The standard electrode potential of Zn^{2+}/Z | vedclass

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(B) The cell reaction for a Daniell cell is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.Here,$n = 2$ (number of electrons transferred).The

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$1.1$ and $-213$

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(B) The cell reaction for a Daniell cell is: $Zn(s) + Cu^{2+}(aq) ightarrow Zn^{2+}(aq) + Cu(s)$.Here,$n = 2$ (number of electrons transferred).The standard cell potential is calculated as: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34\, V - (-0.76\, V) = 1.1\, V$.The standard Gibbs free energy change is given by: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Substituting the values: $\Delta G^{\circ} = -2 	imes 96500\, C\, mol^{-1} 	imes 1.1\, V = -212300\, J\, mol^{-1} = -212.3\, kJ\, mol^{-1} \approx -213\, kJ\, mol^{-1}$.Thus,the emf is $1.1\, V$ and the free energy change is $-213\, kJ\, mol^{-1}$.

The standard electrode potential of $Zn^{2+}/Zn$ is $-0.76\, V$ and that of $Cu^{2+}/Cu$ is $0.34\, V$. The emf $(V)$ and the free energy change $(kJ\, mol^{-1})$,respectively for a Daniell cell will be

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