Electrochemistry - Result Question 42
####44. On the basis of the following $E^{\circ}$ values, the strongest oxidizing agent is :
[2008]
$[Fe(CN)_6]^{4-} \to[Fe(CN)_6]^{3-}+e^{-} ; E^{\circ}=-0.35 V$
$Fe^{2+} \to Fe^{3+}+e^{-} \quad E^{\circ}=-0.77 V$
(a) $[Fe(CN)_6]^{4}$
(b) $Fe^{2+}$
(c) $Fe^{3+}$
(d) $[Fe(CN)_6]^{3-}$
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Solution:
- (c) From the given data we find $Fe^{3+}$ $(E _{(Fe^{3+} / Fe^{2+})}^{\circ}=+0.77 V)$ is strongest oxidizing agent.
More the positive value of reduction potential, more is the tendency to get reduced (better oxidising agent).
