E^{circ}_{M^{3+} mid M^{2+}} text{ (in } Vtex | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The standard electrode potential $E^{\circ}_{M^{3+} \mid M^{2+}}$ represents the ease of reduction from the $+3$ oxidation state to the $+2$ oxida

Vedclass · Accepted Answer

$Mn$

Vedclass · Answer

(B) The standard electrode potential $E^{\circ}_{M^{3+} \mid M^{2+}}$ represents the ease of reduction from the $+3$ oxidation state to the $+2$ oxidation state. For the $3d$ series elements,the values are: $V^{3+} \mid V^{2+} = -0.26 \ V$ $Cr^{3+} \mid Cr^{2+} = -0.41 \ V$ $Mn^{3+} \mid Mn^{2+} = +1.57 \ V$ $Fe^{3+} \mid Fe^{2+} = +0.77 \ V$ Comparing these values,$Mn^{3+} \mid Mn^{2+}$ has the highest positive value because $Mn^{2+}$ $(3d^5)$ is a stable half-filled configuration,making the reduction of $Mn^{3+}$ to $Mn^{2+}$ highly favorable. Therefore,the correct option is $B$.

$E^{\circ}_{M^{3+} \mid M^{2+}} \text{ (in } V\text{) is highest for}$

Similar Questions

Passivity of iron is due to the formation of

$Cu^{2+}$ salt reacts with potassium iodide to give :

Which of the following is the correct order of negative values of $E^o\,M^{2+} | M$ for the elements $Cr, Mn, Fe,$ and $Co$?

Transition metal with low oxidation number will act as

All $Cu(II)$ halides are known,except the iodide. The reason for this is that:

Vedclass Test Series

Exam Paper Generator

Online Exam Module