Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cellreactions and their standard potentials are given below:
${MnO}_{4}^{-} (aq.) + 8 H^{+} (aq.) + 5 e^{-} \to {Mn}^{2 +} (aq.) + 4 H_{2} O (l), E^{o} = 1.51 V \begin{array}{r} {Cr}_{2} O_{7}^{2 -} (aq.) + 14 H^{+} (aq .) + 6 e^{-} \to 2 {Cr}^{3 +} (aq .) + 7 H_{2} O (l) \\ E^{\circ} = 1 .38 V \end{array} {Fe}^{3 +} (aq .) + e^{-} \to {Fe}^{2 +} (aq .) E^{\circ} = 0 .77 V {Cl}_{2} (g) + 2 e^{-} \to 2 {Cl}^{-} (aq.) E^{\circ} = 1 .40 V$
Identify the only incorrect statement regarding the quantitative estimation of aqueous $Fe {({NO}_{3})}_{2}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${MnO}_{4}^{-}$ can be used in aqueous HCl

${Cr}_{2} O_{7}^{2 -}$ can be used in aqueous HCl

${MnO}_{4}^{-}$ can be used in aqueous $H_{2} {SO}_{4}$

${Cr}_{2} O_{7}^{2 -}$ can be used in aqueous $H_{2} {SO}_{4}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

${MnO}_{4}^{-} will oxidise {Cl}^{-}$ ion according to the equation
$2 {MnO}_{4}^{-} + 16 H^{+} + 10 {Cl}^{-} ⟶ 2 {Mn}^{2 +} + 8 H_{2} O + 5 {Cl}_{2} ↑$
The cell corresponding to this reaction is as follows:
$Pt, {Cl}_{2} (1 atm) |{Cl}^{-} ∥ {MnO}_{4}^{-}, {Mn}^{2 +}, H^{+}| Pt$
$E_{cell}^{o} = 1 .51 - 1 .40 = 0 .11 V$
$E_{cell}^{0} being +ve, {ΔG}^{\circ} will be -ve and hence the above$