Question

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox Titration. Some half – cell reactions and their standard potential are given below:
$Mn{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)+5e^{-}\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(I\right);E^{\circ }=1.51V$
$C{r}_2{O}_7^{2-}\left(aq\right)+14H^{+}\left(aq\right)+6e^{-}\to 2C{r}^{3+}\left(aq\right)+7H_{2}O\left(I\right);E^{\circ }=1.38V$
$F{e}^{3+}\left(aq\right)+e^{-}\to F{e}^{2+}\left(aq\right);E^{\circ }=0.77V$
$C{l}_2\left(g\right)+2e^{-}\to 2C{l}^{-}\left(aq\right);E^{\circ }=1.40V$

Identify the only incorrect statement regarding the quantitative estimation of aqueous $Fe(N{O}_3{)}_2$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$Mn{O}_4^{-}$ will oxidize $C{l}^{-}$ into $C{l}_2$ according to the following equation:
$2Mn{O}_4^{-}+16H^{+}10C{l}^{-}\to 2M{n}^{2+}+8H_{2}O+5C{l}_2$
The corresponding cell is
$PtC{l}_2\left(1atm\right)|C{l}^{-}||Mn{O}_4^{-},M{n}^{2+},H^{+}|Pt$
$E_{cell}^{\circ }=+1.51-1.40=+0.11V$
A positvie value confirms that the above process is spontaneous. Thus, $Mn{O}_4^{-}$ will oxidize $C{l}^{-}$ as well as $Fe(N{O}_3{)}_2$ simultaneously.