Before You Begin

Question #44148

Difficulty: 8/10 | 🟢 Free

The Question

Consider the redox chemistry of manganese in aqueous media at 298 K. The standard reduction potentials are given as: $E^\circ(\ce{MnO4- / Mn^2+}) = +1.51\text{ V}$ $E^\circ(\ce{MnO2 / Mn^2+}) = +1.23\text{ V}$ $E^\circ(\ce{Mn^{3+} / Mn^2+}) = +1.51\text{ V}$

A student investigates the comproportionation of $\ce{MnO4-}$ and $\ce{Mn^2+}$ to form $\ce{MnO2(s)}$ (Reaction I) and the disproportionation of $\ce{Mn^{3+}(aq)}$ to $\ce{MnO2(s)}$ and $\ce{Mn^2+}$ (Reaction II): Reaction I: $2\ce{MnO4-} + 3\ce{Mn^2+} + 2\ce{H2O} \rightleftharpoons 5\ce{MnO2(s)} + 4\ce{H+}$ Reaction II: $2\ce{Mn^{3+}} + 2\ce{H2O} \rightleftharpoons \ce{MnO2(s)} + \ce{Mn^2+} + 4\ce{H+}$

Determine the standard cell potentials $E^\circ_{\text{I}}$ and $E^\circ_{\text{II}}$ for Reactions I and II, respectively. Furthermore, identify the effect of increasing the pH from 0 to 6 on the thermodynamic stability of $\ce{Mn^{3+}(aq)}$ against disproportionation.

Options

A.
$E^\circ_{\text{I}} = +0.467\text{ V}$ , $E^\circ_{\text{II}} = +0.56\text{ V}$ ; $\ce{Mn^{3+}}$ becomes more unstable at pH 6
B.
$E^\circ_{\text{I}} = +0.280\text{ V}$ , $E^\circ_{\text{II}} = -0.28\text{ V}$ ; $\ce{Mn^{3+}}$ becomes more stable at pH 6
C.
$E^\circ_{\text{I}} = +0.467\text{ V}$ , $E^\circ_{\text{II}} = +0.56\text{ V}$ ; $\ce{Mn^{3+}}$ becomes more stable at pH 6
D.
$E^\circ_{\text{I}} = +0.840\text{ V}$ , $E^\circ_{\text{II}} = +0.28\text{ V}$ ; $\ce{Mn^{3+}}$ becomes more unstable at pH 6

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Comproportionation Reversibility|KMnO₄ Redox pH Dependency|Equivalent Concept in Redox

Why This Question Is Tricky

Common Mistakes

This problem tests two major concepts: the non-additivity of standard reduction potentials and the pH-dependency of redox stability.

A common trap is directly subtracting potentials without converting to Gibbs free energy ( $\Delta G^\circ = -nFE^\circ$ ), leading to incorrect $E^\circ$ values (e.g., $1.51 - 1.23 = 0.28$ V).

Additionally, students often overlook that the disproportionation of $\ce{Mn^{3+}}$ releases $\ce{H+}$ ions.

Consequently, in less acidic media (higher pH), the reaction is driven forward, making $\ce{Mn^{3+}}$ highly unstable unless stabilized by complexing agents.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Comproportionation Reversibility
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free