Question

Adding powdered Pb and Fe to a solution containing 1M each of $P{b^{2 + }}$ and $F{e^{2 + }}$ ions would result in the formation of : ($Pb^ \circ_{P{b^{2 + }}/Pb}$ = -0.126V , $Fe^ \circ_{F{e^{2 + }}/Fe}$ = -0.44V).
A) More of $Pb$ and $F{e^{2 + }}$ ions.
B) More of $P{b^{2 + }}$ and $Fe$ ions.
C) More of \[Pb\] and $Fe$ ions.
D) More of $P{b^{2 + }}$ and $F{e^{2 + }}$ ions.

Answer 1

Hint: This type of arrangement occurs in an electrolytic cell where electricity is conducted through a solution by the movement of ions when two electrodes are connected to a conductor flow of electrons takes place from high potential to low potential.

Complete answer:
> The potential difference of the joined electrode provides force for the movements of the electrons which give the EMF (electromotive force).
> In the question we are given SRP or standard reduction potential of lead and iron. SRP is a value that predicts the ability of a cation to move towards the negative electrode and get reduced. The cation with the higher value of SRP gets reduced in preference to a cation with a lower value of SRP.
\[P{b^{2 + }} + 2{e^ - } \to Pb \\
F{e^{2 + }} + 2{e^ - } \to Fe \\ \]
> The standard reduction potential of $Pb^ \circ_{P{b^{2 + }}/Pb}$ is - 0.126 V which is less negative than the $Fe^ \circ_{F{e^{2 + }}/Fe}$.
> The given standard reduction potential value or ${E^ \circ }$ we see that Fe is more electropositive than Pb. The more electropositive metal displaces the less electropositive metal in the solution. Therefore iron displaces lead and hence iron is a better reducing agent than lead, lead is a better oxidizing agent some $P{b^{2 + }}$ ions are reduced to Pb and some Fe ions are oxidized to $F{e^{2 + }}$ .
> In addition to the powdered Pb and Fe to a solution there would result in the formation of more $Pb$ and $Fe^{2+}$ ions.

Hence the correct option is option A.

Note: A negative electrode potential means that the redox couple is a stronger reducing agent than hydrogen.