Question

Question #43843

Answer 1

See my comments below...

Explanation:

We did not get the specific reactions that are to be worked out, but here is how you do it:

All spontaneous reactions will have a positive reaction potential. This potential is calculated by subtracting the potential for the oxidation from that of the reduction.

When you select any two electrode potentials from your list, and attempt a reaction based on these metal | metallic ion pairs, the higher value will always serve as the reduction in a spontaneous reaction (and the lower value, the oxidation of course).

However, I have serious concerns about the values you are reporting in this problem. It becomes unnecessarily confusing to use both oxidation potentials and reduction potentials (to say nothing of redundant). You should stick to a table of reduction potentials only.

Your first two values are okay, (Zn and Fe) but the two Cu values that follow should be positive. The fifth and sixth values are incorrect, as they do not relate to the metal in each case, but to the change between the two oxidation states for the ions `(Cu^+ | Cu^(2+) and Fe^(2+) | Fe^(3+))`. Finally, the value for Ag is +0.80 V.

Using these corrected values, `Ag | Ag^+` is +0.80 V, and for `Cu | Cu^(2+)` is +0.33 V.

Now for an example, in the combination of `Cu | Cu^(2+)` with `Ag | Ag^+`, we obtain a positive result be subtracting as follows:

(+0.80 V) - (+0.33 V) = +0.47 V

Since this positive result is obtained by subtracting the Cu electrode potential from the Ag, it must be that `Ag+` is reducing, while `Cu` is oxidizing:

`Ag^+ + e^- rarr Ag`

`Cu rarr Cu^(2+) + 2e^-`

This reaction (`2Ag^+ + Cu rarr Cu^(2+) + 2Ag`) is spontaneous, and the reaction potential is +0.47 V as determined above.