A) Fe > Mn > Co > Cr
B) Co > Mn > Fe > Cr
C) Cr > Mn > Co > Fe
D) Mn > Fe > Cr > Co
Correct Answer: D
Solution :
This can be understood on the basis of E ° values for\[{{M}^{2+}}/M\]. \[\begin{align} & \begin{matrix} {{E}^{o}}/V & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cr & \,\,\,\,\,\,Mn & \,\,\,\,Fe & \,\,\,\,\,\,Co \\ \end{matrix} \\ & \begin{matrix} {{M}^{2+}}/M & -0.90 & -1.18 & -0.44 & -0.28 \\ \end{matrix} \\ \end{align}\] \[{{E}^{o}}\] value for Mn is more negative than expected from general trend due to extra stability of half-filled \[\text{M}{{\text{n}}^{\text{2+}}}\] ion. Thus the correct order should be Mn > Cr > Fe > Co An examination of E° values for redox couple \[\text{M}{{\text{n}}^{\text{3+}}}/\text{M}{{\text{n}}^{\text{2+}}}\]shows that \[\text{C}{{\text{r}}^{\text{2+}}}\] is strong reducing agent (\[{{E}^{o}}_{{{M}^{3+}}/{{M}^{2+}}}=0.41\,V\]) and liberates \[{{\text{H}}_{\text{2}}}\]from dilute acids. \[2C{{r}^{2+}}(aq)+2{{H}^{+}}(aq)\xrightarrow[{}]{{}}2C{{r}^{3+}}(aq)+{{H}_{2}}\uparrow (g)\]\[\therefore \]The correct order is Mn > Fe > Cr > Co.
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