Chemistry - Electrochemistry Question with Solution | TestHub

For a reaction to be spontaneous, the standard cell potential $E_{\text{cell}}^{\circ }$ must be positive. The standard cell potential is given by:

$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$

Let us calculate $E_{\text{cell}}^{\circ }$ for each given reaction:

For ${\mathrm{Zn}}_{(\mathrm{s})}+{\mathrm{Mg}}_{(\mathrm{aq})}^{+2}⟶{\mathrm{Zn}}_{(\mathrm{aq})}^{+2}+{\mathrm{Mg}}_{(\mathrm{s})}$:
$\mathrm{Mg}$ is reduced (cathode) and $\mathrm{Zn}$ is oxidized (anode).
$E_{\text{cell}}^{\circ }=-2.37-(-0.76)=-1.61\text{ V}$ (Non-spontaneous)

For $2{\mathrm{Ag}}_{(\mathrm{s})}+{\mathrm{Zn}}_{(\mathrm{aq})}^{+2}⟶2{\mathrm{Ag}}_{(\mathrm{aq})}^{+}+{\mathrm{Zn}}_{(\mathrm{s})}$:
$\mathrm{Zn}$ is reduced (cathode) and $\mathrm{Ag}$ is oxidized (anode).
$E_{\text{cell}}^{\circ }=-0.76-0.79=-1.55\text{ V}$ (Non-spontaneous)

For ${\mathrm{Zn}}_{(\mathrm{s})}+{\mathrm{Cu}}_{(\mathrm{aq})}^{+2}⟶{\mathrm{Zn}}_{(\mathrm{aq})}^{+2}+{\mathrm{Cu}}_{(\mathrm{s})}$:
$\mathrm{Cu}$ is reduced (cathode) and $\mathrm{Zn}$ is oxidized (anode).
$E_{\text{cell}}^{\circ }=0.34-(-0.76)=+1.10\text{ V}$ (Spontaneous)

For ${\mathrm{Cu}}_{(\mathrm{s})}+{\mathrm{Mg}}_{(\mathrm{aq})}^{+2}⟶{\mathrm{Cu}}_{(\mathrm{aq})}^{+2}+{\mathrm{Mg}}_{(\mathrm{s})}$:
$\mathrm{Mg}$ is reduced (cathode) and $\mathrm{Cu}$ is oxidized (anode).
$E_{\text{cell}}^{\circ }=-2.37-0.34=-2.71\text{ V}$ (Non-spontaneous)

Since $E_{\text{cell}}^{\circ }$ is positive only for the reaction between $\mathrm{Zn}$ and ${\mathrm{Cu}}^{+2}$, it is the spontaneous reaction.

Answer: ${\mathrm{Zn}}_{(\mathrm{s})}+{\mathrm{Cu}}_{(\mathrm{aq})}^{+2}⟶{\mathrm{Zn}}_{(\mathrm{aq})}^{+2}+{\mathrm{Cu}}_{(\mathrm{s})}$