$E^{\circ}$ 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 $(-0.76 \ V)$.
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Similar Questions
The standard electrode potentials $E^{\circ} (V)$ for $Li^{+} / Li$ and $Na^{+} / Na$ respectively are:
According to standard electrode potential,which is the strongest reducing agent and oxidising agent?
$Al (-1.66 \ V), Cu (+0.34 \ V), Li (-3.05 \ V), Ag (+0.80 \ V)$
$Al (-1.66 \ V), Cu (+0.34 \ V), Li (-3.05 \ V), Ag (+0.80 \ V)$
Easy
View SolutionElectrode potential data are given below :
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; \, E^o = +0.77 \, V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; \, E^o = -1.66 \, V$
$Br_{2(aq)} + 2e^- \to 2Br^{-}_{(aq)}; \, E^o = +1.08 \, V$
Based on the data given above,the reducing power of $Fe^{2+}$,$Al$ and $Br^{-}$ will increase in the order:
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; \, E^o = +0.77 \, V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; \, E^o = -1.66 \, V$
$Br_{2(aq)} + 2e^- \to 2Br^{-}_{(aq)}; \, E^o = +1.08 \, V$
Based on the data given above,the reducing power of $Fe^{2+}$,$Al$ and $Br^{-}$ will increase in the order:
Easy
View SolutionThe standard electrode potentials of $Zn^{2+}/Zn$ and $Ag^{+}/Ag$ are $-0.763 \ V$ and $+0.799 \ V$ respectively. The standard potential of the cell is ............ $V$.
Easy
View Solution$A$ reaction,$Ni_{(s)} + Cu^{2+}_{(aq)} \rightarrow Ni^{2+}_{(aq)} + Cu_{(s)}$ occurs in a cell. Calculate $E^0_{cell}$ if $E^0_{Cu} = 0.337 \ V$ and $E^0_{Ni} = -0.257 \ V$. (in $V$)
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