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Question #45024

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following electrochemical concentration cell at $298\text{ K}$ :

$\ce{Pt(s) | H2(g, 1 bar) | NaOH(aq, 1.0 \times 10^{-7} M) || HCl(aq, 1.0 \times 10^{-7} M) | H2(g, 1 bar) | Pt(s)}$

Calculate the electromotive force ( $E_{\text{cell}}$ ) of this cell.

(Given: $\frac{2.303 RT}{F} = 0.059\text{ V}$ , $K_w = 1.0 \times 10^{-14}$ , $\sqrt{5} = 2.24$ , $\log_{10}(1.62) = 0.21$ )

Options

A.
+0.025 V
B.
0.000 V
C.
+0.035 V
D.
-0.025 V

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Electrode Potential Sign Conventions|Electrolysis Product Prediction|pH of Extremely Dilute Solutions

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Hidden Variable' of water auto-ionization in extremely dilute solutions.

Students often calculate the pH of dilute solutions by directly taking the negative logarithm of the solute concentration.

For concentrations $\le 10^{-7}\text{ M}$ , the auto-ionization of water ( $K_w = 10^{-14}$ ) provides a comparable amount of $\ce{H+}$ or $\ce{OH-}$ ions.

Failing to account for this common ion effect leads to an assumed $[H^+]$ of $10^{-7}\text{ M}$ in both half-cells, resulting in the trap answer of $0.000\text{ V}$ .

Another trap involves adding exactly $10^{-7}\text{ M}$ from water without accounting for the suppression of water dissociation by the common ion effect, leading to $+0.035\text{ V}$ .

Finally, it rigorously tests Nernstian sign conventions for concentration cells.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Electrode Potential Sign Conventions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free