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The $E_{{M^{3 + }}/{M^{2 + }}}^\circ $ values for $Cr,Mn,Fe\,and\,Co$ are $ - 0.41, + 1.57, + 0.77$ and $ + 1.97$ respectively. For which one of the following metals the change in oxidation state from $ + 2$ to $ + 3$ is easiest?
A. $Co$
B. $Mn$
C. $Fe$
D. $Cr$

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Last updated date: 16th Jun 2024
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Answer
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Hint: In an electrolytic cell when a metal electrode is placed in solution of its ions of unit molarity then the potential difference generated between the electrode and solution is called as standard electrode potential. In an electrolytic cell we know that reduction takes place at the cathode while oxidation takes place at the anode.

Complete step by step answer:
According to the IUPAC convention, standard reduction potential is taken as the electrode potential. The given value of electrode potential can be considered as the standard reduction potential unless it is mentioned that it is oxidation potential.
Standard reduction potential means that reduction is taking place at the electrode. Standard oxidation potential means that oxidation is taking place at the electrode. An electrolytic cell consists of cathode and anode. Now the standard electrode potential of any element can be measured by preparing a cell having half-cell of that element and the other half cell is the standard hydrogen electrode. The potential of SHE is arbitrarily taken as zero. So it can be used to find out the electrode potential .
The standard electrode potentials are very important and we can extract a lot of information from them. Suppose if standard electrode potential of an electrode is greater than zero then its reduced form is more stable as compared to hydrogen gas. (since the electrode of SHE is zero). Similarly if the electrode potential of an electrode is negative then hydrogen gas is more stable than the reduced form of the electrode.
Among the elements, fluorine has the highest positive value of electrode potential. It means that the fluorine has the highest tendency to reduce to ${F^ - }$ ions. So we can say that fluorine is the strongest oxidizing agent and weakest reducing agent.
Lithium has the lowest electrode potential indicating that lithium has the highest tendency to oxidise to lithium ion $(L{i^ + })$ and reduce the other species. It means that lithium is the strongest reducing agent.
Suppose two values of electrode potential are given then the electrode having highest positive value of potential will undergo reduction and the other electrode will undergo oxidation.
Now in the given question, we can see that cobalt $(Co)$ has the highest positive value of electrode potential. It indicates that cobalt has the highest tendency to get reduced among the given ions.
Chromium ion has the lowest electrode potential indicating that it has the highest tendency to get oxidized from $ + 2$ to $ + 3$ state. It means that among the given ions, chromium will be the strongest reducing agent since it has the highest tendency to get itself oxidized. So for chromium ion, the change in oxidation state from $ + 2$ to $ + 3$ is easiest.

So, the correct answer is Option D.

Note: Among the following given ions, the increasing order of strength of reducing agent is- $Cr > Fe > Mn > Co$ .The increasing order of strength of oxidizing agent is- $Co > Mn > Fe > Cr$ .As electrode potential increases (positive), oxidizing power increases and as electrode potential decreases reducing power increases.