Given, that E°(O2/H2O) = +1.23 V; E°(S2O8^2-/SO4^2-) = 2.05 V; E°(Br2/Br^-) = +1.09 V, E°(Au^3+/Au)

Q: What if oxidation potentials were given instead of reduction potentials?

If oxidation potentials are given, you must first convert them to reduction potentials by reversing their signs ($E^\circ_{\text{red}} = -E^\circ_{\text{ox}}$), and then compare. The highest reduction potential still indicates the strongest oxidising agent.

Q: Why does a higher reduction potential mean a stronger oxidising agent?

A higher reduction potential means the species has a very strong affinity for electrons. By readily accepting electrons, it gets reduced itself, thereby forcing another species to lose electrons (oxidising them).

Q: Can a species with a negative reduction potential be an oxidising agent?

Yes, but it would be a very weak oxidising agent. It would prefer to act as a reducing agent (getting oxidised) rather than gaining electrons.

Q: How do we know these are reduction potentials?

The notation $E^\circ_{A^{n+}/A}$ represents the process $A^{n+} + ne^- \rightarrow A$, which is a reduction process (gain of electrons). Hence, these are standard reduction potentials.

Q: What is the strongest reducing agent among these?

The strongest reducing agent would be the product species of the half-reaction with the lowest reduction potential. Here, $Br^-$ would be the strongest reducing agent among the product species.

Question

Given, that $E_{O_2/H_2O}^{\circ} = +1.23	ext{ V}$; $E_{S_2O_8^{2-}/SO_4^{2-}}^{\circ} = 2.05	ext{ V}$; $E_{Br_2/Br^-}^{\circ} = +1.09	ext{ V}$, $E_{Au^{3+}/Au}^{\circ} = +1.4	ext{ V}$. The strongest oxidising agent is

Options:

(A) $Au^{3+}$

(B) $O_2$

(C) $S_2O_8^{2-}$

(D) $Br_2$

Sigma Percentile · Accepted Answer

### Decoding the Electrochemical Series

Imagine you are standing in front of a lineup of chemical species, each eager to grab electrons. Some are desperate, while others are relatively indifferent. How do we measure this desperation? We use the **Standard Reduction Potential ($E^\circ$)**.

In this problem, we are given the $E^\circ$ values for four different half-cells. Our mission is to identify the strongest oxidising agent among them. To do this, we must first understand what an oxidising agent truly is.

### The Master Concept: Oxidising Power

An oxidising agent is a chemical ninja that steals electrons from other species. By taking those electrons, it oxidises the other substance while getting **reduced** itself.

Therefore, the strength of an oxidising agent is directly proportional to its own tendency to undergo reduction. This tendency is exactly what the Standard Reduction Potential measures.

**The Golden Rule:** The higher the positive value of the Standard Reduction Potential ($E^\circ_{	ext{SRP}}$), the greater the tendency of the species to accept electrons, and consequently, the stronger it acts as an oxidising agent.

### Comparing the Contenders

Let's line up the given $E^\circ$ values and see who comes out on top:

$$E^\circ_{Br_2/Br^-} = +1.09	ext{ V}$$

$$E^\circ_{O_2/H_2O} = +1.23	ext{ V}$$

$$E^\circ_{Au^{3+}/Au} = +1.40	ext{ V}$$

$$E^\circ_{S_2O_8^{2-}/SO_4^{2-}} = +2.05	ext{ V}$$

When we arrange these values in decreasing order, the hierarchy becomes crystal clear:

$$+2.05	ext{ V} > +1.40	ext{ V} > +1.23	ext{ V} > +1.09	ext{ V}$$

### The Final Verdict

The peroxodisulfate ion ($S_2O_8^{2-}$) boasts the highest standard reduction potential at $+2.05	ext{ V}$. This massive positive value indicates an overwhelming thermodynamic drive to accept electrons and convert into sulfate ions ($SO_4^{2-}$).

Because it is the most eager to get reduced, $S_2O_8^{2-}$ is undeniably the **strongest oxidising agent** in this group.

If we were to rank them all, the decreasing order of oxidising power would perfectly mirror their $E^\circ$ values:

$$S_2O_8^{2-} > Au^{3+} > O_2 > Br_2$$

Always remember to verify that the potentials provided in the exam are indeed reduction potentials before making your comparison!

Animated Solution for Chemistry - Electrochemistry: Given, that $E_{O_{2} / H_{2} O}^{\circ} = + 1.23 V$ ; $E_{S_{2} O_{8}^{2 -} / S O_{4}^{2 -}}^{\circ} = 2.05 V$ ; $E_{B r_{2} / B r^{-}}^{\circ} = + 1.09 V$ , $E_{A u^{3 +} / A u}^{\circ} = + 1.4 V$ . The strongest oxidising agent is

Select Answer:

Visualized Solution

$E^{\circ} Values$

$Oxidising Power \propto E_{SRP}^{\circ}$

$Comparing E^{\circ} Values$

$Identifying the Strongest$

$Conclusion$

$Decreasing Order$

The Sigma Insight: Electrochemical Series, Electrode Potential and EMF

Decoding the Electrochemical Series

The Master Concept

Comparing the Contenders

The Final Verdict

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