Question

Standard electrode potentials are:
${\text{F}}{{\text{e}}^{2 + }}{\text{/Fe}}\left( {{{\text{E}}^ \circ } = - 0.44{\text{ V}}} \right),{\text{ F}}{{\text{e}}^{3 + }}{\text{/F}}{{\text{e}}^{2 + }}\left( {{{\text{E}}^ \circ } = 0.77{\text{ V}}} \right)$
${\text{F}}{{\text{e}}^{2 + }}$, ${\text{F}}{{\text{e}}^{3 + }}$ and ${\text{Fe}}$ blocks are kept together, then:
A) ${\text{F}}{{\text{e}}^{3 + }}$ increases
B) ${\text{F}}{{\text{e}}^{3 + }}$ decreases
C) ${\text{F}}{{\text{e}}^{2 + }}{\text{/F}}{{\text{e}}^{3 + }}$ remains unchanged
D) ${\text{F}}{{\text{e}}^{2 + }}$ decreases

Answer 1

Hint: We know that the potential of the half-reaction or the half-cell reaction measured against the standard hydrogen electrode under standard conditions is known as standard electrode potential. The standard conditions suggest that the temperature is $298{\text{ K}}$, the pressure is $1{\text{ atm}}$ and the concentration of the electrolyte is $1{\text{ M}}$.

Complete step by step answer:We know that the potential of the half-reaction or the half-cell reaction measured against the standard hydrogen electrode under standard conditions is known as standard electrode potential. The standard conditions suggest that the temperature is $298{\text{ K}}$, the pressure is $1{\text{ atm}}$ and the concentration of the electrolyte is $1{\text{ M}}$.
If the standard potential of the reaction is positive then the reaction is spontaneous and occurs under standard state conditions. If the standard potential of the reaction is negative then the reaction is non-spontaneous.
For the conversion of ${\text{F}}{{\text{e}}^{2 + }}$ to ${\text{Fe}}$, the standard electrode potential is $ - 0.44{\text{ V}}$ i.e. negative. Thus, the conversion of ${\text{F}}{{\text{e}}^{2 + }}$ to ${\text{Fe}}$ is non-spontaneous.
For the conversion of ${\text{F}}{{\text{e}}^{3 + }}$ to ${\text{F}}{{\text{e}}^{2 + }}$, the standard electrode potential is $0.77{\text{ V}}$ i.e. positive. Thus, the conversion of ${\text{F}}{{\text{e}}^{3 + }}$ to ${\text{F}}{{\text{e}}^{2 + }}$ is spontaneous.
If the standard reduction potential is higher than the species is a strong oxidising agent. Thus, the species with higher standard reduction potential has a higher tendency to get reduced. Thus, the reaction that occurs is as follows:
${\text{Fe}} + 2{\text{F}}{{\text{e}}^{3 + }} \to 3{\text{F}}{{\text{e}}^{2 + }}$
From the reaction, we can say that the concentration of ${\text{F}}{{\text{e}}^{3 + }}$ ions decreases and the concentration of ${\text{F}}{{\text{e}}^{2 + }}$ ions increases.
Thus, when ${\text{F}}{{\text{e}}^{2 + }}$, ${\text{F}}{{\text{e}}^{3 + }}$ and ${\text{Fe}}$ blocks are kept together then ${\text{F}}{{\text{e}}^{3 + }}$ decreases.

Thus, the correct option is (B) ${\text{F}}{{\text{e}}^{3 + }}$ decreases.

Note: Remember that if the standard reduction potential is higher than the species is a strong oxidising agent. Thus, the species with higher standard reduction potential has a higher tendency to get reduced. And if the standard reduction potential is lower than the species is a strong reducing agent. Thus, the species with lower standard reduction potential has a higher tendency to get oxidised.