Question

$Zn^{2+} \to Zn_{(s)}; E^° = - 0.76\, V$
$Cu^{2+} \to Cu_{(s)}; E^°=- 0.34\, V$
Which of the following is spontaneous?

• A. $Zn^{2+} + Cu \to Zn + Cu^{2+}$
• B. $Cu^{2+} + Zn \to Cu + Zn^{2+}$
• C. $Zn^{2+} + Cu^{2+} \to Zn + Cu$
• D. none of these

Answer 1

Answer: (B)

$zn^{2+} \to Zn_{s} ; E^{\circ}=-0.76\,V $
$Cu^{2+} \rightarrow Cu_{\left(s\right)} ; E^{\circ}=-0.34\,V $
A redox reaction is feasible if $E_{cell}$ is Positive.
$\frac{\begin{matrix}zn^{2+}+2e^{-} \to Zn_{\left(s\right)} ; E^{\circ}=-0.76\,V\\ Cu^{2+}+2e^{-}\to Cu_{\left(s\right)}; E^{\circ}=-0.34\,V\end{matrix}}{\therefore Cu^{2+}+zn \to zn^{2+}+Cu ; }$
$E_{cell}=E^{\circ}_{\left(Cu^{2+} /cu\right)}-E^{\circ}_{\left(zn^{2+}/zn\right)}$
Therefore, $E_{cell}=-0.34-\left(-0.76\right)$
$=-0.34+0.76=+0.42\,V$