A) \[{{\Delta }_{O}}\] is oxidised to \[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
B) \[[Fe{{(CN)}_{2}}{{({{H}_{2}}O)}_{4}}]\] is oxidised to \[{{[Fe{{(CN)}_{4}}({{H}_{2}}{{O}_{2}})]}^{2-}}\]
C) \[{{\Delta }_{O}}(II)<{{\Delta }_{O}}(III)<{{\Delta }_{O}}(I)\] is oxidised to \[{{\Delta }_{O}}(III)<{{\Delta }_{O}}(II)<{{\Delta }_{O}}(I)\]
D) Zn is reduced to \[{{\Delta }_{O}}(II)<{{\Delta }_{O}}(I)<{{\Delta }_{O}}(III)\]
Correct Answer: C
Solution :
In the reaction \[{{25}^{o}}C\] oxidation number of \[C{{H}_{3}}COOH\] changes from zero to \[N{{H}_{4}}OH\] . Hence, \[({{10}^{-5}})\] is oxidised from \[({{10}^{-5}})\] to \[N{{H}_{4}}OH\].
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