Chemistry - Electrochemistry Question with Solution | TestHub

For a reaction to be non-spontaneous, the standard cell potential $E_{\text{cell}}^0$ must be negative, where $E_{\text{cell}}^0=E_{\text{cathode}}^0-E_{\text{anode}}^0$.

For the reaction $2{\text{Ag}}_{(s)}+{\text{Fe}}_{(aq)}^{+2}\to 2{\text{Ag}}_{(aq)}^{+1}+{\text{Fe}}_{(s)}$:

$\text{Ag}$ undergoes oxidation (anode) and ${\text{Fe}}^{+2}$ undergoes reduction (cathode).

$E_{\text{cell}}^0=E_{{\text{Fe}}^{+2}/\text{Fe}}^0-E_{{\text{Ag}}^{+}/\text{Ag}}^0$

$E_{\text{cell}}^0=-0.44-0.79=-1.23\text{ V}$

Since $E_{\text{cell}}^0<0$, this reaction is non-spontaneous.

For the other reactions, $E_{\text{cell}}^0$ is positive, making them spontaneous:

$2{\text{Al}}_{(s)}+3{\text{Cu}}_{(aq)}^{+2}\to 2{\text{Al}}_{(aq)}^{+3}+3{\text{Cu}}_{(s)}\Rightarrow E_{\text{cell}}^0=0.34-(-1.66)=+2.00\text{ V}$

${\text{Fe}}_{(s)}+{\text{Cu}}_{(aq)}^{+2}\to {\text{Fe}}_{(aq)}^{+2}+{\text{Cu}}_{(s)}\Rightarrow E_{\text{cell}}^0=0.34-(-0.44)=+0.78\text{ V}$

$2{\text{Al}}_{(s)}+3{\text{Fe}}_{(aq)}^{+2}\to 2{\text{Al}}_{(aq)}^{+3}+3{\text{Fe}}_{(s)}\Rightarrow E_{\text{cell}}^0=-0.44-(-1.66)=+1.22\text{ V}$

Answer: $2{\text{Ag}}_{(s)}+{\text{Fe}}_{(aq)}^{+2}\to 2{\text{Ag}}_{(aq)}^{+1}+{\text{Fe}}_{(s)}$