Addition of powdered lead and iron to a solut | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The standard reduction potentials are $E^{\circ}_{Pb^{2+}/Pb} = -0.13 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. Since $E^{\circ}_{Pb^{2+}/P

Vedclass · Accepted Answer

$Fe^{2+}$ concentration increases and $Pb^{2+}$ concentration decreases

Vedclass · Answer

(D) The standard reduction potentials are $E^{\circ}_{Pb^{2+}/Pb} = -0.13 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. Since $E^{\circ}_{Pb^{2+}/Pb} > E^{\circ}_{Fe^{2+}/Fe}$,$Pb^{2+}$ ions will act as the oxidizing agent and $Fe$ metal will act as the reducing agent. The spontaneous cell reaction is: $Pb^{2+}(aq) + Fe(s) \rightarrow Fe^{2+}(aq) + Pb(s)$. In this reaction,$Pb^{2+}$ ions are reduced to $Pb$ metal,so the concentration of $Pb^{2+}$ decreases. Simultaneously,$Fe$ metal is oxidized to $Fe^{2+}$ ions,so the concentration of $Fe^{2+}$ increases.

Addition of powdered lead and iron to a solution which is $1.0 \, M$ in both $Pb^{2+}$ and $Fe^{2+}$ would result in [$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$ and $E^{\circ}_{Pb^{2+}/Pb} = -0.13 \, V$]

Similar Questions

The standard reduction potentials for $Zn^{2+}/Zn$,$Ni^{2+}/Ni$ and $Fe^{2+}/Fe$ are $-0.76 \ V$,$-0.23 \ V$ and $-0.44 \ V$ respectively.
The reaction $X + Y^{2+} \rightarrow X^{2+} + Y$ will be spontaneous when:

The $E^0_{cell}$ of the cell $Al(s)|Al^{3+}(1M)||Pb^{2+}(1M)|Pb(s)$ is $1.5 \text{ V}$. If $E^0_{Pb^{2+}/Pb}$ is $-0.14 \text{ V}$, then $E^0_{Al^{3+}/Al}$ will be: (in $\text{ V}$)

If the standard reduction potentials of four electrodes $A$, $B$, $C$, and $D$ are $+2.5 \text{ V}$, $+3.0 \text{ V}$, $-2.0 \text{ V}$, and $-1.5 \text{ V}$ respectively, in which of the following cases is the standard $emf$ of the cell maximum?

Iron displaces copper from its salt solution,because

Vedclass Test Series

Exam Paper Generator

Online Exam Module