Before You Begin

Question #45012

Difficulty: 8/10 | 🟢 Free

The Question

The standard reduction potentials for the permanganate ion at $298 \text{ K}$ are given as:

$\ce{MnO4^- (aq) + 8H^+ (aq) + 5e^- -> Mn^{2+} (aq) + 4H2O (l)}$ , $E^\circ = 1.51 \text{ V}$ $\ce{MnO4^- (aq) + 4H^+ (aq) + 3e^- -> MnO2 (s) + 2H2O (l)}$ , $E^\circ = 1.69 \text{ V}$

A student constructs two half-cells at $298 \text{ K}$ : Half-cell 1: $0.01 \text{ M } \ce{KMnO4}$ and $0.1 \text{ M } \ce{MnSO4}$ in an acidic buffer of $\text{pH} = 2$ . Half-cell 2: $0.1 \text{ M } \ce{KMnO4}$ and excess solid $\ce{MnO2}$ in a neutral buffer of $\text{pH} = 7$ .

Assuming the specified half-reactions are the sole determinants of the electrode potentials, what is the difference in the reduction potentials ( $E_1 - E_2$ ) of the two half-cells? (Take $\frac{2.303 RT}{F} = 0.06 \text{ V}$ )

Options

A.
$0.196 \text{ V}$
B.
$-0.172 \text{ V}$
C.
$0.188 \text{ V}$
D.
$-0.036 \text{ V}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Half-Reaction Media Sensitivity|KMnO₄ Redox pH Dependency|Electrode Potential Dependencies

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Nernstian Trap' in redox chemistry.

Students often memorize standard reduction potentials ( $E^\circ$ ) and assume that a higher $E^\circ$ implies a stronger oxidant under all conditions.

Here, $E^\circ$ for $\ce{MnO4^-} \rightarrow \ce{MnO2}$ ( $1.69 \text{ V}$ ) is higher than for $\ce{MnO4^-} \rightarrow \ce{Mn^{2+}}$ ( $1.51 \text{ V}$ ).

However, $E^\circ$ assumes $[\ce{H^+}] = 1 \text{ M}$ ( $\text{pH} = 0$ ), which is physically contradictory for a 'neutral' medium reaction.

By forcing the calculation of the formal potential at the actual pH ( $\text{pH } 7$ vs $\text{pH } 2$ ), the student must account for the Nernstian shift.

The trap also includes changing the n-factor (from 5 to 3), which alters the slope of the potential drop.

Failing to account for the pH dependency or the n-factor change leads to incorrect conclusions about the oxidizing power of $\ce{KMnO4}$ in different media.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Half-Reaction Media Sensitivity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free