutions, let's say we immerse them in a solution of $B^{n+}$ ( assuming A is more reducing). Then would there be the same scenario as you explained ? $\endgroup$ – On the way to success Jun 17 '15 at 0:22
  • $\begingroup$ I have one more doubt about what you have said.How can you say "so the reaction potential is negative and the reaction is not feasible." $\endgroup$ – On the way to success Jun 17 '15 at 0:28
  • $\begingroup$ For the first comment, do you mean what happens if you immerse both rods in one solution at the same time? $\endgroup$ – bon Jun 17 '15 at 10:12
  • $\begingroup$ For the second, the reaction (or cell) potential is related to the Gibbs energy change. If the reaction potential is negative, $\Delta G$ is positive, which means that the equilibrium constant is less than one. Technically this means that some precipitation will still occur, particularly if the cell potential is close to zero but broadly speaking, if the reaction potential is negative, the reaction is unfeasible. This gives the relationship between the three quantities. $\endgroup$ – bon Jun 17 '15 at 10:13

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