4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.