Animated Solution for Chemistry - Electrochemistry: Given, $E_{C{r}^{3+}/Cr}^{\circ }=-0.74\text{ V}$; $E_{Mn{O}_4^{-}/M{n}^{2+}}^{\circ }=1.51\text{ V}$ $E_{C{r}_2{O}_7^{2-}/C{r}^{3+}}^{\circ }=1.33\text{ V}$; $E_{Cl/C{l}^{-}}^{\circ }=1.36\text{ V}$ Based on the data given above, strongest oxidising agent will be

In the fascinating world of electrochemistry, chemical species are constantly battling for electrons. Some are desperate to give them away, while others are incredibly hungry to snatch them up. This "electron hunger" is mathematically quantified by a property known as the Standard Reduction Potential ($E^{\circ }$).

By definition, an oxidising agent is a substance that oxidises another substance. But how does it do that? It does so by stealing electrons from the other substance. In the process of stealing electrons, the oxidising agent itself gets reduced. Therefore, the strength of an oxidising agent is directly proportional to its tendency to get reduced.

The rule of thumb in electrochemistry is beautifully simple:

Conversely, a lower (or more negative) standard reduction potential means the substance hates getting reduced and would much rather give away electrons, making it a strong reducing agent.

Let's look at the data provided in the question and arrange our contenders on a vertical scale of electron hunger:

1. Chromium Couple: $E_{C{r}^{3+}/Cr}^{\circ }=-0.74\text{ V}$ 2. Dichromate Couple: $E_{C{r}_2{O}_7^{2-}/C{r}^{3+}}^{\circ }=1.33\text{ V}$ 3. Chlorine Couple: $E_{Cl/C{l}^{-}}^{\circ }=1.36\text{ V}$ 4. Permanganate Couple: $E_{Mn{O}_4^{-}/M{n}^{2+}}^{\circ }=1.51\text{ V}$

When we plot these values, it becomes immediately obvious who the heavyweight champion of electron stealing is. The permanganate couple ($Mn{O}_4^{-}/M{n}^{2+}$) sits at the very top with a massive potential of $1.51\text{ V}$.

Because $1.51\text{ V}$ is the highest positive value among the choices, the permanganate ion ($Mn{O}_4^{-}$) has the greatest thermodynamic drive to accept electrons and undergo reduction to $M{n}^{2+}$.

This makes $Mn{O}_4^{-}$ the strongest oxidising agent in the given list.

A quick thought experiment: What if the question had asked for the strongest reducing agent? You would look for the lowest value. At $-0.74\text{ V}$, solid Chromium ($Cr$) would be the most eager to give up electrons, making it the strongest reducing agent here.