Question

The values of $E^0$ of some of the reactions are given below:
$I_2\left(g\right)+2e^{-}\to 2I^{-}\left(aq\right);E^0=+0.54V$
$C{l}_2\left(g\right)+2e^{-}\to 2C{l}^{-}\left(aq\right);E^0=+1.36V$
$F{e}^{3+}\left(aq\right)+e^{-}\to F{e}^{2+}\left(aq\right);E^0=+0.76V$
$C{e}^{4+}\left(aq\right)+e^{-}\to C{e}^{3+}\left(aq\right);E^0=+1.60V$
$S{n}^{4+}\left(aq\right)+2e^{-}\to S{n}^{2+}\left(aq\right);E^0=+0.15V$
Which of the following statement is/are true?

• A. $F{e}^{3+}$ will oxidise $C{e}^{3+}$
• B. $I_2$ will displace $C{l}_2$ from $KCl$
• C. $2FeC{l}_3+SnC{l}_2\to 2FeC{l}_2+SnC{l}_4$ is a spontaneous reaction.
• D. $C{e}^{4+}$ will displace $I_2$ from $KI$

Answer 1

Answer: (C), (D)

$C{e}^{4+}$ will displace $I_2$ from $KI$

$F{e}^{3+}/F{e}^{2+}$ couple has lower $E^0$ value than $C{e}^{4+}/C{e}^{3+}$, so $F{e}^{2+}$ will reduce $C{e}^{4+}$ to $C{e}^{3+}$. It won't oxidise $C{e}^{3+}$ to $C{e}^{4+}$.

$I_2/I^{-}$ couple has lower $E^0$ value than $C{l}_2/C{l}^{-}$couple. Hence, $C{l}_2$ has more tendency to get reduced and exist as $C{l}^{-}$ in solution so $I_2$ will not displace $C{l}_2$ from $KCl$.

In redox reaction,
$FeC{l}_3+SnC{l}_2\to FeC{l}_2+SnC{l}_4$
$F{e}^{3+}$ is reduced to $F{e}^{2+}$ and $S{n}^{2+}$ is oxidised to $S{n}^{4+}$.

$E_{cell}^0=\text{SRP of substance reduced}-\text{SRP of substance oxidised}$
(SRP - Standard reduction potential)

$E_{cell}^0=E_{cathode\left(red\right)}^0-E_{anode\left(red\right)}^0$

$E_{cell}^0=0.76-0.15$
$E_{cell}^0=0.61V$
Thus, the given redox reaction is spontaneous.

$C{e}^{4+}/C{e}^{3+}$ couple has higher $E^0$ value than $I_2/I^{-}$ couple so $C{e}^{4+}$ will oxidise $I^{-}$ to $I_2$.
Thus, $C{e}^{4+}$ will displace $I_2$ from $KI$ solution.