At 298 K, the standard electrode potentials of $\mathrm{Cu}{}^{2+}/\mathrm{Cu}$, $\mathrm{Zn}{}^{2+}/\mathrm{Zn}$, $\mathrm{Fe}{}^{2+}/\mathrm{Fe}$ and $\mathrm{Ag}{}^{+}/\mathrm{Ag}$ are $0.34\text{V}$, $-0.76\text{V}$, $-0.44\text{V}$ and $0.80\text{V}$, respectively. On the basis of standard electrode potential, predict which of the following reactions cannot occur?

Solution & Explanation

\textbf{Answer:} (D) $2\mathrm{CuSO}{}_4(\mathrm{aq})+2\mathrm{Ag}(\mathrm{s})\to 2\mathrm{Cu}(\mathrm{s})+\mathrm{Ag}{}_2\mathrm{SO}{}_4(\mathrm{aq})$ cannot occur. \textbf{Solution:} A metal can displace another metal ion from solution only if it is a stronger reducing agent, i.e. it has the more negative (lower) reduction potential. Given $E^{\circ }$: $\mathrm{Zn}{}^{2+}/\mathrm{Zn}$ $=-0.76$, $\mathrm{Fe}{}^{2+}/\mathrm{Fe}$ $=-0.44$, $\mathrm{Cu}{}^{2+}/\mathrm{Cu}$ $=+0.34$, $\mathrm{Ag}{}^{+}/\mathrm{Ag}$ $=+0.80\text{V}$. For option (D), $\mathrm{Ag}$ ($+0.80\text{V}$) is a weaker reducing agent than $\mathrm{Cu}$ ($+0.34\text{V}$), so $\mathrm{Ag}$ cannot reduce $\mathrm{Cu}{}^{2+}$. $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0.34-0.80=-0.46\text{V}<0$ A negative $E_{cell}^{\circ }$ means ${\Delta }_r{G}^{\circ }>0$, so the reaction is non-spontaneous and cannot occur. Reactions (A), (B), (C) all have a more reactive metal displacing a less reactive one, so they are feasible.