$E^{\circ}$ of $Cu$ is $+0.34 \ V$ while that of $Zn$ is $-0.76 \ V$. Explain.
(N/A) The standard electrode potential $(E^{\circ})$ of a metal depends on the sum of its enthalpy of sublimation,ionization enthalpy,and hydration enthalpy.
For $Cu$,the sum of sublimation and ionization enthalpies is very high,which is not compensated by the hydration enthalpy of $Cu^{2+}$. Thus,$Cu$ has a positive $E^{\circ}$ value $(+0.34 \ V)$.
For $Zn$,the second ionization enthalpy is relatively low because the removal of electrons from the $4s$-orbital leads to a stable $3d^{10}$ configuration. This low energy requirement makes the overall process favorable,resulting in a negative $E^{\circ}$ value $(-0.76 \ V)$.
For $Cu$,the sum of sublimation and ionization enthalpies is very high,which is not compensated by the hydration enthalpy of $Cu^{2+}$. Thus,$Cu$ has a positive $E^{\circ}$ value $(+0.34 \ V)$.
For $Zn$,the second ionization enthalpy is relatively low because the removal of electrons from the $4s$-orbital leads to a stable $3d^{10}$ configuration. This low energy requirement makes the overall process favorable,resulting in a negative $E^{\circ}$ value $(-0.76 \ V)$.
Explore More
Similar Questions
For the cell $Zn_{(s)} | Zn^{2+}_{(aq)} || M^{x+}_{(aq)} | M_{(s)}$,different half cells and their standard electrode potentials are given below:
$M^{x+}_{(aq)} / M_{(s)}$ $Au^{3+}_{(aq)} / Au_{(s)}$ $Ag^{+}_{(aq)} / Ag_{(s)}$ $Fe^{3+}_{(aq)} / Fe^{2+}_{(aq)}$ $Fe^{2+}_{(aq)} / Fe_{(s)}$ $E^o M^{x+} / M (V)$ $1.40$ $0.80$ $0.77$ $-0.44$
If $E^o Zn^{2+}/Zn = -0.76 \ V$,which cathode will give a maximum value of $E^o_{cell}$ per electron transferred?
| $M^{x+}_{(aq)} / M_{(s)}$ | $Au^{3+}_{(aq)} / Au_{(s)}$ | $Ag^{+}_{(aq)} / Ag_{(s)}$ | $Fe^{3+}_{(aq)} / Fe^{2+}_{(aq)}$ | $Fe^{2+}_{(aq)} / Fe_{(s)}$ |
| $E^o M^{x+} / M (V)$ | $1.40$ | $0.80$ | $0.77$ | $-0.44$ |
If $E^o Zn^{2+}/Zn = -0.76 \ V$,which cathode will give a maximum value of $E^o_{cell}$ per electron transferred?
DifficultJEE MAIN 2019
View SolutionThe e.m.f. of a cell whose half cells are given below is .............. $V$
$Mg^{2+} + 2e^- \to Mg_{(s)}$; $E^o = - 2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}$; $E^o = + 0.34 \ V$
$Mg^{2+} + 2e^- \to Mg_{(s)}$; $E^o = - 2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}$; $E^o = + 0.34 \ V$
Medium
View SolutionThe $E^0_{cell}$ 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 the standard electrode potential $E^0_{Al^{3+}/Al}$ will be: (in $\text{ V}$)
DifficultMHT CET 2026
View SolutionWhich statement is correct for $\Delta G$ and $E_{cell}$ for a cell reaction?
MediumGUJCET 2025
View SolutionWhen $E^o_{Ag^{+}/Ag} = 0.8 \ V$ and $E^o_{Zn^{2+}/Zn} = -0.76 \ V$,which of the following is correct?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo