Ncert Exercises & Solutions

The most stable oxidized species among the following is:
\(E_{{\mathrm{Cr}_2 \mathrm{O}_7^2}/ \mathrm{Cr}^{3+}}^{o} =1.33 \mathrm{~V} ; E_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{o}=1.36 \mathrm{~V} \)
\( E_{\mathrm{MnO_{4}}^{-} / \mathrm{Mn}^{2+}}^{o}=1.51 \mathrm{~V} ; E_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{o}=-0.74 \mathrm{~V}\)

1.	Cr³⁺	2.	MnO₄^-
3.	Cr₂O₇^2-	4.	Mn²⁺

HINT: Lower the value of reduction potential, the higher will be the stability of oxidized species.

Explanation:

{E^{°}}_{{Cr}^{3 +} / Cr}

has most - ve value equal to -0.74 among the given four choices. So, Cr³⁺ is the most stable oxidized species.

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