Question

For which of these oxidation/reduction pairs will the reduction potential vary with pH?
I. $AmO_2^{2 + }/AmO_2^ +$ II. $AmO_2^{2 + }/A{m^{4 + }}$ III. $A{m^{4 + }}/A{m^{2 + }}$.
A) I only
B) II only
C) I and II only
D) I, II, and III

Answer 1

In the question, the three reaction terms are given and one should analyze the reaction conversions and balance the reaction accordingly. The pH of a reaction depends only on the concentration of the hydrogen ions and one needs to find out in which reaction the hydrogen ion term is present.

Complete step by step answer:

1. First of all we will analyze each oxidation/reduction pair one by one and decide which reaction will vary in pH with the reduction potential. Let's see the first reaction $AmO_2^{2 + }/AmO_2^{2 + }$ and write a balanced reaction for the same as follows,
   $Am{O_2}^{2 + } + 1{e^ - } \to Am{O_2}^ +$
   In the above reaction, there is one gain of the electron which makes the ${\text{ + 2}}$ of the reactant in ${\text{ + 1}}$ charge which is on the product. In this reaction, there is no involvement of the hydrogen ion that is ${H^ + }$ which means this oxidation/reduction pair reduction potential will not vary with the change in the pH which shows this option is an incorrect choice of answer.
2. Now let's analyze the reaction $AmO_2^{2 + }/A{m^{4 + }}$ and write a balanced reaction for the same as follows,
   $Am{O_2}^{2 + } + 4{H^ + } + 2{e^ - } \to A{m^{4 + }} + 2{H_2}O$
   In the above reaction, there is a gain of two electrons and the formation of the two water molecules. To balance out the formed water molecules there are four hydrogen ions added to the reactant side. As there are four hydrogen ions present that is ${H^ + }$ which makes this oxidation/reduction pair reduction potential will vary with the change in the pH hence this option is the correct choice of answer.
3. Now let's write a balanced equation for the $A{m^{4 + }}/A{m^{2 + }}$ as follows,
   $A{m^{4 + }} + 2{e^ - } \to A{m^{2 + }}$
   In the above reaction, there is a gain of two electrons which makes the ${\text{ + 4}}$ of the reactant in ${\text{ + 2}}$ charge which is on the product. In this reaction, there is no involvement of the hydrogen ion that is ${H^ + }$ which means this oxidation/reduction pair reduction potential will not vary with the change in the pH which shows this option is an incorrect choice of answer.
4. Therefore, only the II pair varies with the pH

Therefore, the correct option is B

Note:
According to the logarithmic scale of pH as the ${H^ + }$ ion concentration is dependent on the power of ${\text{10}}$. This relativity between the ion and the pH is due to the effect of hydrogen ions and hydroxyl ions on pH. As the higher the concentration of hydrogen ions the lower will be the pH, and the lower the concentration of hydrogen ions, the lower will be the pH.