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Question #44095

Difficulty: 8/10 | 🟢 Free

The Question

An electrochemical cell is constructed at $298 \text{ K}$ :

$\ce{Pt(s) | H2(g, 1 bar) | 0.1 M Na2HPO4(aq) || 0.1 M NaH2PO4(aq) | H2(g, 1 bar) | Pt(s)}$

Given the acid dissociation constants for phosphoric acid ( $\ce{H3PO4}$ ) at $298 \text{ K}$ : $pK_{a1} = 2.2$ , $pK_{a2} = 7.2$ , $pK_{a3} = 12.4$ .

Assume $\frac{2.303 RT}{F} = 0.06 \text{ V}$ .

Calculate the cell potential ( $E_{cell}$ ).

Options

A.
+0.306 V
B.
-0.306 V
C.
+0.156 V
D.
0.000 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

Salt Hydrolysis Logic|Nernst Equation Concentration Dependence|Electrode Potential Sign Conventions

Why This Question Is Tricky

Common Mistakes

This problem bridges Electrochemistry and Ionic Equilibrium.

The core trap lies in the concentration dependence of the Nernst equation.

Students often assume the nominal salt concentration ( $0.1 \text{ M}$ ) dictates the cell potential, leading to an erroneous $0.000 \text{ V}$ .

The hidden variable is the local $[H^+]$ at the electrode, which is governed by the hydrolysis of amphiprotic salts.

A secondary trap is applying the standard weak acid/strong base hydrolysis formula ( $pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C$ ) instead of the amphiprotic formula ( $pH = \frac{pK_{a1} + pK_{a2}}{2}$ ).

Recognizing that the pH of an amphiprotic salt is independent of its concentration is crucial for solving this efficiently.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Salt Hydrolysis Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free