Chemistry - Electrochemistry Question with Solution | TestHub

For a cell reaction to be spontaneous under standard conditions, the standard cell potential $E_{\text{cell}}^{\circ }$ must be positive.

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

Evaluating the given options:

For $\text{Ca(s)}+{\text{Cd}}^{2+}\text{(aq)}\to {\text{Ca}}^{2+}\text{(aq)}+\text{Cd(s)}$:
Cathode (reduction): ${\text{Cd}}^{2+}\to \text{Cd}$
Anode (oxidation): $\text{Ca}\to {\text{Ca}}^{2+}$
$E_{\text{cell}}^{\circ }=-0.403-(-2.866)=+2.463\text{ V}$ (Spontaneous)

For $2{\text{Br}}^{-}\text{(s)}+{\text{Sn}}^{2+}\text{(aq)}\to {\text{Br}}_2\text{(l)}+\text{Sn(s)}$:
Cathode: ${\text{Sn}}^{2+}\to \text{Sn}$
Anode: $2{\text{Br}}^{-}\to {\text{Br}}_2$
$E_{\text{cell}}^{\circ }=-0.136-1.08=-1.216\text{ V}$ (Non-spontaneous)

For $2\text{Ag(s)}+{\text{Ni}}^{2+}\text{(aq)}\to 2{\text{Ag}}^{+}\text{(aq)}+\text{Ni(s)}$:
Cathode: ${\text{Ni}}^{2+}\to \text{Ni}$
Anode: $\text{Ag}\to {\text{Ag}}^{+}$
$E_{\text{cell}}^{\circ }=-0.257-0.799=-1.056\text{ V}$ (Non-spontaneous)

For $2\text{Au(s)}+{\text{Zn}}^{2+}\text{(aq)}\to 2{\text{Au}}^{+}\text{(aq)}+\text{Zn(s)}$:
Cathode: ${\text{Zn}}^{2+}\to \text{Zn}$
Anode: $\text{Au}\to {\text{Au}}^{+}$
$E_{\text{cell}}^{\circ }=-0.763-1.68=-2.443\text{ V}$ (Non-spontaneous)

The only reaction with a positive standard cell potential is the first one.

Answer: $\text{Ca(s)}+{\text{Cd}}^{2+}\text{(aq)}\to {\text{Ca}}^{2+}\text{(aq)}+\text{Cd(s)}$, $[E_{\text{Ca}}^0=-2.866$ V & $E_{\text{Cd}}^0=-0.403$ V$]$