Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below: MnO 4 − (aq

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 potentials are given below:

$MnO_{4}^{-}$ (aq.) $+ 8 H^{+}$ (aq.) $+ 5 e^{-} \to Mn^{2 +}$ (aq.) $+ 4 H_{2} O (l)$

$E^{\circ} = 1.51 V$

$Cr_{2} O_{7}^{2 -}$ (aq.) $+ 14 H^{+}$ (aq.) $+ 6 e^{-} \to 2 Cr^{3 +}$ (aq.) $+ 7 H_{2} O (l)$

$E^{\circ} = 1.38 V$

$Fe^{3 +}$ (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 $Fe (NO_{3})_{2}$

TestHub · Accepted Answer

The incorrect statement is A. M n O 4 − ​ CANNOT be used in aqueous HCl. Concept: M n O 4 − ​ is a strong oxidizing agent. In the presence of HCl, M n O 4 − ​ will oxidize C l − to C l 2 ​ . The E ∘ for C l 2 ​ / C l − is 1.40 V. Since the E ∘ for M n O 4 − ​ reduction is 1.51 V, it can oxidize C l − to C l 2 ​ . This would interfere with the redox titration of F e ( N O 3 ​ ) 2 ​ , leading to inaccurate results. C r 2 ​ O 7 2 − ​ with E ∘ = 1.38 V can be used because it does not oxidize C l − to C l 2 ​ . Using H 2 ​ S O 4 ​ does not have this issue.

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