Question

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

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

Answer 1

Answer: (A)

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

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

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

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