The standard electrode potential (E^o) of Cu^ | vedclass

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(C) The standard electrode potential $(E^o)$ of a metal is determined by the sum of three energy factors: enthalpy of sublimation (atomization),ioniza

Vedclass · Accepted Answer

The high energy required to transform $Cu(s)$ to $Cu^{2+}(aq)$ is not compensated by its hydration enthalpy.

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(C) The standard electrode potential $(E^o)$ of a metal is determined by the sum of three energy factors: enthalpy of sublimation (atomization),ionization enthalpy,and enthalpy of hydration. For $Cu$,the sum of the enthalpy of sublimation and the first two ionization enthalpies is very high. This high energy requirement is not fully compensated by the hydration enthalpy of $Cu^{2+}$ ions. As a result,the overall $E^o$ value for $Cu^{2+}/Cu$ is positive $(+0.34 \, V)$. In contrast,for $Zn$,the energy required to convert $Zn(s)$ to $Zn^{2+}(aq)$ is relatively low and is well-compensated by the high hydration enthalpy of $Zn^{2+}$,leading to a negative $E^o$ value $(-0.76 \, V)$.

The standard electrode potential $(E^o)$ of $Cu^{2+}/Cu$ is $+0.34 \, V$,while that of $Zn^{2+}/Zn$ is $-0.76 \, V$. Explain the reason for this difference.

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