Given : E^o_ Cr^ 3+ /Cr =-0.74 , V, , E^o_ MnO_4^-/Mn^ 2+ =1.51 ,… - Vidyadip

MCQ

Given :

$E^o_{Cr^{3+}/Cr} =-0.74\, V,\,$ $E^o_{MnO_4^-/Mn^{2+}} =1.51 \,V$

$E^o_{Cr_2O_7^{2-}/Cr^3+} =1.33\,V:$ $E^o_{Cl/Cl^-} =1.36\,V$

Based on the data given above, strongest oxidising agent will be :

A
$Cl$
B
$Cr^{3+}$
C
$Mn^{2+}$
✓
$MnO_4^-$

✓

Answer

Correct option: D.

$MnO_4^-$

d
Higher the value of standard reduction potential stronger will be the oxidising agent, hence $M n O_{4}^-$ is the strongest oxidising agent.

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