Question

Peroxodisulphate($N{a_2}{S_2}{O_8}$) salts are strong oxidizing agents used as bleaching agents for fats, oils and fabrics. Can oxygen gas oxidise sulphate ion to peroxodisulphate ion (${S_2}O_8^{2 - }$) in acidic solution with the oxygen ${O_2}$ being reduced to water? Given.
1. ${O_2}(g) + 4{H^ + }(aq) + 4e \to 2{H_2}O$ ${E^{\circ}} = 1.23V$
2. ${S_2}O_8^{2 - }(aq) + 2e \to 2SO_4^{2 - }(aq)$ ${E^{\circ}} = 2.01V$

Answer 1

The standard reduction potential is in a category known as the standard cell potentials or standard electrode potentials. The standard cell potential is the potential difference between the cathode and anode. A solution with a higher (more positive) reduction potential than the new species will have a tendency to gain electrons from the new species (to be reduced by oxidizing the new species) and a solution with a lower (more negative) reduction potential will have a tendency to lose electrons to the new species (to be oxidized by reducing the new species).

Complete answer:
$E_{cell}^{\circ}$ is the electromotive force (also called cell voltage or cell potential) between two half-cells. The greater the $E_{cell}^{\circ}$ of a reaction the greater the driving force of electrons through the system, the more likely the reaction will proceed (more spontaneous). $E_{cell}^{\circ}$ is measured in volts (V). The overall voltage of the cell = the half-cell potential of the reduction reaction + the half-cell potential of the oxidation reaction. To simplify,
A redox reaction to be spontaneous or feasible. $E_{cell}^{\circ}$ must be greater than zero .
Since the value of the reduction potential of the second redox reaction (that is of oxygen) is more than of the first reaction, so the second one is the reduction reaction. The first one is the oxidation reaction.
Let’s write anode and cathode reactions for the oxidation of sulphate ions by oxygen.
Reaction at Cathode: ${O_2}(g) + 4{H^ + }(aq) + 4e \to 2{H_2}O$
Reaction of Anode: ${S_2}O_8^{2 - }(aq) + 2e \to 2SO_4^{2 - }(aq)$
Now, to write the total reaction, we will need to multiply anodic reaction by 2 because it involves change of 2 electrons while cathodic reaction involves change of 4 electrons.
Total Reaction: ${O_2}(g) + 4{H^ + }(aq) + 4SO_4^{2 - }(aq) \to 2{H_2}O(l) + 2{S_2}O_8^{2 - }(aq)$
Now, for this reaction we can use the below equation to predict $E_{cell}^{\circ}$.
$E_{cell}^{\circ} = {E^ \oplus } - {E^ - }$
Now, we are given with the standard reduction potential of Cathode as ${E^ \oplus } = 1.23V$ and that of Anode as ${E^ - } = 2.01V$.
So, for this cell we can write that $E_{cell}^{\circ} = 1.23 - 2.01 = - 0.78V$
So $E_{cell}^{\circ} = - 0.78V$
Thus, the resultant potential of this cell will be in less than zero and thus we can say that this reaction will not occur without giving any external energy.
So, we can conclude that oxygen will not oxidise sulphate ion to peroxodisulphate ion in acidic solution.

Note:
The potential of an oxidation reduction (loss of electron) is the negative of the potential for a reduction potential (gain of electron). Most tables only record the standard reduction half-reactions. In other words, most tables only record the standard reduction potential; so in order to find the standard oxidation potential, simply reverse the sign of the standard reduction potential. The more positive reduction potential of reduction reactions are more spontaneous. When viewing a cell reduction potential table, the higher the cell is on the table, the higher potential it has as an oxidizing agent.