Question

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

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

Answer 1

Answer: (D)

For the reaction,
$2F{e}^{3+}+2I^{-}\to 2F{e}^{2+}+I_2$
$E_{cell}^{\circ }=E_{F{e}^{3+}/F{e}^{2+}}^{\circ }-E_{I_2/I^{-}}^{\circ }=0.77-\left(0.54\right)=+0.23V$
Here, $E_{cell}^{\circ }$ is $+ve$ so, reaction is feasible.
For the reaction,
$Cu+2A{g}^{+}\to C{u}^{2+}+2Ag$
$E_{cell}^{\circ }=E_{A{g}^{3+}/Ag}^{\circ }-E_{C{u}^{2+}/Cu}^{\circ }=0.80-\left(0.34\right)=+0.46V$
Here, $E_{cell}^{\circ }$ is $+ve$ so, the reaction is feasible.
For the reaction, $2F{e}^{3+}+Cu\to 2F{e}^{2+}+C{u}^{2+}$
$E_{cell}^{\circ }=E_{F{e}^{3+}/F{e}^{2+}}^{\circ }-E_{C{u}^{2+}/Cu}^{\circ }=0.77-\left(0.34\right)=+0.43V$
Here, $E_{cell}^{\circ }$ is $+ve$ so, the reaction is feasible.
For the reaction, $Ag+F{e}^{3+}\to A{g}^{+}+F{e}^{2+}$
$E_{cell}^{\circ }=E_{F{e}^{3+}/F{e}^{2+}}^{\circ }-E_{A{g}^{3+}/Ag}^{\circ }=0.77-\left(0.80\right)=-0.03V$
Here, $E_{cell}^{\circ }$ is negative so, the reaction is not feasible.