A) \[{{[Fe{{(CN)}_{6}}]}^{4-}}\]
B) \[F{{e}^{2+}}\]
C) \[F{{e}^{3+}}\]
D) \[{{[Fe{{(CN)}_{6}}]}^{3-}}\]
Correct Answer: C
Solution :
Key Idea: Oxidised form \[+n{{e}^{-}}\to \]Reduced form The substances which have' lower reduction potential are stronger reducing agent while the substances which have higher reduction potential are stronger oxidising agent. \[{{[Fe{{(CN)}_{6}}]}^{3-}}+{{e}^{-}}\to {{[Fe{{(CN)}_{6}}]}^{4-}};\]\[{{E}^{o}}=0.35V\] \[F{{e}^{3+}}+{{e}^{-}}\to F{{e}^{2+}};\]\[{{E}^{o}}=0.77V\] The reduction potential of \[F{{e}^{3+}}/F{{e}^{2+}}\] is higher, hence \[F{{e}^{3+}}\]is strongest oxidising agent.
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