Question

Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
$E^{\circ}$ values : $Fe^{3+}/Fe^{2+} = + 0.77; I_{2\left(s\right)}/I^- = + 0.54$;
$Cu^{2+}/Cu = + 0.34$; $Ag^{+}/Ag = + 0.80\, V$

• A. $Fe^{3+}$ and $I^-$
• B. $Ag^{+}$ and $Cu$
• C. $Fe^{3+}$ and $Cu$
• D. $Ag$ and $Fe^{3+}$

Answer 1

Answer: (D)

For the reaction,
$2Fe^{3+} + 2I^{-} \to 2Fe^{2+} + I_{2}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{I_2/I^{-}} = 0.77 - \left(0.54\right) = + 0.23 \,V$
Here, $E^{\circ }_{cell}$ is $+ve$ so, reaction is feasible.
For the reaction,
$Cu + 2Ag^{+} \to Cu^{2+} + 2Ag$
$E^{\circ }_{cell} = E^{\circ }_{Ag^{3+}/Ag} - E^{\circ }_{Cu^{2+}/Cu}= 0.80 - \left(0.34\right) = + 0.46 \,V$
Here, $E^{\circ }_{cell}$ is $+ve$ so, the reaction is feasible.
For the reaction, $2Fe^{3+ }+ Cu \to 2Fe^{2+} + Cu^{2+}$
$E^{\circ }_{cell} = E^{\circ }_{Fe^{3+}/Fe^{2+}} - E^{\circ }_{Cu^{2+}/Cu} = 0.77 - \left(0.34\right) = + 0.43 \,V$
Here, $E^{\circ }_{cell}$ is $+ve$ so, the reaction is feasible.
For the reaction, $Ag + Fe^{3+} \to Ag^{+} + Fe^{2+}$
$E^{\circ }_{cell} = E^{\circ }_{Fe^{3+}/Fe^{2+}} -E^{\circ }_{Ag^{3+}/Ag} = 0.77 - \left(0.80\right) = - 0.03\, V$
Here, $E^{\circ }_{cell}$ is negative so, the reaction is not feasible.