For the cell Cu_{(s)} | Cu^{2+}_{(aq)} (1 M) | vedclass

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(B) The standard cell potential $E^{\circ}_{cell}$ is calculated as $E^{\circ}_{cathode} - E^{\circ}_{anode}$.Here,the cathode is $Zn^{2+}/Zn$ and the

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Non-spontaneous

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(B) The standard cell potential $E^{\circ}_{cell}$ is calculated as $E^{\circ}_{cathode} - E^{\circ}_{anode}$.Here,the cathode is $Zn^{2+}/Zn$ and the anode is $Cu^{2+}/Cu$.$E^{\circ}_{cell} = E^{\circ}_{Zn^{2+}/Zn} - E^{\circ}_{Cu^{2+}/Cu} = -0.76 \ V - 0.34 \ V = -1.10 \ V$.Since $E^{\circ}_{cell}$ is negative,the Gibbs free energy change $\Delta G^{\circ} = -nFE^{\circ}_{cell}$ is positive.Therefore,the cell reaction is non-spontaneous.

For the cell $Cu_{(s)} | Cu^{2+}_{(aq)} (1 \ M) || Zn^{2+}_{(aq)} (1 \ M) | Zn_{(s)}$,given $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$,the cell reaction is:

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