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Question #44133

Difficulty: 8/10 | 🟢 Free

The Question

Hydrogen peroxide (\ce{H2O2}) contains a peroxy linkage and can act as both an oxidizing and reducing agent. Consider the following non-standard electrochemical cell at $298\text{ K}$ :

$\ce{Pt(s) | O2(g, 1 bar) | H2O2(aq, 0.1 M), pH = 5 || pH = 1, H2O2(aq, 0.1 M) | Pt(s)}$

Given the standard reduction potentials: $E^\circ_{\ce{O2/H2O2}} = +0.68\text{ V}$ $E^\circ_{\ce{H2O2/H2O}} = +1.77\text{ V}$

Calculate the cell potential ( $E_{\text{cell}}$ ) at $298\text{ K}$ . (Assume $\frac{2.303RT}{F} = 0.059\text{ V}$ )

Options

A.
$1.267\text{ V}$
B.
$1.090\text{ V}$
C.
$0.913\text{ V}$
D.
$1.326\text{ V}$

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

Peroxy Linkage Identification|Nernst Equation Concentration Dependence|Electrode Potential Sign Conventions

Why This Question Is Tricky

Common Mistakes

This problem tests the application of the Nernst equation to a compartmentalized disproportionation reaction.

The primary trap is the 'Hidden Variable' of pH dependency.

In a single-beaker disproportionation of H2O2, the H+ ions cancel out, making the overall standard potential seemingly independent of pH.

However, in a divided electrochemical cell, the H+ concentrations at the anode and cathode do not cancel.

Students often erroneously use the standard potential (1.09 V) or fail to account for the [H2O2] concentration from both compartments in the reaction quotient, leading to incorrect Nernstian shifts.

Quick Facts About This Question

Question Information

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