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This reaction occurs in a cell with \(\mathrm{H}_{2}(\mathrm{~g})\) pressure of \(1.0 \mathrm{~atm}\) and (conc. \(\left.\mathrm{Cl}^{-}\right)=1.0 \mathrm{M}\) at \(25^{\circ} \mathrm{C} ;\) the measured \(E_{\text {cell }}=0.34 \mathrm{~V}\). Calculate the \(\mathrm{pH}\) of the solution. $$ \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{AgCl}(\mathrm{s}) \longrightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) $$

Short Answer

Expert verified
The pH of the solution is approximately 1.74.

Step by step solution

01

Write the Nernst Equation

For the given cell reaction, we use the Nernst equation:\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \]Since the cell is operating at standard temperature (25°C or 298 K), \( R = 8.314 \ \text{J} \ \text{mol}^{-1} \ \text{K}^{-1} \), and \( F = 96485 \ \text{C} \ \text{mol}^{-1} \). The reaction involves 2 moles of electrons (\( n = 2 \)).
02

Identify the Standard Cell Potential

The reaction can be split into two half-reactions: the reduction of \( \mathrm{AgCl} \) to \( \mathrm{Ag} \), and the oxidation of \( \mathrm{H}_2 \).- For the reduction of \( \mathrm{AgCl} \), standard potential \( E^0 = +0.222 \ \text{V} \) for one \( \mathrm{AgCl} \) per electron, thus for 2 electrons and 2 \( \mathrm{AgCl} \), it's the same.- For the oxidation of \( \mathrm{H}_2 \), standard potential \( E^0 = 0.000 \ \text{V} \).The overall standard cell potential \( E^0_{cell} = 0.222 \ \text{V} \).
03

Calculate the Reaction Quotient (Q)

The reaction quotient \( Q \) is:\[ Q = \frac{[\mathrm{H}^+]^2 [\mathrm{Cl}^-]^2}{P_{\mathrm{H}_2}} \]Given \([\mathrm{Cl}^-] = 1.0 \ \text{M}\) and \(P_{\mathrm{H}_2} = 1.0 \ \text{atm}\), this simplifies to:\[ Q = [\mathrm{H}^+]^2 \]
04

Rearrange the Nernst Equation to Solve for [H+]

Rearrange the Nernst equation to solve for \([\mathrm{H}^+]\):\[ 0.34 = 0.222 - \frac{0.0591}{2} \ln([\mathrm{H}^+]^2) \]Solving gives:\[ 0.118 = -0.02955 \ln([\mathrm{H}^+]^2) \]\[ \ln([\mathrm{H}^+]^2) = -3.992 \]\[ [\mathrm{H}^+]^2 = e^{-3.992} \]
05

Calculate [H+] Concentration

Solve for \([\mathrm{H}^+]\):\[ [\mathrm{H}^+] = \sqrt{e^{-3.992}} \approx 0.0181 \ \text{M} \]
06

Calculate pH

Finally, calculate the pH using the formula:\[ \text{pH} = -\log([\mathrm{H}^+]) \]\[ \text{pH} = -\log(0.0181) \approx 1.74 \]

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Key Concepts

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

Electrochemistry
Electrochemistry explores the relationship between chemical reactions and electricity. It's the study of how chemical energy is converted into electrical energy and vice versa. In electrochemical cells, a chemical reaction occurs by the transfer of electrons through an external circuit.
These cells can be either:
  • Galvanic cells: Produce electric energy through spontaneous reactions.
  • Electrolytic cells: Use electric energy to drive non-spontaneous reactions.
In our example, the reaction in a cell involves hydrogen gas and silver chloride, converting chemical energy into measurable voltage (cell potential). Understanding the flow of electrons and the reactions at the electrodes is essential to grasp how these cells work. The Nernst equation is pivotal in predicting the behavior of electrochemical cells based on concentrations and cell potential.
Cell Potential
Cell potential, denoted as \(E_{cell}\), is the electrical potential difference between two electrodes in an electrochemical cell. It measures how much energy per electron can be extracted from the cell reaction. The potential depends on:
  • The nature of the substances involved.
  • The concentrations of the ionic species.
  • The temperature and pressure of the gases involved.
Standard cell potential \(E^0_{cell}\) is measured under standard conditions: 1 M concentration, 1 atm pressure, and 25°C temperature. In the given exercise, it was crucial to distinguish between the standard potential and the measured potential to understand the system's behavior.
The standard reduction potentials for the half-reactions provide insight into the electron flow. The Nernst equation then allows the non-standard situation (actual conditions) to be analyzed by considering concentration changes from standard conditions.
pH Calculation
Calculating pH involves determining the hydrogen ion concentration in a solution. pH is defined as:\[\text{pH} = -\log([\mathrm{H}^+])\]This equation transforms the concentration of hydrogen ions into a more manageable scale. In the exercise, we need to find \([\mathrm{H}^+]\) using the relationship given by the Nernst equation. Rearranging the equation helps uncover the ion concentration based on cell potential.
The steps included solving for hydrogen ion concentration \(([\mathrm{H}^+]))\), extracting it from the logarithmic form given by the Nernst equation, and then calculating the pH. This provides a direct measure of the acidity of the solution under the conditions of the reaction. Understanding this process lets us connect electrochemical properties with chemical equilibrium and acidity.

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Most popular questions from this chapter

Chlorine gas can oxidize zinc metal in a reaction that has been suggested as the basis of a battery. Write the half reactions involved. Label which is the oxidation half reaction and which is the reduction half-reaction.

In a mercury battery, the anode reaction is $$ \mathrm{Zn}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{ZnO}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{e}^{-} $$ and the cathode reaction is $$ \mathrm{HgO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(\ell)+2 \mathrm{OH}^{-}(\mathrm{aq}) $$ The cell potential is \(1.35 \mathrm{~V}\). Calculate how many hours such a battery can provide power at a rate of \(4.0 \times 10^{-4}\) watt \(\left(1\right.\) watt \(\left.=1 \mathrm{~J} \mathrm{~s}^{-1}\right)\) if \(1.25 \mathrm{~g} \mathrm{HgO}\) is available.

Describe the principal parts of an \(\mathrm{H}_{2} \mid \mathrm{O}_{2}\) fuel cell. Write a balanced equation for the reaction at the cathode; at the anode. Give the formula of the product of the fuel cell reaction.

Write half-reactions for these changes: (a) Reduction of \(\mathrm{MnO}_{4}^{-}\) ion to \(\mathrm{Mn}^{2+}\) ion in acid solution (b) Reduction of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) ion to \(\mathrm{Cr}^{3+}\) ion in acid solution (c) Oxidation of chlorine gas to \(\mathrm{ClO}^{-}\) ions (d) Reduction of hydrogen peroxide to water in acidic solution (e) Oxidation of nitrous acid to nitrate ions in acidic solution

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) can be used as the reducing agent in a fuel cell. $$ \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ (a) If \(\Delta_{t} G^{\circ}\) for the reaction is \(-598 \mathrm{~kJ},\) calculate the value of \(E^{\circ}\) expected for the reaction. (b) Suppose the equation is written with all coefficients doubled. Determine \(\Delta_{\mathrm{r}} G^{\circ}\) and \(E^{\circ}\) for this new reaction.

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