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(C) The given reactions are oxidation half-reactions. To determine the oxidizing agent,we look at the reduction potentials $(E^o_{red})$.For the first

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$Fe^{3+}$

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(C) The given reactions are oxidation half-reactions. To determine the oxidizing agent,we look at the reduction potentials $(E^o_{red})$.For the first reaction: $[Fe(CN)_6]^{3-} + e^- 	o [Fe(CN)_6]^{4-}$,$E^o_{red} = +0.35 \ V$.For the second reaction: $Fe^{3+} + e^- 	o Fe^{2+}$,$E^o_{red} = +0.77 \ V$.$A$ higher reduction potential indicates a stronger tendency to get reduced,making the species a stronger oxidizing agent.Comparing the two,$Fe^{3+}$ has a higher reduction potential $(+0.77 \ V > +0.35 \ V)$.Therefore,$Fe^{3+}$ is the strongest oxidizing agent. The correct option is $(C)$.

On the basis of the following $E^o$ values,the strongest oxidizing agent is:
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$

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On the basis of the following $E^o$ values,the strongest oxidizing agent is:$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$