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

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.

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.