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The hydrogen electrode is dipped in a solution of \(\mathrm{pH}=\) \(3.0\) at \(25^{\circ} \mathrm{C}\). The potential of hydrogen electrode would be (a) \(-0.177 \mathrm{~V}\) (b) \(0.177 \mathrm{~V}\) (c) \(1.77 \mathrm{~V}\) (d) \(0.277 \mathrm{~V}\)

Short Answer

Expert verified
(b) \(0.177 \mathrm{~V}\)

Step by step solution

01

Understanding the Hydrogen Electrode Potential Formula

The potential of a hydrogen electrode can be calculated using the Nernst equation, which is given by: \( E = E^0 - \frac{0.0591}{n} \log [H^+] \). Here, \( E^0 \) for the hydrogen electrode is 0 V, \( n \) is the number of electrons transferred in the reaction (which is 1 for the hydrogen electrode), and \( [H^+] \) is the hydrogen ion concentration in the solution.
02

Converting pH to Hydrogen Ion Concentration

Since the \( \mathrm{pH} = 3.0 \), we can find the hydrogen ion concentration using the formula \( [H^+] = 10^{-\mathrm{pH}} \). Thus, \( [H^+] = 10^{-3} = 0.001 \text{ M} \).
03

Applying the Nernst Equation

Substitute the values into the Nernst equation: \( E = 0 - \frac{0.0591}{1} \log (0.001) \). That simplifies to \( E = -0.0591 \times \log (10^{-3}) \).
04

Calculating the Logarithm

Calculate the logarithm: \( \log (10^{-3}) = -3 \). Substitute this back into the Nernst equation: \( E = -0.0591 \times (-3) \).
05

Solving for the Electrode Potential

Simplifying, \( E = -0.0591 \times (-3) = 0.1773 \text{ V} \). The potential is therefore \( 0.177 \text{ V} \), as rounding off gives it to three decimal places.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Electrode
A hydrogen electrode is a reference electrode based on the half-reaction of hydrogen gas. It consists of a platinum electrode coated with platinum black, dipped in a solution where hydrogen gas is bubbled at a given pressure, and it is immersed in a solution with a known activity of hydrogen ions, usually expressed through the pH of the solution.
The hydrogen electrode is considered the standard or reference electrode because it provides a standard potential of 0 V. This doesn't mean it has no voltage; instead, it serves as a neutral point in a system to make comparisons with other electrodes. The potential of the electrode can vary depending on the pH of the solution, making it a crucial part of electrochemical experiments, especially when involving calculations with the Nernst equation.
Electrode Potential
Electrode potential, also known as redox potential, refers to the measure of the tendency of a chemical species to gain or lose electrons, essentially being oxidized or reduced. In the context of the hydrogen electrode, this potential can be calculated using the Nernst equation. This equation adjusts the standard electrode potential to the conditions under which the measurement is made.
The specific formula is \[ E = E^0 - \frac{0.0591}{n} \log [H^+] \]where \( E^0 \) is the standard electrode potential (0 V for hydrogen), \( n \) is the number of moles of electrons exchanged, and \([H^+]\) is the concentration of hydrogen ions. It demonstrates how potential changes with varying concentrations.
In practice, electrode potential informs on the reactivity and energy transformations occurring in an electrochemical cell, and is fundamental to understanding electrochemistry.
pH and Concentration Conversion
The pH of a solution is a measure of its acidity or alkalinity and has a direct relation to the concentration of hydrogen ions, \([H^+]\). Converting pH values to hydrogen ion concentrations is a fundamental skill for evaluating chemical reactions and electrode potentials in electrochemistry.
  • The relationship is expressed mathematically as \([H^+]=10^{-\text{pH}}\).
  • In contexts like the hydrogen electrode, the pH value directly impacts the measured potential, as seen with its calculation using the Nernst equation.
For example, a solution with \(\text{pH} = 3.0\) yields a hydrogen ion concentration of \(10^{-3} = 0.001 \text{ M}\). This precise conversion is crucial for accurate predictions and experiments involving electrochemical cells. Knowing how to manipulate and convert these values can enhance a student's ability to solve diverse electrochemical problems effectively.

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Most popular questions from this chapter

Given \(\mathrm{E}^{\circ} \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.72 \mathrm{~V}, \mathrm{E}^{\circ} \mathrm{Fe}^{2+} / \mathrm{Fe}=-0.42 \mathrm{~V}\). The potential for the cell \(\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0.1 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.01 \mathrm{M})\right| \mathrm{Fe}\) is (a) \(0.26 \mathrm{~V}\) (b) \(0.399 \mathrm{~V}\) (c) \(-0.339 \mathrm{~V}\) (d) \(-0.26 \mathrm{~V}\)

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The electrical conductivity of the flowing aqueous solutions is highest for [2008] (a) \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) (b) \(0.1 \mathrm{M} \mathrm{CH}_{2} \mathrm{FCOOH}\) (c) \(0.1 \mathrm{M} \mathrm{CHF}_{2} \mathrm{COOH}\) (d) \(0.1 \mathrm{M} \mathrm{CH}_{2} \mathrm{ClCOOH}\)

In the electrolytic cell, flow of electrons is from (a) cathode to anode in solution (b) cathode to anode through external supply (c) cathode to anode through internal supply (d) anode to cathode through internal supply

Identify correct statements for the following incomplete reactions. (i) \(\mathrm{KI}+\mathrm{Cl}_{2} \rightarrow\) (ii) \(\mathrm{KClO}_{3}+\mathrm{I}_{2} \rightarrow\) (iii) \(\mathrm{KNO}_{3(3)}+{ }^{3} \mathrm{~K}_{(s)} \rightarrow\) (iv) \(\mathrm{P}+\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow\) (Red) (a) All are redox reactions (b) Only I, II and III have one of the product in its element form (c) I and II are displacement reactions (d) III is disproportionation reaction

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