Question

Standard electrode potentials for Fe electrodes are given as
$Fe ^{2+}+2 e^{-} \rightarrow Fe ; \,\,\,\,\, E^{\circ}=-0.44 \,V$
$Fe ^{3+}+e^{-} \rightarrow Fe ^{2+} ; \,\,\,\,\, E^{\circ}=+0.77 \,V$
$Fe ^{2+}, Fe ^{3+}$ and $Fe$ blocks are kept together then

• A. $\left[ Fe ^{3+}\right]$ decreases
• B. $\left[ Fe ^{3+}\right]$ increases
• C. $\left[ Fe ^{2+}\right] /\left[ Fe ^{3+}\right]$ remains unchanged
• D. $\left[ Fe ^{2+}\right]$ decreases.

Answer 1

Answer: (A)

$Fe ^{2+} / Fe$ acts as anode because its standard reduction potential is low i.e., oxidation occurs at this electrode and $Fe ^{3+} / Fe$ acts as cathode because its standard reduction potential is high i.e., reduction occurs at this electrode.

$\therefore Fe \rightarrow Fe ^{2+}+2 e^{-}($ Anode $)$

and $\frac{\left( Fe ^{3+}+e^{-} \rightarrow Fe ^{2+}\right) \times 2(\text { Cathode })}{ Fe +2 Fe ^{3+} \rightarrow 3 Fe ^{2+}}$

Thus, if $Fe ^{2+}, Fe ^{3+}$ and $Fe$ blocks are kept together, $Fe ^{3+}$ get reduced to $Fe ^{2+}$ i.e., concentration of $Fe ^{3+}$ decreases.