Question

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

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

Answer 1

Answer: (A)

$\left[Fe^{3+}\right]$ decreases

Answer 2

$Fe^{2+}$/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.
$\therefore\quad Fe \rightarrow Fe^{2+}+2e^{-} \left(Anode\right)$
$and \, \frac{\left(Fe^{3+}+e^{-}\rightarrow Fe^{2+}\right)\times2 \left(Cathode\right)}{Fe+2Fe^{3+}\rightarrow3Fe^{2+}}$
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.