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(A) reaction is spontaneous if the standard cell potential $E^{\circ}_{cell}$ is positive,which corresponds to a negative Gibbs free energy change $\D

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Copper-silver nitrate interface

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(A) reaction is spontaneous if the standard cell potential $E^{\circ}_{cell}$ is positive,which corresponds to a negative Gibbs free energy change $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.For system $(A)$: $Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Ag^{+}/Ag} - E^{\circ}_{Cu^{2+}/Cu} = 0.80 \ V - 0.34 \ V = 0.46 \ V$.Since $E^{\circ}_{cell} > 0$,the reaction is spontaneous.For system $(B)$: $Ag(s) + Cu^{2+}(aq) \rightarrow Ag^{+}(aq) + Cu(s)$.$E^{\circ}_{cell} = E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Ag^{+}/Ag} = 0.34 \ V - 0.80 \ V = -0.46 \ V$.Since $E^{\circ}_{cell} Therefore,the copper-silver nitrate interface shows a spontaneous reaction.

Consider the systems having liquid-solid interface,$(A)$ copper wire in silver nitrate solution and $(B)$ silver wire in copper sulphate solution. Predict which interface will show spontaneous reaction,if $E_{Cu^{2+}/Cu}^{\circ} = 0.34 \ V$ and $E_{Ag^{+}/Ag}^{\circ} = 0.80 \ V$?

Similar Questions

Given the electrode potentials:
$Fe^{3+} + e^- \to Fe^{2+}; E^o = 0.771 \ V$
$I_2 + 2e^- \to 2I^{-}; E^o = 0.536 \ V$
The $E^o_{\text{cell}}$ for the cell reaction $2Fe^{3+} + 2I^{-} \to 2Fe^{2+} + I_2$ is:

Which of the following reactions will be spontaneous in an electrochemical cell constructed from the given standard electrode potentials: $E_{Cl_2 \mid 2Cl^-}^0 = 1.36 \ V$ and $E_{Br_2 \mid 2Br^-}^0 = 1.09 \ V$?

Give the symbolic representation of the following half-cells (electrodes):
$(i)$ $Zn_{(s)} | Zn^{2+}_{(aq)} (1M)$
$(ii)$ $Cu^{2+}_{(aq)} (1M) | Cu_{(s)}$
$(iii)$ $Pt_{(s)} | H_{2(g)} (1 \text{ bar}) | H^+_{(aq)} (1M)$

Which statement is correct for $\Delta G$ and $E_{cell}$ for a cell reaction?

The standard electrode potentials at $25\,^oC$ for the following half-reactions are given:
$Zn^{2+} + 2e^- \to Zn, E^o = -0.762\,V$
$Mg^{2+} + 2e^- \to Mg, E^o = -2.37\,V$
When zinc dust is added to a solution of $MgCl_2$,what happens?

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