A) +2
B) +1
C) 0
D) -1
Correct Answer: C
Solution :
Vanadate ion \[VO_{4}^{3-}\](O. N. = 5) is reduced to \[{{V}^{x}}\] (species Z) by \[{{C}_{2}}O_{4}^{2-}\] ion in acidic medium.\[{{V}^{x}}\]is oxidised by \[MnO_{4}^{-}\] to \[{{V}^{5+}}\] \[\therefore \] \[5{{V}^{x}}+(5-x)MnO_{4}^{-}\to (5-x)M{{n}^{2+}}+5{{V}^{5+}}\]?(i) \[S{{O}_{2}}\]reduces\[{{V}^{5+}}\]to\[{{V}^{4+}}\]which in turn isoxidised to \[{{V}^{5+}}\] by \[MnO_{4}^{-}\] \[\therefore \,\,\,5{{V}^{4+}}+MnO_{4}^{-}\to \,M{{n}^{2+}}+5{{V}^{5+}}\] Volumes of \[MnO_{4}^{-}\] used in Eqs. (i), (ii) are in the ratio, of \[(5-x):1\] \[\therefore \] \[\frac{5-x}{1}=\frac{5}{1}\] (given) \[x=0\]You need to login to perform this action.
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