Question

Using the standard electrode potential, find out the pair between which redox reaction is not feasible.

$E\ominus$values :

$\frac{F{e}^{3+}}{F{e}^{2+}}=+0.77\frac{I_2}{I^{-}}=+0.54;\frac{C{u}^{2+}}{Cu}=+0.34\frac{A{g}^{+}}{Ag}=+0.80V$

(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

(a) $2F{e}^{3+}2e^{-}\to 2F{e}^{2+};E^{\ominus }=+0.77$
$2I^{-}\to I_2+2e^{-};E^{\ominus }=-0.54V$ (sign of $E^{\ominus }$ is reversed)
$2f{e}^{3+}+2I^{-}\to 2F{e}^{2+}+I_2;E_{cell}^{\ominus }=+0.23V$
This reaction is fessible since $E_{cell}^{\ominus }$ is positive

(b) $Cu\to C{u}^{2+}+2e^{-};E^{\ominus }=-0.34V$ (sign of $E^{\ominus }$ has)
$2A{g}^{+}2e^{-}\to 2Ag;E^{\ominus }=+0.80V$
$Cu+2A{g}^{+}\to 2C{u}^{2+}+2Ag;E^{\ominus }=+0.46V$
This reaction is fessible since $E_{cell}^{\ominus }$ is positive.

(c) $2F{e}^{3+}+2e^{-}\to 2F{e}^{2+};E^{\ominus }=+0.77V$
$\frac{Cu\to C{u}^{2+}+2e^{-};E^{\ominus }=-0.34V}{2F{e}^{3+}+Cu\to 2F{e}^{2+}+C{u}^{2+};E^{\ominus }=+0.43V}$ (sign of $E^{\ominus }$ is reversed)
This reaction is feasible since $E_{cell}^{\ominus }$ is positive

(d) $Ag\to A{g}^{+}+3^{-};E^{\ominus }=-0.80V$ (sign of $E^{\ominus }$ is reversed)
$F{e}^{3+}+e^{-}\to F{e}^{2+};E^{\ominus }=+0.77vAg+F{e}^{3+}\to A{g}^{+}F{e}^{2+};E^{\ominus }=-0.003V$
This reaction is not feasible since $E_{cell}^{\ominus }$ is negative.