A) Oxidation of Fc and Cr both and reduction of \[{{O}_{2}}\]
B) oxidation of Fe and reduction of Cr
C) oxidation of Fe and Cr both reduction of barbon
D) No redox reaction
Correct Answer: A
Solution :
[a] \[\underset{\begin{smallmatrix} \uparrow \\ +2 \end{smallmatrix}}{\mathop{F{{e}^{2+}}}}\,+\underset{\begin{smallmatrix} \uparrow \\ +3 \end{smallmatrix}}{\mathop{2CrO_{2}^{-}}}\,+\underset{\begin{smallmatrix} \uparrow \\ 0 \end{smallmatrix}}{\mathop{{{O}_{2}}}}\,\to \underset{\begin{smallmatrix} \uparrow \\ +3 \end{smallmatrix}}{\mathop{F{{e}_{2}}}}\,\underset{\begin{smallmatrix} \uparrow \\ -2 \end{smallmatrix}}{\mathop{{{O}_{3}}}}\,+\underset{\begin{smallmatrix} \uparrow \\ +6 \end{smallmatrix}}{\mathop{CrO_{4}^{2-}}}\,\] \[F{{e}^{2+}}\] and \[C{{r}^{3+}}\] are oxidized (increase in oxidation number) \[{{O}_{2}}\]is reduced,
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